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11.3Resistors in electric circuits

Simple circuits

We have so far defined emf, voltage, resistance, current and power dissipated in a resistor. We are now ready to put all these ideas together and discuss the main topic of this chapter—electric circuits.

These circuits include cells and batteries, connecting wires, ammeters (to measure current) and voltmeters (to measure voltage) and various resistors. The symbols used for these circuit components are shown in Table 11.1.

Symbol | Component name
| connecting wire
| cell
| battery of cells
| resistor
| dc power supply
| ac power supply
| junction of conductors
| crossing conductors (no connection)
| lamp
| voltmeter
| ammeter
| switch
| galvanometer
| potentiometer
| variable resistor
| heating element
| Light dependent resistor (LDR)
| Thermistor

Table 11.1: Names of electrical components and their circuit symbols.

We start with the circuit shown in Figure 11.18. The cell is assumed to have zero internal resistance. The current enters the circuit from the positive terminal of the cell. The direction of the current is shown by the blue arrow. The terminals of the cell are directly connected to the ends of the resistor (there is no intervening internal resistor). Therefore, the potential difference at the ends of the resistor is 12 V. Using the definition of resistance we write R=V I.

Substituting values gives 24=12 I, and rearranging gives I=12 24=0.5.

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So, the current I in the circuit is 0.5 A.

Figure 11.18: A simple one-loop circuit with one cell with negligible internal resistance and one resistor.

A very useful and convenient way to understand circuits is to think in terms of a quantity called potential rather than potential difference. We will not define potential but will assume that if a cell has an emf ε then we can assign zero potential to the negative terminal and potential ε to the positive terminal. As we follow the current in the circuit starting from the positive terminal the rule is that the potential stays unchanged unless we come to a resistance. This is why points B, C and D have potential 12 V. The potential difference between B and C or D is then zero. If we now follow the circuit starting from G, the negative terminal, to F and then E we again see that the potential stays the same, i.e. 0 V, until we reach the resistor at E. So we see that the potential difference across the resistor is 12 V. So as we go through the resistor in the direction of the current the potential drops.

Resistors

Resistors in series

Consider the circuit in Figure 11.19 that contains two resistors. The resistors are connected in series wire connecting resistors does not have any junctions so both resistors take the same current which means the wire joining the two does is a single wire without any junctions. The two resistors take the same current. What is this current?

Figure 11.19: Finding the total resistance of the circuit.

Notice that the potential between the resistors is unknown. If we call it x, then the potential difference across R 2 is 12 – x and the potential difference across R 1 is x – 0.

But, if we add these potential differences, we get 12 V, in other words, the emf of the circuit.

So, if we call the voltage across R 1, V 1 and the voltage across R 2, V 2 we have V 1 + V 2 = 12.

But V 1 = I R 1 and V 2 = I R 2, so I R 1 + I R 2 = 12.

In other words, 2.0I + 4.0I = 12, so the current is 2.0 A.

Can we generalise so we can find the answer faster?

From I R 1 + I R 2 = 12 we find I=12 R 1+R 2.

This is the same answer we would get if we replaced the two resistors in the dotted box by a single resistor of value R 1 + R 2.

In other words, there is an equivalent circuit that replaces the given circuit with a circuit with only one resistor of value R 1 + R 2. This is called the total resistance of the original circuit with two resistors.

But the equivalent circuit with just one resistor is very easy. We immediately find I = 2.0 A. Having found the current, we can find the voltages across each resistor:

●across the 2.0 Ω it is 2.0 × 2.0 = 4.0 V

●across the 4.0 Ω it is 2.0 × 4.0 = 8.0 V.

The sum of the voltages is the emf as expected. (The potential in between the two resistors is, therefore, 4.0 V.)

Exam tip

The voltages across all resistors in series add up to the emf of the source.

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For any number of resistors in series, the total resistance is their sum: R T = R 1 + R 2 + R 3 + ⋯.

Adding resistors in series increases the total resistance.

