Physics · Secondary School · Electric circuits

Electric Circuits

The light that comes on as soon as we press the switch, the electric shower that runs hotter in winter and the electricity bill that arrives every month all depend on electric charges in motion. We start from current and potential difference, make sense of resistance and power, build series and parallel circuits and finish with the real cell and with staying safe around the electricity at home.

1Electric current
2Potential difference and energy
3Ohm's laws
4Power
5Series and parallel
6Mixed circuits
7Real cells and safety
✓Challenges

STEP 1

What moves inside a wire?

In a metal, some of the electrons are not bound to any atom and wander freely through the material. We call the ordered motion of these charges an electric current, and we measure its size by the charge that crosses a section of the wire each second.

The unit of current is the ampere, equal to one coulomb per second. Each electron carries a tiny charge, $e = 1.6 \cdot 10^{-19}\ \text{C}$, and so, as we can check by dividing 1 C by $e$, a current of 1 A corresponds to more than six billion billion electrons passing each second.

By a convention that dates from before the discovery of the electron, the conventional direction of the current is the one in which positive charges would move, from the positive terminal to the negative one, outside the cell. We keep this convention to this day, and the electrons move in the opposite direction to it.

Something that may come as a surprise is how slowly the electrons move. In an ordinary copper wire they advance fractions of a millimetre per second, and the light still comes on the instant we press the switch, because the wire is already full of electrons and the signal that sets them moving travels at almost the speed of light.

$i = \dfrac{\Delta q}{\Delta t}$$\Delta q = n\,e$$1\ \text{A} = 1\ \text{C/s}$$e = 1.6 \cdot 10^{-19}\ \text{C}$ is the elementary charge, the magnitude of the charge of the electron, and $n$ is the number of electrons. In the simulation, the speed of the dots is greatly exaggerated and each dot stands for an enormous number of electrons; the drift velocity is worked out for a 1 mm² copper wire.

Let's discuss

With the switch closed, watch the electrons for a few seconds. Which way do they move, and where does the arrow of the conventional current point?
Reset the counter, set the current to 2 A and wait about 10 s. Does the charge shown agree with $i \cdot \Delta t$?
Open the switch. Do the electrons stop moving? What happens to the current?
Take the current up to 5 A and read the drift velocity. Roughly how long would an electron take to travel along a 1 m wire?

Current

Time

Charge that crossed

Electrons that crossed

Drift velocity

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Over 4 s, a charge of 12 C crosses a section of a wire. What is the electric current in the wire?

Show solution
The current is the charge that crosses the section divided by the time, $i = \dfrac{\Delta q}{\Delta t} = \dfrac{12}{4}$.
$i = 3\ \text{A}$
basic
A current of 2 A flows in a wire for 1 minute. How much charge crosses a section of the wire in that time?

Show solution
Isolating the charge in $i = \Delta q/\Delta t$, and remembering that 1 min has 60 s, we have $\Delta q = i\,\Delta t = 2 \cdot 60$.
$\Delta q = 120\ \text{C}$
basic
In a copper wire connected to a cell, the electrons move from the negative terminal to the positive terminal. What is the conventional direction of the current in this wire, and why is it opposite to the motion of the electrons?

Show solution
The conventional direction was chosen before the discovery of the electron, as if the current were made of positive charges.
Positive charges would move, outside the cell, from the positive terminal to the negative one, which is the opposite of what the electrons do.
Outside the cell, the conventional current goes from the positive terminal to the negative one, against the motion of the electrons, by a historical convention.
basic
How many electrons, together, add up to a charge of 1 C? Use $e = 1.6 \cdot 10^{-19}\ \text{C}$.

Show solution
Each electron carries $1.6 \cdot 10^{-19}\ \text{C}$, and the number of electrons is the total charge divided by the charge of each one, $n = \dfrac{\Delta q}{e} = \dfrac{1}{1.6 \cdot 10^{-19}}$.
$n = 6.25 \cdot 10^{18}$ electrons
intermediate
A current of 0.32 A flows in a wire for 10 s. How many electrons cross a section of the wire in that interval? Use $e = 1.6 \cdot 10^{-19}\ \text{C}$.

Show solution
The charge that passes is $\Delta q = i\,\Delta t = 0.32 \cdot 10 = 3.2\ \text{C}$.
Dividing by the charge of one electron, $n = \dfrac{3.2}{1.6 \cdot 10^{-19}}$.
$n = 2.0 \cdot 10^{19}$ electrons
intermediate
In a copper wire, the electrons advance, on average, less than a millimetre per second. Why, then, does the living-room light come on practically the instant we press the switch, several metres away from it?

Show solution
The wire is already full of free electrons before we press the switch, including inside the lamp.
When we close the circuit, what travels along the wire is the electrical signal, at almost the speed of light, and the electrons in the whole circuit start moving practically together.
It is like a hose already full of water, in which water comes out of the end as soon as we turn on the tap.
The light comes on at once because the signal travels very fast, even though each electron moves slowly.
intermediate
The current in an appliance is 4 A for 5 s and then falls uniformly to zero over the next 2 s. What is the total charge that passed through the appliance?

Show solution
On a graph of $i$ against $t$, the charge is the area under the curve.
In the first 5 s the area is a rectangle, $4 \cdot 5 = 20\ \text{C}$, and during the fall it is a triangle, $\dfrac{4 \cdot 2}{2} = 4\ \text{C}$.
Adding the two parts, $\Delta q = 20 + 4$.
$\Delta q = 24\ \text{C}$
intermediate
A mobile phone battery is marked 4000 mAh. How much charge, in coulombs, does it deliver before it runs out? If the phone draws 0.25 A on average, how long does the charge last?

Show solution
The milliampere-hour is a unit of charge, and $4000\ \text{mAh} = 4\ \text{A} \cdot 3600\ \text{s}$.
With an average draw of 0.25 A, $\Delta t = \dfrac{\Delta q}{i} = \dfrac{4\ \text{Ah}}{0.25\ \text{A}}$.
The battery holds 14,400 C and lasts about 16 h.
challenge
A copper wire with a cross-section of $1\ \text{mm}^2$ carries 2 A. Copper has about $8.5 \cdot 10^{28}$ free electrons per cubic metre. Show that $i = n\,e\,A\,v$, where $v$ is the average speed at which the electrons advance, and work out $v$. How long would an electron take to travel along 1 m of wire? Use $e = 1.6 \cdot 10^{-19}\ \text{C}$.

Show solution
In $\Delta t$, the electrons that cross the section are those contained in a cylinder of length $v\,\Delta t$, which number $n\,A\,v\,\Delta t$. Their charge divided by $\Delta t$ gives $i = n\,e\,A\,v$.
With $A = 10^{-6}\ \text{m}^2$, the product $n\,e\,A$ $= 8.5 \cdot 10^{28} \cdot 1.6 \cdot 10^{-19} \cdot 10^{-6}$ $\approx 1.36 \cdot 10^4\ \text{C/m}$, and $v = \dfrac{i}{n\,e\,A} = \dfrac{2}{1.36 \cdot 10^4}$ $\approx 1.5 \cdot 10^{-4}\ \text{m/s}$.
To travel 1 m, the electron takes $\dfrac{1}{1.5 \cdot 10^{-4}} \approx 6800\ \text{s}$.
$v \approx 0.15\ \text{mm/s}$, and the electron would take almost 2 hours to move 1 m.
challenge
In a solution of salt in water, the current is carried by ions. In 2 s, $3 \cdot 10^{18}$ sodium ions (charge $+e$) cross a section of the solution to the right and $3 \cdot 10^{18}$ chloride ions (charge $-e$) cross the same section to the left. What is the current, and in which direction? Use $e = 1.6 \cdot 10^{-19}\ \text{C}$.