CHECK YOURSELF 4

A circuit contains two resistors connected in series. A third resistor is added in series to the other two. What happens to the current in the circuit?

Figure 11.20 shows a circuit with three resistors R 1 = 2.0 Ω, R 2 = 6.0 Ω and R 3 = 4.0 Ω in series.

Figure 11.20: The top circuit is replaced by the equivalent circuit containing just one resistor.

We replace the three resistors by the equivalent resistor of R T = 2.0 + 6.0 + 4.0 = 12 Ω. The current through the equivalent resistor is found as follows:

R=V I⇒I=V R=24 12=2.0 A

This current, therefore, is also the current that enters the dotted box. In other words, it is the current in each of the three resistors of the original circuit. We may thus deduce that the voltages across each of the three resistors are:

V 1=I R 1=4.0 V V 2=I R 2=12 V V 3=I R 3=8.0 V

Notice that the sum of the voltages equals the emf of the cell.

Worked Example 11.8

What is the voltage across each resistor in Figure 11.21?

Figure 11.21: For worked example 11.8.

Answer

Method 1

The two voltages must add up to 12 V. The voltage across 3R will be three times larger, so the voltages are 3 V and 9 V.

Method 2

The total resistance for the circuit is 4R.

Hence the current is I=12 4 R=3 R

The voltage across R is V=I R=3 R×R=3 V.

The voltage across 3R is V=I(3 R)=3 R×3 R=9 V.

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Worked Example 11.9

In the circuit in Figure 11.22, the intensity of light incident on the LDR is increased. What will happen to the potential difference across the LDR?

Figure 11.22: For worked example 11.9.

Answer

The sum of the potential differences across the LDR and R is the emf. The resistance of the LDR will decrease and so the current in the circuit will increase. Hence the potential difference across R will increase and so that across the LDR will decrease.

Resistors in parallel

Consider now part of another circuit, in which the current splits into three other currents that flow in three resistors, as shown in Figure 11.23 The current that enters the junction at A must equal the current that leaves the junction at B, by the law of conservation of charge. The left ends of the three resistors are connected at the same point and the same is true for the right ends. This means that three resistors have the same potential difference across them. This is called a parallel connection resistors connected so that they have the same potential difference across them. We must then have that:

I=I 1+I 2+I 3

This is again a consequence of charge conservation. The current entering the junction is I and the currents leaving the junction are I 1, I 2 and I 3. Whatever charge enters the junction must exit the junction and so the sum of the currents into a junction equals the sum of the currents leaving the junction.

Figure 11.23: Three resistors connected in parallel.

Let V be the common potential difference across the resistors. Then:

I 1=V R 1,I 2=V R 2 and I 3=V R 3

and so:

I=V R 1+V R 2+V R 3

I=V(1 R 1+1 R 2+1 R 3)

If we replace the three resistors in the dotted box with a single resistor R T, the potential difference across it would be V and the current through it would be I. Thus:

I=V R T

Comparing with the last equation, we find:

1 R T=1 R 1+1 R 2+1 R 3

The formula shows that the total resistance is smaller than any of the individual resistances being added.

We have thus learned how to replace resistors that are connected in series or parallel by a single resistor in each case, thus greatly simplifying the circuit.

Exam tip

Adding resistors in series increases the total resistance of a circuit (and so decreases the current leaving the battery). Adding resistors in parallel decreases the total resistance of the circuit (and so increases the current leaving the battery).

CHECK YOURSELF 5

The total resistance of three identical resistors is 1 Ω when they are connected in parallel. What is the total resistance when they are connected in series?

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Worked Example 11.10

In the part of the circuit shown in Figure 11.24 the current is 3.0 A.

What is the current in each resistor?

Figure 11.24: For worked example 11.10.

Answer

Method 1

The total resistance is found from 1 R T=1 6.0+1 12=1 4.0.

So, R T = 4.0 Ω.

Hence the common potential difference across the two resistors is:

V=I R T=3.0×4.0=12 V

And the currents are:

I 1=12 6.0=2.0 A (top) and I 2=12 12=1.0 A (bottom).

Method 2

The two resistors are connected in parallel so they have the same potential difference across them.

The potential difference across the top resistor is 6.0I 1 and that across the lower resistor is 12I 2.

Hence 6.0I 1 = 12I 2 ⇒ I 1 = 2I 2.

(The smaller resistor gets more current.)

But also, I 1 + I 2 = 3.0 A, so I 1 = 2.0 A and I 2 = 1.0 A.

Resistors in series and parallel

A typical circuit will contain both parallel and series connections. In Figure 11.25, the two top resistors are in series. They are equivalent to a single resistor of 8.0 Ω. This resistor and the 24 Ω resistor are in parallel, so together they are equivalent to a single resistor of:

1 R T=1 8.0+1 24=1 6⇒R T=6.0 Ω

Figure 11.25: Part of a circuit with both series and parallel connections.

Consider now Figure 11.26. The two top 6.0 Ω resistors are in series, so they are equivalent to a 12 Ω resistor. This, in turn, is in parallel with the other 6.0 Ω resistor, so the left block is equivalent to:

1 R T=1 12+1 6.0=1 4⇒R T=4.0 Ω

Let us go to the right block. The lower 12 Ω and the 24 Ω resistors are in series, so they are equivalent to 36 Ω. This is in parallel with the top 12 Ω, so the equivalent resistor of the right block is:

1 R T=1 36+1 12=1 9⇒R T=9.0 Ω

The overall resistance is thus:

4.0+9.0=13 Ω

Figure 11.26: A complicated part of a circuit containing many parallel and series connections.

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Suppose now that this part of the circuit is connected to a source of emf 156 V (and negligible internal resistance). The current that leaves the source is:

I=156 13=12 A

When it arrives at point A, it will split into two parts. Let the current in the top part be I 1 and that in the bottom part I 2. We have I 1 + I 2 = 12 A. We also have that 12I 1 = 6I 2, since the top and bottom resistors of the block beginning at point A are in parallel and so have the same potential difference across them. Thus, I 1 = 4.0 A and I 2 = 8.0 A. Similarly, in the block beginning at point B the top current is 9.0 A and the bottom current is 3.0 A.

CHECK YOURSELF 6

Two identical resistors are connected in series to a cell of negligible internal resistance. The total power dissipated is 16 W.

What is the power dissipated when the two resistors are connected to the same cell in parallel?

Worked Example 11.11

aDetermine the total resistance of the circuit shown in Figure 11.27.

bHence calculate the current and power dissipated in each of the resistors.

Figure 11.27: For worked example 11.11.

Answer

aThe resistors of 2.0 Ω and 3.0 Ω are connected in parallel and are equivalent to a single resistor of resistance R that may be found from:

1 R=1 2+1 3=5 6⇒R=6 5=1.2 Ω

In turn, this is in series with the resistance of 1.8 Ω, so the total equivalent circuit resistance is 1.8 + 1.2 = 3.0 Ω.

bThe current that leaves the cell is thus:

I=6.0 3.0=2.0 A

The potential difference across the 1.8 Ω resistor is V = 1.8 × 2.0 = 3.6 V, leading to a potential difference across the two parallel resistors of V = 6.0 − 3.6 = 2.4 V. Thus the current in the 2 Ω resistor is:

I=2.4 2.0=1.2 A

This leads to power dissipated of:

P=R I 2=2.0×1.2 2=2.9 W or P=V 2 R=2.4 2 2.0=2.9 W or P=V I=2.4×1.2=2.9 W

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For the 3 Ω resistor:

I=2.4 3.0=0.80 A

which leads to power dissipated of P = R I 2 = 3.0 × 0.802 = 1.9 W.

The power in the 1.8 Ω resistor is P = R I 2 = 1.8 × 2.02 = 7.2 W.