Show solution
Negative charge moving to the left is equivalent, as far as the current is concerned, to positive charge moving to the right, and the two contributions add up.
The charge carried is $\Delta q = 6 \cdot 10^{18} \cdot 1.6 \cdot 10^{-19}$ $= 0.96\ \text{C}$, and $i = \dfrac{0.96}{2}$.
$i = 0.48\ \text{A}$, to the right, the direction in which the positive ions move.

STEP 2

How much energy does each charge carry?

We can think of the cell as something that pushes the charges already in the circuit and gives each of them a certain amount of energy, which they pass on to the lamp, the motor or the resistor they flow through.

We call the energy that each coulomb gains or loses between two points the potential difference (p.d.), or voltage, which we write as U, where many British books use V. A 1.5 V cell gives 1.5 J to each coulomb, and the sockets at home in Brazil, where all the values we use come from, supply 127 V or 220 V, depending on the region and the circuit.

To build intuition, we can think of the circuit as a water system. The cell would be a pump that keeps up a difference in level, the current would be the flow rate and the lamp a water wheel driven by the fall.

This model may help us picture what happens in the wire, and the simulation also shows a case in which it fails, which reminds us that it is only an analogy.

$U = \dfrac{E}{q}$$E = q\,U$$1\ \text{V} = 1\ \text{J/C}$In the simulation, the lamp is treated as a $6\ \Omega$ resistor that does not change as it heats up, a simplification we discuss in step 3. In the water model, the difference in level is proportional to the potential difference.

Let's discuss

Start with one cell and go up to three. What happens to the difference in water level, to the current and to the brightness of the lamp?
Open the switch. Does the cell seem to lose its voltage when the current stops? Compare with the closed valve in the water model.
Click 'Cut the wire and the pipe' and compare the two sides. At what point does the water model stop describing the circuit?
At 12 V, how many joules does the lamp receive each second? Check against the power reading.

Voltage

Current

Energy per coulomb

Power in the lamp

Energy in 1 minute

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
An ordinary cell has a voltage of 1.5 V. How much energy does it supply to each coulomb that passes through it? And to a charge of 20 C?

Show solution
The voltage is the energy per unit charge, and 1.5 V means 1.5 J for each coulomb.
For 20 C, $E = q\,U = 20 \cdot 1.5$.
1.5 J per coulomb, and 30 J for the 20 C.
basic
The 12 V battery of a car drives 300 C through the starter motor while the engine is being started. How much electrical energy does the battery deliver to the motor?

Show solution
Each coulomb carries 12 J, and the total energy is $E = q\,U = 300 \cdot 12$.
$E = 3600\ \text{J}$
basic
A small bird lands on a bare high-voltage cable and nothing happens to it. Why?

Show solution
For a current to flow through the bird's body, there has to be a potential difference between its two feet.
Both feet are on the same cable, a few centimetres apart, and the p.d. between them is practically zero.
With no p.d. between its feet, practically no current flows through the bird's body.
basic
Three 1.5 V cells are connected in series, with the positive terminal of one touching the negative terminal of the next, as in a torch. What is the voltage of the set? And if one of them is put in the wrong way round?

Show solution
In series, the voltages add up, $1.5 + 1.5 + 1.5 = 4.5\ \text{V}$.
The reversed cell now subtracts its voltage, $1.5 + 1.5 - 1.5 = 1.5\ \text{V}$.
4.5 V with all three the right way round, and only 1.5 V with one reversed.
intermediate
In the water analogy for a circuit, what do the cell, the current, the voltage and the resistance correspond to? Give an example of a situation in which the analogy fails.

Show solution
The cell plays the role of the pump, which raises the water, and the voltage corresponds to the difference in level, or the difference in pressure, that the pump keeps up.
The current corresponds to the flow rate, and the resistance to a narrow stretch of pipe that hinders the flow.
The analogy fails when we cut the circuit, because a cut pipe spills water, whereas a cut wire does not spill electrons, and the current simply stops in the whole circuit.
intermediate
A mobile phone battery is rated at 3.8 V and 4000 mAh. How much energy does it store, in joules?

Show solution
The charge is $q = 4\ \text{A} \cdot 3600\ \text{s} = 14\,400\ \text{C}$.
Each coulomb carries 3.8 J, and $E = q\,U = 14\,400 \cdot 3.8$.
$E \approx 5.5 \cdot 10^4\ \text{J}$, about 55 kJ.
intermediate
An electron is accelerated from rest through a p.d. of 100 V, as used to happen in old cathode-ray televisions. How much kinetic energy does it gain? Use $e = 1.6 \cdot 10^{-19}\ \text{C}$.

Show solution
The energy the charge receives is $E = q\,U = 1.6 \cdot 10^{-19} \cdot 100$.
$E = 1.6 \cdot 10^{-17}\ \text{J}$
intermediate
A 220 V socket and a 127 V socket supply different appliances. If the same charge of 50 C passes through an appliance on each socket, which one receives more energy, and how much more?

Show solution
The energy is $E = q\,U$. On the 220 V socket, $E = 50 \cdot 220 = 11\,000\ \text{J}$, and on the 127 V one, $E = 50 \cdot 127 = 6350\ \text{J}$.
The difference is $11\,000 - 6350$.
The appliance on 220 V receives 4650 J more.
challenge
A 1.5 V AA cell delivers, over its lifetime, a charge of about 2500 mAh and costs R$ 4.00. How much energy does it supply, and how much would a kilowatt-hour bought in cells cost? Compare with the mains tariff, which we take as R$ 0.80 per kWh. Remember that $1\ \text{kWh} = 3.6 \cdot 10^6\ \text{J}$.

Show solution
The charge is $q = 2.5 \cdot 3600 = 9000\ \text{C}$, and the energy, $E = 9000 \cdot 1.5 = 13\,500\ \text{J}$.
In kilowatt-hours, $\dfrac{13\,500}{3.6 \cdot 10^6} = 3.75 \cdot 10^{-3}\ \text{kWh}$.
The price per kWh is the price of the cell divided by this energy, $\dfrac{4.00}{3.75 \cdot 10^{-3}}$ reais.
A kWh bought in cells comes to about R$ 1067, some 1300 times the mains tariff.
challenge
In a lightning strike, a charge of 20 C passes through a p.d. estimated at $1 \cdot 10^8\ \text{V}$. How much energy is released? For how many months would it supply a home that uses 150 kWh a month? Use $1\ \text{kWh} = 3.6 \cdot 10^6\ \text{J}$.

Show solution
The energy is $E = q\,U = 20 \cdot 10^8 = 2 \cdot 10^9\ \text{J}$.
In kilowatt-hours, $\dfrac{2 \cdot 10^9}{3.6 \cdot 10^6} \approx 556\ \text{kWh}$, and $\dfrac{556}{150} \approx 3.7$.
About $2 \cdot 10^9\ \text{J}$, the home's consumption for some 3.7 months.

STEP 3

What gets in the way of the current?

As they pass through a material, the electrons collide with the atoms of the lattice and lose part of the energy they received. The electrical resistance measures this difficulty, and we define it by $R = U/i$, in ohms (Ω).

In many conductors kept at the same temperature, the current grows roughly in proportion to the voltage and $R$ stays practically constant. These are the ohmic conductors, which obey Ohm's first law, $U = R\,i$, and whose $U \times i$ graph is a straight line through the origin.

$U = R\,i$$R = \rho\,\dfrac{L}{A}$$\rho$ is the resistivity, in Ω·m: copper $1.7 \cdot 10^{-8}$, aluminium $2.8 \cdot 10^{-8}$, iron $1.0 \cdot 10^{-7}$ and nichrome $1.1 \cdot 10^{-6}$. The area goes in in m², with $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$. The lamp in the simulation follows a simple model, $U = 4\,i + 80\,i^3$, which passes through the 12 V and 0.5 A of a panel lamp.

Ohm's second law tells us what the resistance of a wire depends on. It grows with the length, falls with the cross-sectional area and depends on the material through the resistivity $\rho$. This explains why the electric shower uses much thicker wires than a table lamp.

The filament lamp is the classic example of a non-ohmic conductor. The tungsten filament gets very hot, its resistance grows with temperature, and the $U \times i$ graph curves. At small currents the filament barely heats up, and the start of the curve looks like a straight line.

Let's discuss

Save the line for the wire, double the length and save the new line. How has the slope changed, and what does that tell us about $R$?
With the length fixed, increase the cross-sectional area. At the same voltage, does the current go up or down?
Swap the nichrome for copper without touching the dimensions. Why do we use copper in household wiring and nichrome in the elements of heaters?
Swap the wire for the lamp and change the voltage little by little. Does the ratio $U/i$ stay constant, as it did for the wire?

Length

Cross-sectional area

Supply voltage

Resistivity

Resistance

Voltage

Current

Ratio U/i

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A current of 2 A flows through a $20\ \Omega$ resistor. What is the voltage across its terminals?

Show solution
By Ohm's first law, $U = R\,i = 20 \cdot 2$.
$U = 40\ \text{V}$
basic
A lamp connected to 127 V carries 0.5 A. What is the resistance of the lit filament?

Show solution
Isolating $R$ in $U = R\,i$, we have $R = \dfrac{U}{i} = \dfrac{127}{0.5}$.
$R = 254\ \Omega$
basic
A wire has resistance $R$. Another wire, of the same material, has twice the length and twice the cross-sectional area. What is the resistance of the second wire?

Show solution
By Ohm's second law, $R = \rho\,\dfrac{L}{A}$, and doubling $L$ doubles $R$, while doubling $A$ halves $R$.
The two effects cancel out, $R^{\prime} = \rho\,\dfrac{2L}{2A}$.
The resistance is still equal to $R$.
basic
What is the resistance of 10 m of copper wire with a cross-section of $1\ \text{mm}^2$? The resistivity of copper is $1.7 \cdot 10^{-8}\ \Omega \cdot \text{m}$.

Show solution
We convert the area to square metres, $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$, and use $R = \rho\,\dfrac{L}{A} = 1.7 \cdot 10^{-8} \cdot \dfrac{10}{10^{-6}}$.
$R = 0.17\ \Omega$
intermediate
On the $U \times i$ graph of an ohmic resistor, the line passes through the origin and through the point (0.5 A; 6 V). What is the resistance, and what will the current be when the voltage is 18 V?

Show solution
In an ohmic resistor, $R = \dfrac{U}{i}$ is the slope of the line, $R = \dfrac{6}{0.5}$.
At 18 V, the current is $i = \dfrac{18}{R}$.
$R = 12\ \Omega$ and $i = 1.5\ \text{A}$
intermediate
A 60 W filament lamp, connected to 127 V, carries about 0.47 A. With the lamp cold, an ohmmeter reads about $20\ \Omega$. Work out the resistance with the lamp lit and explain the difference. Is the lamp an ohmic resistor?

Show solution
Lit, $R = \dfrac{U}{i} = \dfrac{127}{0.47} \approx 270\ \Omega$, more than ten times the value measured cold.
The tungsten filament reaches temperatures above 2000 °C, and the resistivity of the metal grows a great deal with temperature.
Since $U/i$ changes with the current, the $U \times i$ graph of the lamp is a curve.
The lamp is not ohmic, because the resistance of the filament grows as it heats up.
intermediate
A wire is stretched until it is twice as long, without losing any material, so that its volume stays the same. What happens to its resistance?

Show solution
With the volume $L\,A$ constant, doubling the length halves the area.
Then $R^{\prime} = \rho\,\dfrac{2L}{A/2} = 4\,\rho\,\dfrac{L}{A}$.
The resistance becomes four times greater.
intermediate
We want to make a $22\ \Omega$ resistor out of nichrome wire with a cross-section of $0.5\ \text{mm}^2$. The resistivity of nichrome is $1.1 \cdot 10^{-6}\ \Omega \cdot \text{m}$. What length of wire do we need?

Show solution
Isolating the length in $R = \rho\,L/A$, we have $L = \dfrac{R\,A}{\rho} = \dfrac{22 \cdot 0.5 \cdot 10^{-6}}{1.1 \cdot 10^{-6}}$.
$L = 10\ \text{m}$
challenge
Two wires, A and B, are made of the same material. Wire B has twice the length and twice the diameter of wire A. Connected one at a time to the same cell, which one carries more current, and how many times more?

Show solution
The area grows with the square of the diameter, and wire B has $A_B = 4\,A_A$.
So $R_B = \rho\,\dfrac{2L}{4A} = \dfrac{R_A}{2}$.
At the same voltage, $i = U/R$, and the current is inversely proportional to the resistance.
Wire B carries twice the current of wire A.
challenge
We measured the voltage and current in a small panel lamp and obtained the table below. Is the lamp ohmic? Work out $U/i$ at 2 V and at 12 V.
U (V): 2 · 4 · 6 · 12
i (A): 0.20 · 0.30 · 0.36 · 0.50

Show solution
The ratios are $\dfrac{2}{0.20} = 10\ \Omega$, $\dfrac{4}{0.30} \approx 13.3\ \Omega$, $\dfrac{6}{0.36} \approx 16.7\ \Omega$ and $\dfrac{12}{0.50} = 24\ \Omega$.
If the lamp were ohmic, the ratio would be the same in every row, and the points on the graph would lie on a straight line through the origin.
The lamp is not ohmic, because $U/i$ grows along with the voltage.

STEP 4

What does the electricity bill charge for?

The power of an appliance is the energy it receives per second. Each coulomb delivers $U$ joules and $i$ coulombs pass each second, so that $P = U\,i$, in watts. Combining this with Ohm's law, we arrive at the forms $R\,i^2$ and $U^2/R$, which are worth keeping to hand.

In a resistor, all of this energy turns into heat, the Joule effect, which warms the electric shower, the iron and the toaster. The shower has a curious feature, because on the winter setting it uses a shorter stretch of the wire, of lower resistance, and with the mains voltage fixed the power $U^2/R$ becomes greater.

The electricity company charges for the energy used. We measure this energy in kilowatt-hours, the energy of a 1000 W appliance switched on for one hour, which is equal to $3.6 \cdot 10^6\ \text{J}$. A powerful appliance switched on for a few minutes may therefore cost less than a low-power appliance left on all day.

$P = U\,i = R\,i^2 = \dfrac{U^2}{R}$$E = P\,\Delta t$$1\ \text{kWh} = 3.6 \cdot 10^6\ \text{J}$We take the tariff as R$ 0.80 per kWh, in Brazilian reais, an approximate value that changes with the distribution company, taxes and the seasonal surcharge. In the simulation, the shower runs on 220 V, and the fridge appears with its average power over the day, since the motor switches on and off.

Let's discuss

Switch the shower from winter to summer and compare the resistance and the power. Which of the two settings uses the shorter wire?
Turn the shower off and see which appliance now weighs most. Why does the fridge, with its small power, seem expensive?
Cut the shower from 40 to 20 minutes a day. How much does the family save over 30 days, in kWh and in reais?
Switch on the microwave and the iron, both high-power, and compare what they cost with what the TV costs, left on for longer.

Shower time per day

Period

Tariff per kWh

Shower power

Shower resistance

Shower current

Energy in the period

Cost

In joules

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A 9 W LED lamp stays on for 5 hours a day for 30 days. How much energy does it use, in kWh?

Show solution
The energy is the power times the time, $E = P\,\Delta t = 9\ \text{W} \cdot 150\ \text{h}$ $= 1350\ \text{Wh}$.
$E = 1.35\ \text{kWh}$
basic
A 5500 W electric shower is connected to 220 V. What current flows through it?