(The total power is 2.9 + 1.9 + 7.2 = 12 W and equals emf × current = 6.0 × 2.0 = 12 W as expected.)

Worked Example 11.12

In the circuit of Figure 11.28 the three lamps are identical and may be assumed to have a constant resistance. Discuss what happens to the brightness of:

alamp A

blamp B

when the switch is closed.

(Take ‘brightness’ to mean the same as power dissipated.)

Figure 11.28: For worked example 11.12.

Answer

aWhen the switch is closed the total resistance of the circuit decreases because we have a parallel connection for lamps B and C. Hence the current through A increases and the brightness increases.

b Method 1

With the switch open, A and B are connected in series and have the same resistance, so the potential difference across A and across B is ε 2 (half the emf of the cell).

This means that A and B are equally bright. When the switch is closed, B and C have a total resistance that is half that of A. So, the potential difference across A is double that across B.

This means that:

●the potential difference across A is 2 ε 3

●the potential difference across B is ε 3.

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So, A increases in brightness and B gets dimmer.

Method 2 (a mathematical answer)

Let the emf of the cell be ε and the resistance of each lamp be R.

Before the switch is closed, A and B take equal current ε 2 R and so are equally bright (the total resistance is 2R).

When the switch is closed, the total resistance of the circuit changes and so the current changes as well. The new total resistance is 3 R 2 (lamps B and C in parallel and the result in series with A) so the total current is now 2 ε 3 R, larger than before.

The current in A is greater and so the power (that is, the brightness) is greater than before.

The current of 2 ε 3 R is divided equally between B and C.

B now takes a current ε 3 R, which is smaller than before.

So, B is dimmer.

Worked Example 11.13

Look at Figure 11.29. Determine the current in the 2.0 Ω resistor and the potential difference across the two marked points, A and B, when the switch is:

aopen

bclosed.

Figure 11.29: For worked example 11.13.

Answer

aWhen the switch is open, the total resistance is 4.0 Ω, so the total current is 3.0 A.

The potential difference across the 2.0 Ω resistor is 2.0 × 3.0 = 6.0 V.

The potential difference across points A and B is the same as that across the 2.0 Ω resistor (= 6.0 V).

bWhen the switch is closed, no current flows through the 2.0 Ω resistor, since the current takes the path through the switch, which offers no resistance. (The 2.0 Ω resistor has been ‘shorted out’.)

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The resistance of the circuit is then 2.0 Ω and the current leaving the cell is 6.0 A.

The potential difference across points A and B is now zero. (There is current flowing from A to B,but the resistance from A to B is zero, hence the potential difference is I × R = 6.0 × 0 = 0 V.)

Worked Example 11.14

Four lamps each of constant resistance 60 Ω are connected as shown in Figure 11.30.

aDetermine the power in each lamp.

bLamp A burns out. Calculate the power in each lamp and the potential difference across the burnt-out lamp.

Figure 11.30: For worked example 11.14.

Answer

aWe know the resistance of each lamp, so to find the power we need to find the current in each lamp.

Lamps A and B are connected in series so they are equivalent to one resistor of value R AB = 60 + 60 = 120 Ω. This is connected in parallel to C, giving a total resistance of:

1 R ABC=1 120+1 60 1 R ABC=1 40⇒R ABC=40 Ω

Finally, this is in series with D, giving a total circuit resistance of:

R T=40+60=100 Ω

The current leaving the cell is thus:

I=30 100=0.30 A

The current through A and B is 0.10 A, and that through C is 0.20 A. The current through D is 0.30 A.

So, the power in each lamp is:

P A=P B

P A=60×(0.10)2=0.6 W

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P C=60×(0.20)2=2.4 W

P D=60×(0.30)2=5.4 W

bWith lamp A burnt out, the circuit is as shown in Figure 11.31.

Figure 11.31: For worked example 11.14 part b.

Lamp B gets no current, so we are left with only C and D connected in series, giving a total resistance of:

R T=60+60=120 Ω

The current is:

I=30 120=0.25 A.