Show solution
From the relation $P = U\,i$, we have $i = \dfrac{P}{U} = \dfrac{5500}{220}$.
$i = 25\ \text{A}$
basic
On the 'winter' setting, the shower heats the water more. At the socket, the voltage is the same on both settings. Is the resistance in use in winter greater or smaller than in summer?

Show solution
With the voltage fixed, the power is $P = \dfrac{U^2}{R}$, and a smaller resistance gives a greater power.
On the winter setting, the switch connects only a shorter stretch of the resistance wire, and the current increases.
The winter resistance is smaller than the summer one.
basic
A 2000 W electric oven runs for 30 minutes. How much energy does it use, in kWh, and how much does that cost at the tariff of R$ 0.80 per kWh?

Show solution
In kilowatts and hours, $E = 2\ \text{kW} \cdot 0.5\ \text{h} = 1\ \text{kWh}$.
The cost is $1 \cdot 0.80$ reais.
1 kWh, which costs R$ 0.80.
intermediate
A 220 V electric shower has a power of 2200 W on the summer setting and 4400 W on the winter setting. What is the resistance on each setting?

Show solution
From $P = U^2/R$, we have $R = \dfrac{U^2}{P}$. In summer, $R = \dfrac{220^2}{2200}$, and in winter, $R = \dfrac{220^2}{4400}$.
$22\ \Omega$ in summer and $11\ \Omega$ in winter, half the resistance for twice the power.
intermediate
A current of 3 A flows through a $10\ \Omega$ resistor. What is the power dissipated, and how much heat does it give off in 1 minute?

Show solution
By the Joule effect, $P = R\,i^2 = 10 \cdot 3^2$.
In 60 s, the heat given off is $E = P\,\Delta t$.
$P = 90\ \text{W}$ and $5400\ \text{J}$ of heat per minute.
intermediate
A 1000 W hairdryer was made for 127 V and is plugged, by mistake, into a 220 V socket. Assuming the resistance does not change, what does the power become, and what will probably happen to the hairdryer?

Show solution
With $R$ fixed, $P = U^2/R$ grows with the square of the voltage, $\dfrac{P^{\prime}}{P} = \left(\dfrac{220}{127}\right)^2 \approx 3.0$.
The hairdryer would now dissipate about $3 \cdot 1000\ \text{W}$, three times what it was designed for.
About 3000 W, and the hairdryer is likely to overheat and burn out.
intermediate
A family replaces 10 filament lamps of 60 W with 8 W LED lamps, which give roughly the same light. The lamps are on for 5 h a day. How much does the family save over 30 days, in kWh and in reais, at the tariff of R$ 0.80 per kWh?

Show solution
Each replacement saves $60 - 8 = 52\ \text{W}$, and all ten together, 520 W.
Over 30 days of 5 h, that is 150 h, and $E = 0.52\ \text{kW} \cdot 150\ \text{h}$.
In reais, the cost is this energy times R$ 0.80.
78 kWh a month, or R$ 62.40.
challenge
A 2000 W electric kettle heats 1 L of water from 20 °C to 100 °C. Assuming all the electrical energy turns into heat in the water, how long does this take? Use the specific heat capacity of water, $4.2\ \text{J/(g} \cdot {}^\circ\text{C)}$, and a mass of 1000 g.

Show solution
The heat needed is $Q = m\,c\,\Delta T = 1000 \cdot 4.2 \cdot 80$ $= 336\,000\ \text{J}$.
With $P = Q/\Delta t$, we have $\Delta t = \dfrac{336\,000}{2000}$.
168 s, about 2.8 minutes. In a real kettle, part of the heat escapes, and the time tends to be a little longer.
challenge
The resistance wire of a 220 V electric shower, on the winter setting, has $11\ \Omega$. Someone cuts off a quarter of the length of the wire, thinking the shower will get hotter. What do the power and the current become? The circuit breaker on the circuit is rated at 25 A.

Show solution
By Ohm's second law, the resistance falls in the same proportion as the length, $R^{\prime} = \dfrac{3}{4} \cdot 11 = 8.25\ \Omega$.
The power becomes $P = \dfrac{220^2}{8.25}$, and the current, $i = \dfrac{220}{8.25}$.
The shower gets hotter, at about 5900 W, but the current of 26.7 A exceeds the 25 A of the circuit breaker, which should trip.

STEP 5

One path or several paths?

When we connect resistors one after the other, along a single path, we say they are in series. The current is the same in all of them, the voltage of the supply is shared among them and the equivalent resistance is the sum of the resistances.

In parallel, the resistors are connected to the same two points, each with its own path. All of them receive the same voltage, the currents in the branches add up and the equivalent resistance is smaller than the smallest of them, since each new branch opens one more way through for the current.

The sockets and lights at home are in parallel, and so each appliance receives the full 127 V or 220 V of the mains and works on its own. Old strings of fairy lights connected the bulbs in series, and one blown bulb was enough to put out the whole string, and you may have seen someone testing bulb after bulb in search of the culprit.

$R_{eq} = R_1 + R_2 + \dots$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$$R_{eq} = \dfrac{R_1\,R_2}{R_1 + R_2}$The first formula holds in series, the other two in parallel (the last one, for two resistors). In the simulation, the supply is 12 V, lamps L2 and L3 are $12\ \Omega$ and we assume the resistances do not change as they heat up. The brightness follows the power of each lamp.

Let's discuss

In series, with three identical lamps, compare the brightness with that of a single lamp on its own. Then switch to parallel.
Blow lamp L2 in each of the two connections. Why is the result so different?
In series, increase the resistance of L1. Does it shine more or less brightly than the others? And in parallel?
Go from two to three lamps in parallel and follow the current in the cell. What might this suggest about plugging many appliances into the same circuit?

Resistance of L1

Equivalent resistance

Current in the cell

Total power

Power in L1

Power in L2

Power in L3

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Three resistors, of $2\ \Omega$, $3\ \Omega$ and $5\ \Omega$, are connected in series to a 20 V supply. What is the equivalent resistance and what is the current?

Show solution
In series, the resistances add up, $R_{eq} = 2 + 3 + 5$.
The current is the same in all of them and equals $i = \dfrac{U}{R_{eq}}$.
$R_{eq} = 10\ \Omega$ and $i = 2\ \text{A}$
basic
What is the equivalent resistance of two $6\ \Omega$ resistors in parallel? And of a $12\ \Omega$ resistor in parallel with a $6\ \Omega$ one?

Show solution
For two resistors in parallel, $R_{eq} = \dfrac{R_1\,R_2}{R_1 + R_2}$. In the first case, $R_{eq} = \dfrac{6 \cdot 6}{12}$, and in the second, $R_{eq} = \dfrac{12 \cdot 6}{18}$.
$3\ \Omega$ and $4\ \Omega$, always less than the smallest of the resistors.
basic
Why are the sockets and lights in a house connected in parallel, and not in series?

Show solution
In parallel, each appliance is connected directly to the two wires of the mains and receives the full voltage, 127 V or 220 V.
That way, switching off an appliance, or one burning out, does not interrupt the others, which go on working independently.
basic
In an old string of fairy lights, the bulbs are in series. Why, when one of them blows, does the whole string go out?

Show solution
In series, there is a single path for the current, which passes through every bulb.
The blown bulb breaks this path, and the current drops to zero in the whole circuit.
With the single path broken, no bulb receives any current.
intermediate
Two lamps, of $20\ \Omega$ and $40\ \Omega$, are in series on a 12 V supply. Work out the current, the voltage and the power in each one. Which shines more brightly? Assume constant resistances.