The power in C and D is:

P C=P D=60×(0.25)2=3.8 W

We see that D becomes dimmer and C brighter.

The potential difference across lamp C is:

V=I R

V=0.25×60

V=15 V

Lamp B takes no current, so the potential difference across it is zero. The potential difference across points X and Y is the same as that across lamp C (= 15 V).

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TEST YOUR UNDERSTANDING

32A wire has resistance R. The wire is cut into two pieces of equal length. The two pieces are connected in parallel. What is the resistance of the combination?

33A wire has resistance R. The wire is cut into two pieces. One piece has twice the length of the other. The two pieces are connected in parallel. What is the resistance of the combination?

34Determine the total resistance for each of the circuit parts in Figure 11.32.

Figure 11.32: For question 34.

35Six identical resistors each of resistance R are connected as shown in Figure 11.33. The total resistance is 11 Ω.

Figure 11.33: For question 35.

aWhat is the value of R?

bFor mathematicians only: if the pattern of resistors shown in the figure were to continue forever by adding 4 and 5 (and so on) parallel resistors, would it make sense to ask for the total resistance?

36Another problem for mathematicians: what is the total resistance between X and Y in the arrangement that extends forever, shown in Figure 11.34? Each resistance is 1.0 Ω.

Figure 11.34: For question 36.

37Look at Figure 11.35. What is the resistance between points X and Y?

Figure 11.35: For question 37.

38In the circuit shown in Figure 11.36, the cell has emf ε and negligible internal resistance. Derive an expression for the voltage V 1 across resistor R 1 and the voltage V 2 across resistor R 2.

Figure 11.36: For question 38.

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39Look at Figure 11.37. The cell has negligible internal resistance and emf 9.0 V.

Figure 11.37: For question 39.

What is the current in each resistor?

40For the potential divider circuit (Figure 11.38), state one change in R 1 and one change in R 2 so that when both changes are made V out will definitely increase.

Figure 11.38: For question 40.

41Three identical lamps are connected to a cell of negligible internal resistance as shown in Figure 11.39.

Figure 11.39: For question 41.

Lamp X burns out. What happens to the voltage across lamp L?

42Three identical lamps are connected to a cell of negligible internal resistance as shown in Figure 11.40.

Figure 11.40: For question 42.

A fourth lamp, identical to the others, is connected in parallel to the lamps X and Y. What happens to the brightness of lamp Z?

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43Figure 11.41 shows a simple circuit with three identical lamps.

Figure 11.41: For question 43.

aThe total power dissipated in the circuit when the switch is open is P. The internal resistance of the cell is negligible. What is the total power dissipated when the switch is closed?

bWhat happens to the brightness of the lamps X and Y when the switch is closed?

44Look in Figure 11.42. The emf of the cell is 24 V and its internal resistance is negligible. The voltmeter reads zero. What is the resistance of resistor X?

Figure 11.42: For question 44.

45Look at Figure 11.43. All cells and resistors are identical. All cells have zero internal resistance.

Figure 11.43: For question 45.

The power dissipated in circuit a is P. What is the power dissipated in circuits b and c?

46Look at Figure 11.44.

The graph shows the I–V characteristics of two resistors, X and Y.

Circuit A shows the resistors X and Y connected in parallel to a cell of emf 1.5 V and negligible internal resistance. Circuit B shows the resistors X and Y connected in series to the same cell.

Figure 11.44: For question 46.

aCalculate the total current leaving the cell in circuit A.

bEstimate the total current leaving the cell in circuit B.

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47Look at Figure 11.45. A cell of emf 3.00 V is connected to two filament lamps and an ohmic resistor R. The I–V characteristics of X and Y are shown in the graph.

Figure 11.45: For question 47.

aThe current in X is 0.15 A. Find the current in Y.

bDetermine the resistance R.

cThe filament of Y breaks.

State, without calculation, the effect of this on the current leaving the cell.

dHence, suggest the effect on the resistance of X.