Show solution
The equivalent resistance is $60\ \Omega$, and $i = \dfrac{12}{60} = 0.2\ \text{A}$ in both.
The voltages are $U_1 = 20 \cdot 0.2 = 4\ \text{V}$ and $U_2 = 40 \cdot 0.2 = 8\ \text{V}$, and the powers, $P_1 = 4 \cdot 0.2 = 0.8\ \text{W}$ and $P_2 = 8 \cdot 0.2 = 1.6\ \text{W}$.
In series, the $40\ \Omega$ lamp shines more brightly, with twice the power.
intermediate
The same lamps, of $20\ \Omega$ and $40\ \Omega$, are now connected in parallel on the 12 V supply. Work out the current and the power in each one, the total current and the equivalent resistance.

Show solution
In parallel, both receive 12 V. The currents are $i_1 = \dfrac{12}{20} = 0.6\ \text{A}$ and $i_2 = \dfrac{12}{40} = 0.3\ \text{A}$, and the powers, $P_1 = 12 \cdot 0.6 = 7.2\ \text{W}$ and $P_2 = 12 \cdot 0.3 = 3.6\ \text{W}$.
The total current is the sum $0.6 + 0.3$, and $R_{eq} = \dfrac{20 \cdot 40}{60}$.
Now the $20\ \Omega$ lamp shines more brightly, with a total current of 0.9 A and $R_{eq} \approx 13.3\ \Omega$.
intermediate
A string of fairy lights has 50 identical bulbs, of $10\ \Omega$ each, connected in series to a 127 V socket. What is the voltage across each bulb and what is the current?

Show solution
The voltage is shared equally among the identical bulbs, $U_1 = \dfrac{127}{50} = 2.54\ \text{V}$.
The equivalent resistance is $50 \cdot 10 = 500\ \Omega$, and $i = \dfrac{127}{500}$.
2.54 V across each bulb and $i = 0.254\ \text{A}$
intermediate
Three identical $30\ \Omega$ resistors can be connected all in series or all in parallel. What is the equivalent resistance in each case, and what is the ratio between them?

Show solution
In series, $R_{eq} = 3 \cdot 30$. In parallel, $\dfrac{1}{R_{eq}} = \dfrac{3}{30}$.
$90\ \Omega$ and $10\ \Omega$, nine times less in parallel.
challenge
On a 127 V circuit protected by a 20 A circuit breaker, a 1270 W microwave and a 1016 W kettle are switched on at the same time. Someone also switches on a 635 W hairdryer. Does the circuit breaker trip?

Show solution
The appliances are in parallel, and the currents add up. With $i = P/U$, the microwave draws $\dfrac{1270}{127} = 10\ \text{A}$, the kettle, $\dfrac{1016}{127} = 8\ \text{A}$, and the hairdryer, $\dfrac{635}{127} = 5\ \text{A}$.
The total current in the circuit is the sum $10 + 8 + 5$.
The circuit breaker trips, because 23 A exceeds the limit of 20 A.
challenge
A lamp rated '127 V · 100 W' and another rated '127 V · 25 W' are connected in series to a 127 V socket. Assuming constant resistances, which one shines more brightly? Work out the powers.

Show solution
The resistances come from $R = U^2/P$, $R_1 = \dfrac{127^2}{100} \approx 161\ \Omega$ and $R_2 = \dfrac{127^2}{25} \approx 645\ \Omega$.
In series, $i = \dfrac{127}{161 + 645} \approx 0.157\ \text{A}$, and $P = R\,i^2$ gives $P_1 \approx 161 \cdot 0.157^2$ and $P_2 \approx 645 \cdot 0.157^2$.
The 25 W lamp shines more brightly, with about 16 W against 4 W for the other. In a real filament, cold and with a lower resistance, these numbers would change a little.

STEP 6

Building, solving and measuring a circuit

Many circuits mix stretches in series and in parallel. To solve them, we replace each simple group by its equivalent resistance, from the inside out, until a single resistor is left. With it we find the current from the supply and, going back along the same route, the voltages and currents in each stretch.

The ammeter measures current and has to have the current flowing through it, which is why it goes in series. A good ammeter has almost zero resistance, so as not to change the current it is meant to measure.

$R_{eq} = R_1 + \dfrac{R_2\,R_3}{R_2 + R_3}$$i = \dfrac{U}{R_{eq}}$$U_p = R_p\,i$$R_p$ is the resistance of $R_2$ and $R_3$ in parallel, and $U_p$ the voltage across them. The supply in the simulation is ideal, at 12 V, and the fuse blows above 5 A. We assume ideal meters, with an ammeter of zero resistance and a voltmeter of infinite resistance.

The voltmeter measures the voltage between two points and is connected in parallel with the stretch we want to measure. It should have a very large resistance, so that it diverts a negligible current.

A wire of practically zero resistance connected between two points causes a short circuit. The current prefers the wire, the shorted stretch is left with no voltage and, if the short takes in the whole supply, the current grows so much that it can heat up the wires, and that is where the fuse comes in.

Let's discuss

With $R_1 = 4\ \Omega$, $R_2 = 6\ \Omega$ and $R_3 = 12\ \Omega$, solve the circuit on paper and then check with the meters, moving the ammeter from place to place.
Put the voltmeter across $R_1$ and then across the parallel pair. Does the sum of the two readings remind you of any number in the circuit?
Open switch A. What happens to the current in $R_2$ and to the voltage across the parallel pair?
Close switch B and see what happens to $R_1$. Then open B, close switch C and keep an eye on the fuse.

R₁

R₂

R₃

Equivalent resistance

Total current

Voltage across R₁

Voltage across the parallel pair

Current in R₂

Current in R₃

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
We want to measure the current through and the voltage across a lamp. How should we connect the ammeter and the voltmeter, and what do we expect of the internal resistance of each?

Show solution
The ammeter must have the same current flowing through it as the lamp, and an almost zero resistance of its own keeps it from altering that current.
The voltmeter compares the two terminals of the lamp, and a very large resistance of its own means it diverts a negligible current.
Ammeter in series, with $R \approx 0$, and voltmeter in parallel, with a very large $R$.
basic
A $4\ \Omega$ resistor is in series with a pair of resistors in parallel, of $6\ \Omega$ and $12\ \Omega$. What is the equivalent resistance of the circuit?

Show solution
We solve the parallel pair first, $R_p = \dfrac{6 \cdot 12}{6 + 12} = 4\ \Omega$.
This pair is in series with the $4\ \Omega$ resistor, and $R_{eq} = 4 + 4$.
$R_{eq} = 8\ \Omega$
basic
Two lamps are in series on a cell. A wire with no resistance is connected between the two terminals of one of them. What happens to each lamp?

Show solution
The wire offers a path of practically zero resistance in parallel with the lamp, and almost all the current goes through it.
That lamp is shorted out and goes off. The total resistance of the circuit falls, the current increases and the other lamp shines more brightly.
The shorted lamp goes off, and the other one now shines more brightly.
basic
By mistake, a student connects a voltmeter in series with a lamp, on a cell. Does the lamp light up? What does the voltmeter read?

Show solution
The voltmeter has a very large resistance, and in series it limits the current to a negligible value, as if the circuit were almost open.
The lamp does not light up, and almost all the voltage of the cell ends up across the voltmeter, which reads practically that voltage.
intermediate
The circuit in exercise 2 ($4\ \Omega$ in series with $6\ \Omega$ and $12\ \Omega$ in parallel) is connected to 24 V. What do an ammeter in the main wire, a voltmeter across the $4\ \Omega$ resistor and ammeters in the $6\ \Omega$ and $12\ \Omega$ branches read?

Show solution
With $R_{eq} = 8\ \Omega$, the main current is $i = \dfrac{24}{8} = 3\ \text{A}$.
Across the $4\ \Omega$ resistor, $U_1 = 4 \cdot 3 = 12\ \text{V}$, and $24 - 12 = 12\ \text{V}$ is left for the parallel pair.
In the branches, $i_6 = \dfrac{12}{6}$ and $i_{12} = \dfrac{12}{12}$.
3 A in the main wire, 12 V across the $4\ \Omega$ resistor, 2 A and 1 A in the branches.
intermediate
Two branches are in parallel on a 12 V supply. One branch has resistors of $2\ \Omega$ and $4\ \Omega$ in series, and the other, a $3\ \Omega$ resistor. Work out the current in each branch, the current from the supply and the equivalent resistance.

Show solution
The first branch has $2 + 4 = 6\ \Omega$ and a current of $\dfrac{12}{6}$, and the second, a current of $\dfrac{12}{3}$.
At the supply, the branch currents add up, and $R_{eq} = U/i_{\text{supply}}$.
2 A and 4 A in the branches, 6 A from the supply and $R_{eq} = 2\ \Omega$
intermediate
Two $6\ \Omega$ lamps are in series on a 12 V supply. Work out the power of each one. Next, a wire with no resistance shorts out one of the lamps. What does the power of the other become?

Show solution
Before, $i = \dfrac{12}{12} = 1\ \text{A}$, and each lamp dissipates $P = 6 \cdot 1^2$.
With one lamp shorted out, $6\ \Omega$ is left in the circuit, $i = 2\ \text{A}$ and $P = 6 \cdot 2^2$.
6 W in each before the short, and then 24 W in the remaining lamp, four times as much.
intermediate
Three identical $10\ \Omega$ lamps form a mixed circuit on a 15 V supply. L1 is in series with the pair of L2 and L3 in parallel. Work out the current and the power in each lamp.

Show solution
L2 and L3 in parallel come to $5\ \Omega$, and $R_{eq} = 10 + 5 = 15\ \Omega$, with $i = \dfrac{15}{15} = 1\ \text{A}$ in L1.
The current splits equally between L2 and L3, with 0.5 A in each.
The powers are $P_1 = 10 \cdot 1^2$ and $P_2 = P_3 = 10 \cdot 0.5^2$.
L1 receives 1 A and 10 W, and L2 and L3, 0.5 A and 2.5 W each.
challenge
Two $10\ \text{k}\Omega$ resistors are in series on a 12 V supply. To measure the voltage across one of them, we use a cheap voltmeter with an internal resistance of $10\ \text{k}\Omega$. What does it read, and what should it read? What does this show?

Show solution
Without the voltmeter, each resistor has half the voltage, 6 V.
Connected in parallel, the voltmeter forms with the resistor a pair of $\dfrac{10 \cdot 10}{20} = 5\ \text{k}\Omega$, and the current becomes $\dfrac{12}{15\ \text{k}\Omega} = 0.8\ \text{mA}$.
The voltage across the pair, which the voltmeter shows, is $5\ \text{k}\Omega \cdot 0.8\ \text{mA}$.
It reads 4 V instead of 6 V, which is why a good voltmeter needs a resistance much greater than that of the resistors in the circuit.
challenge
In the circuit of exercise 8 (L1 in series with L2 and L3 in parallel, all of $10\ \Omega$, on 15 V), lamp L3 blows. What happens to the brightness of L1 and L2? Work out the new powers.

Show solution
Without L3, the circuit becomes a simple series of L1 and L2, with $R_{eq} = 20\ \Omega$ and $i = \dfrac{15}{20} = 0.75\ \text{A}$.
Both dissipate $P = 10 \cdot 0.75^2$.
L1 drops from 10 W to about 5.6 W and gets dimmer, and L2 rises from 2.5 W to 5.6 W and gets brighter.

STEP 7

The cell has resistance too

A real cell uses up, inside itself, part of the energy it supplies to the charges. We represent this loss by an internal resistance $r$, and the voltage across the terminals is smaller than the electromotive force (e.m.f.) $\varepsilon$, the energy per unit charge that the cell produces.

The greater the current, the greater the loss $r\,i$, and this is why a car's headlights tend to dim when the starter motor draws an enormous current from the battery. If we join the terminals with a wire of no resistance, the voltage drops to zero and the current reaches its short-circuit value, $\varepsilon/r$.

$U = \varepsilon - r\,i$$i = \dfrac{\varepsilon}{R + r}$$i_{sc} = \dfrac{\varepsilon}{r}$$P_{u,\text{max}} = \dfrac{\varepsilon^2}{4\,r}$ with $R = r$We assume $\varepsilon$ and $r$ constant, a good approximation for a cell that is not nearly flat. With $R = r$, the useful power is at its maximum and the efficiency $U/\varepsilon$ is 50%.

The wiring and appliances at home are protected by fuses and circuit breakers, connected in series with the circuit. When the current goes over the limit, because of too many appliances or a short circuit, they open the circuit before the wires overheat.

In an electric shock, what does the harm is the current that passes through the body. The voltage matters because, together with the resistance of the path, it sets that current, and wet skin resists far less than dry skin. Currents of a few tens of milliamperes through the chest can already kill, and since an ordinary circuit breaker does not trip at such currents, protecting people is the job of the residual current device (RCD).

Let's discuss

Short out the load ($R = 0$) and compare the current with $\varepsilon/r$. Where does the energy go in this case?
Vary $R$ from 0 to 20 Ω and follow the point on the $U \times i$ graph. Does it move along a straight line?
Look for the load that gives the greatest useful power and compare it with the value of $r$. What is the efficiency at that point?
Compare the new cell with the worn-out one. Why does a torch with a worn-out cell go dim, if the electromotive force has hardly changed?

Electromotive force ε

Internal resistance r

Load R

Current

Terminal voltage

Internal loss r·i

Useful power

Efficiency

Short-circuit current

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A cell has an electromotive force of 1.5 V and an internal resistance of $0.5\ \Omega$. What is the voltage across its terminals when it supplies 1 A?

Show solution
Part of the energy is lost inside the cell, and $U = \varepsilon - r\,i = 1.5 - 0.5 \cdot 1$.
$U = 1.0\ \text{V}$
basic
A battery has $\varepsilon = 12\ \text{V}$ and $r = 0.1\ \Omega$. What is the short-circuit current, when the terminals are joined by a wire with no resistance?

Show solution
In a short circuit, the terminal voltage is zero, and the whole electromotive force falls across the internal resistance, $i_{sc} = \dfrac{\varepsilon}{r} = \dfrac{12}{0.1}$.
$i_{sc} = 120\ \text{A}$
basic
What are the fuse and the circuit breaker for, and what is the difference between them?

Show solution
Both are in series with the circuit they protect and cut off the current when it goes over a limit, before the wires get too hot.
The fuse has a thin wire that melts and has to be replaced. The circuit breaker trips a switch, which we can reset once the problem is sorted out.
Both cut off excessive currents; the fuse destroys itself in doing so, and the circuit breaker can be reset.
basic
In an electric shock, what decides the danger, the voltage or the current that passes through the body? Why do wet hands make the shock more serious?

Show solution
The harm to the body depends on the current that passes through it and on the path it takes, for example from one hand to the other, through the heart.
The voltage matters because, with $i = U/R$, it sets the current together with the resistance of the body, and wet skin has a much lower resistance than dry skin.
What matters is the current through the body, and wet skin, with its lower resistance, lets a greater current through.
intermediate
A battery with $\varepsilon = 9\ \text{V}$ and $r = 1\ \Omega$ supplies an $8\ \Omega$ resistor. Work out the current, the terminal voltage, the useful power and the efficiency.

Show solution
The current is $i = \dfrac{\varepsilon}{R + r} = \dfrac{9}{8 + 1}$, and the terminal voltage, $U = \varepsilon - r\,i$.
The useful power is $P_u = U\,i$, and the efficiency, $\eta = \dfrac{U}{\varepsilon}$.
$i = 1\ \text{A}$, $U = 8\ \text{V}$, $P_u = 8\ \text{W}$ and $\eta \approx 89\%$
intermediate
A person touches a 127 V wire. With dry skin, the resistance of the path through the body is estimated at $100\ \text{k}\Omega$; with wet skin, at $1\ \text{k}\Omega$. Work out the current in both cases. Currents of about 10 mA can already stop a person from letting go of the wire.

Show solution
By Ohm's law, with dry skin, $i = \dfrac{127}{100\,000}$, and with wet skin, $i = \dfrac{127}{1000}$.
About 1.3 mA with dry skin, enough for a tingle, and 127 mA with wet skin, a current a hundred times greater and very dangerous.
intermediate
The $U \times i$ graph of a cell is a straight line that goes from 6 V (at $i = 0$) to 0 V (at $i = 3\ \text{A}$). Find $\varepsilon$, $r$ and the maximum useful power.

Show solution
At $i = 0$, there is no internal loss, and the voltage read off the graph is $\varepsilon$ itself. In the short circuit, $i_{sc} = \varepsilon/r$, and $r = \dfrac{6}{3}$.
The useful power is at its maximum when $R = r$, with $i = \dfrac{6}{4} = 1.5\ \text{A}$ and $U = 3\ \text{V}$, and then $P_u = 3 \cdot 1.5$.
$\varepsilon = 6\ \text{V}$, $r = 2\ \Omega$ and $P_{u,\text{max}} = 4.5\ \text{W}$
intermediate
Connected to one resistor, a battery supplies 1 A and has 10 V across its terminals. Connected to another resistor, it supplies 2 A and has 8 V across its terminals. What are the electromotive force and the internal resistance?

Show solution
Both measurements obey $U = \varepsilon - r\,i$, with $10 = \varepsilon - r \cdot 1$ and $8 = \varepsilon - r \cdot 2$.
Subtracting the second from the first, $2 = r \cdot 1$, and going back to the first, $\varepsilon = 10 + r$.
$\varepsilon = 12\ \text{V}$ and $r = 2\ \Omega$
challenge
A cell has $\varepsilon = 12\ \text{V}$ and $r = 2\ \Omega$. Work out the useful power and the efficiency with an external resistor of $2\ \Omega$ and with one of $8\ \Omega$. Which of the two connections makes better use of the energy of the cell?

Show solution
With $R = 2\ \Omega$, $i = \dfrac{12}{4} = 3\ \text{A}$ and $U = 6\ \text{V}$, and then $P_u = 6 \cdot 3$ and $\eta = \dfrac{6}{12}$.
With $R = 8\ \Omega$, $i = \dfrac{12}{10} = 1.2\ \text{A}$ and $U = 9.6\ \text{V}$, and then $P_u = 9.6 \cdot 1.2$ and $\eta = \dfrac{9.6}{12}$.
With $R = r$, the useful power is at its maximum (18 W), but half the energy is lost inside the cell; with $8\ \Omega$, the power falls to 11.5 W and the efficiency rises to 80%.
challenge
A 5500 W electric shower on 220 V is installed on a circuit protected by a 20 A circuit breaker, which trips every time the shower is switched on. Someone suggests swapping it for a 40 A circuit breaker, without touching the wiring. Why does the circuit breaker trip, and why is the suggestion dangerous?

Show solution
The current of the shower is $i = \dfrac{P}{U} = \dfrac{5500}{220} = 25\ \text{A}$, above the 20 A of the circuit breaker.
The circuit breaker protects the wires, and its limit has to match the current they can carry without getting too hot. With a 40 A circuit breaker on a wire made for 20 A, the wire can overheat without anything tripping.
An ordinary circuit breaker does not protect people against shocks either, which happen at currents of milliamperes. That is what the RCD is for.
The shower draws 25 A, and the safe solution is to replace the wire and the circuit breaker together, to suit that current.

WRAP-UP

Challenges

Electric current

In a lightning strike, about 20 C of charge crosses the air in 1 ms. What is the average current?

Show solution

The current is the charge divided by the time, $i = \dfrac{20}{10^{-3}}$.
This gives $i = 2 \cdot 10^4\ \text{A}$, some 20,000 A, thousands of times the current of an electric shower.

Potential difference and energy

A 12 V car battery is marked 60 Ah. How much energy does it store, in joules and in kWh?

Show solution

The charge is $q = 60 \cdot 3600 = 216\,000\ \text{C}$, and $E = q\,U = 216\,000 \cdot 12$ $\approx 2.6 \cdot 10^6\ \text{J}$.
In kilowatt-hours, $\dfrac{2.592 \cdot 10^6}{3.6 \cdot 10^6}$ gives 0.72 kWh.

Ohm's second law

An extension lead has two copper wires of 20 m each, with a cross-section of 1.5 mm². What is the total resistance of the wires, and how much voltage is lost in them at 10 A? ($\rho_{\text{copper}} = 1.7 \cdot 10^{-8}\ \Omega \cdot \text{m}$)

Show solution

The current goes along one wire and comes back along the other, and the total length is 40 m.
$R = 1.7 \cdot 10^{-8} \cdot \dfrac{40}{1.5 \cdot 10^{-6}}$ $\approx 0.45\ \Omega$, and the loss is $U = R\,i \approx 4.5\ \text{V}$.

Power and the electricity bill

A 4400 W electric shower is on for 40 min a day. How much does it cost over 30 days, at the tariff of R$ 0.80 per kWh?

Show solution

Per day, $E = 4.4\ \text{kW} \cdot \dfrac{2}{3}\ \text{h} \approx 2.93\ \text{kWh}$, and over 30 days, 88 kWh.
The cost is $88 \cdot 0.80$, R$ 70.40.

Series and parallel

Two resistors, of 3 Ω and 6 Ω, are connected to 18 V, first in series and then in parallel. What are the equivalent resistance and the current from the supply in each case?

Show solution

In series, $R_{eq} = 9\ \Omega$ and $i = \dfrac{18}{9} = 2\ \text{A}$.
In parallel, $R_{eq} = \dfrac{3 \cdot 6}{9} = 2\ \Omega$ and $i = \dfrac{18}{2} = 9\ \text{A}$.

Mixed circuit

On a 12 V supply, a 1 Ω resistor, a pair of 4 Ω resistors in parallel and a 3 Ω resistor are connected in series. What does a voltmeter across the parallel pair read?

Show solution

The pair comes to $2\ \Omega$, and $R_{eq} = 1 + 2 + 3 = 6\ \Omega$, with $i = \dfrac{12}{6} = 2\ \text{A}$.
Across the pair, $U = 2 \cdot 2$, and the voltmeter reads 4 V.

Real cell

A car battery has $\varepsilon = 12.6\ \text{V}$ and $r = 0.02\ \Omega$. When the engine is started, the starter motor draws 150 A. What is the terminal voltage, and why do the headlights dim at that moment?

Show solution

The internal loss is $r\,i = 0.02 \cdot 150 = 3\ \text{V}$, and $U = 12.6 - 3$, or 9.6 V.
The headlights, connected to the same terminals, receive less voltage while the starting current is large.

Thinking, no sums

Why does a filament lamp tend to blow precisely at the moment it is switched on?

Show solution

When cold, the filament has a much lower resistance than when it is lit.
At the moment we switch it on, the current and the power $U^2/R$ are much greater than in normal running, for a fraction of a second, and the weakest point of the filament tends to break during this peak.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with an everyday context, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on electric circuits, which may well be the best practice once these are done.

Electric current · Step 1
Lúcia bought a fast charger for her phone. The packaging says that, in fast mode, it drives a current of 3 A through the battery, and the phone's battery has a capacity of 4500 mAh. Let us assume the current stays constant throughout the charge and that no charge is lost along the way.

Starting from a flat battery, how long does a full charge take?
1. 40 min
2. 1 h 30 min
3. 1 h 50 min
4. 13.5 h
5. 1500 h
Show solution

Answer: B.
The milliampere-hour measures charge, and the battery holds $4500\ \text{mAh} = 4.5\ \text{Ah}$, or $4.5 \cdot 3600 = 16\,200\ \text{C}$.
With $i = \Delta q/\Delta t$, the time is $\Delta t = \dfrac{4.5\ \text{Ah}}{3\ \text{A}} = 1.5\ \text{h}$, an hour and a half.
Reading 1.5 h as 1 h 50 min leads to C, and dividing the current by the charge, to A. Option D multiplies instead of dividing, and E forgets to convert milliamperes to amperes.
In practice, chargers reduce the current when the battery is nearly full, and charging tends to take a little longer than this calculation.
Resistance of wires · Step 3
Henrique needs to plug a heater that draws 10 A into a distant socket and is going to buy a 25 m extension lead. The shop has two options, one with copper wires of $0.5\ \text{mm}^2$ cross-section and another, more expensive, with wires of $2.5\ \text{mm}^2$. The extension lead has two wires, and the current goes along one and comes back along the other. The resistivity of copper is $1.7 \cdot 10^{-8}\ \Omega \cdot \text{m}$.

What power is dissipated as heat in the wires of the thinner extension lead?
1. 17 W
2. 34 W
3. 85 W
4. 170 W
5. 1270 W
Show solution

Answer: D.
The current runs through 50 m of wire, 25 m out and 25 m back, and the resistance is
$R = \rho\,\dfrac{L}{A}$ $= 1.7 \cdot 10^{-8} \cdot \dfrac{50}{0.5 \cdot 10^{-6}}$ $= 1.7\ \Omega$
By the Joule effect, $P = R\,i^2 = 1.7 \cdot 10^2$, or 170 W.
Option C counts only one of the wires, and B corresponds to the thick extension lead, five times less resistive. Option A uses $R\,i$ in place of $R\,i^2$, and E works out the power of the appliance on a 127 V socket, which is not what is lost in the wires.
The 170 W may seem little next to a heater, but it is concentrated in a thin wire, which can get hot enough to melt its insulation, and that is why thin extension leads are no good for high-power appliances.

Power and energy · Step 4

To understand why the electricity bill went up, Cecília noted down the power and the daily time of use of the main appliances at home. The distribution company's tariff, with taxes, is R$ 0.80 per kWh.

Daily use of the appliances
Appliance	Power (W)	Use per day
Electric shower	5400	40 min
Fridge (average)	50	24 h
TV	100	5 h
Iron	1000	15 min
8 LED lamps	80	5 h

In a 30-day month, how much does the shower add to the bill?

R$ 2.88
R$ 86.40
R$ 129.60
R$ 5184.00
R$ 86,400.00

Show solution

Answer: B.
The energy is the power times the time, with the power in kilowatts and the time in hours, since 40 min is $\dfrac{2}{3}$ of an hour.
$E = 5.4 \cdot \dfrac{2}{3} \cdot 30 = 108\ \text{kWh}$
The cost is $108 \cdot 0.80$, or R$ 86.40.
Option A is the cost of a single day, and C treats the 40 min as a whole hour. Option D uses the minutes as if they were hours, and E leaves the power in watts.
From the same table, the fridge uses 36 kWh in the month and the TV 15 kWh, and the shower, on for only a short time, weighs more than all the others put together.

Series and parallel · Step 5
To decorate the veranda, Rafael has three identical lamps and a 12 V supply, which we will assume to be ideal. He tries two arrangements. In arrangement I, the three lamps are in series; in arrangement II, they are in parallel, each connected directly to the terminals of the supply. With everything lit, he unscrews the middle lamp in each arrangement.

What happens to the other two lamps?
1. In both arrangements, the other lamps go out.
2. In arrangement I, the others go out; in II, they stay lit with the same brightness.
3. In arrangement I, the others go out; in II, they stay lit and shine more brightly.
4. In arrangement I, the others shine more brightly; in II, they go out.
5. In both arrangements, the others stay lit with the same brightness.
Show solution

Answer: B.
In series there is a single path, and the unscrewed lamp breaks it, so that the current drops to zero in all of them.
In parallel, each lamp has its own path to the supply and keeps its 12 V. Since $P = U^2/R$ does not change, the brightness does not change either.
Option C assumes the current of the removed lamp is shared out among the others, which does not happen with the fixed voltage of the ideal supply. Option D swaps the behaviour of the two connections, and A and E treat the two as if they were the same.
With a real supply, of small internal resistance, the remaining lamps in parallel may even get a tiny bit brighter, because the total current falls and the internal loss drops.

Electric shock · Step 7

A safety leaflet gives the table below, with the approximate effects of an alternating current passing through the body of an adult for about one second. The values vary from person to person and serve only as a guide.

Effects of current on the body
Current	Likely effect
up to 1 mA	barely perceptible
from 1 mA to 10 mA	tingling and pain
from 10 mA to 30 mA	muscle contraction
above 30 mA	risk of fibrillation

Ivo, barefoot on the wet floor of the utility room, touches a bare 127 V wire with his hand. With his body wet, the resistance of the path to the ground is around $1\ \text{k}\Omega$. If he were dry and wearing rubber flip-flops, it would be over $100\ \text{k}\Omega$.

According to the table, which effects correspond to the two situations?

Barely perceptible in both, because the voltage of the socket is the same.
Tingling with the body wet and barely perceptible with dry skin.
Muscle contraction with the body wet and tingling with dry skin.
Risk of fibrillation with the body wet and tingling with dry skin.
Risk of fibrillation in both, because what decides is the voltage of 127 V.

Show solution

Answer: D.
The effect depends on the current through the body, which comes from $i = U/R$.
Wet, $i = \dfrac{127}{1000} = 0.127\ \text{A}$, or 127 mA, well above 30 mA.
Dry, $i = \dfrac{127}{100\,000} \approx 1.3\ \text{mA}$, in the tingling range.
Options A and E look only at the voltage. Options B and C get the conversion between amperes and milliamperes wrong and end up in lower bands of the table.
The hundredfold difference between the two currents helps us see why bathrooms and utility rooms call for extra care and an RCD in the consumer unit.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2011, Day 1, blue booklet, question 60. The manual of an electric shower lists two models of the same power, one for 127 V and the other for 220 V, and the question asks for the ratio between the resistances, a direct application of $P = U^2/R$.
ENEM 2013, Day 1, blue booklet, question 72. An electrician wants to measure the voltage across a fridge, the total current of a kitchen and the current in a lamp, and we choose the diagram in which the voltmeter and the two ammeters are connected the right way.
ENEM 2016, Day 1, blue booklet, question 74. Three identical lamps form a mixed circuit with a battery, and we need to say which currents, among those measured at five points of the circuit, are equal.
ENEM 2017, Day 2, blue booklet, question 110. A 500 mA fuse protects one of the branches of a resistor circuit, and we work out the greatest supply voltage it can stand without blowing.
ENEM 2022, Day 2, blue booklet, question 96. A table relates the current through the body to its effects, and with the resistance of wet skin we decide which connection of two 12 V batteries explains the accident described.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Current$i = \Delta q / \Delta t$

Charge and electrons$\Delta q = n\,e$

Potential difference$U = E/q$

Ohm's first law$U = R\,i$

Ohm's second law$R = \rho\,L/A$

Power$P = U\,i = R\,i^2 = U^2/R$

Energy$E = P\,\Delta t$, $1\ \text{kWh} = 3.6 \cdot 10^6\ \text{J}$

Series$R_{eq} = R_1 + R_2 + \dots$

Parallel$1/R_{eq} = 1/R_1 + 1/R_2 + \dots$

Two in parallel$R_{eq} = R_1 R_2/(R_1 + R_2)$

Real cell$U = \varepsilon - r\,i$, $i_{sc} = \varepsilon/r$

Maximum useful power$R = r$, $P = \varepsilon^2/4r$