Equivalent Resistance of Multiple Resistors

Equivalent resistance of multiple resistors

1. N95/I/15; J2000/I/15

Four resistors are connected as shown.

Between which 2 points is the resistance of the combination a maximum?

A P and Q C R and S

B Q and S D S and P

Suggested solution

By connecting across any two terminals (P, Q, R or S), the four resistors are effectively connected in parallel (with some connected in series first). (Refer to the circuit diagram below)

P and Q:

Q and S:

R and S:

S and P:

The effective resistance of resistors in parallel is less than the resistance in any of the ‘branches’. Thus for maximum effective resistance, the choice of the terminal should give resistance of each branch to be approximately similar, i.e. choice B where each branch has a subtotal of 5 Ω. And

Resistance between Q and S:

RQS = 2.5 Ω

(If not, you can work out all the configurations)

Ans: B

2. Five resistors of resistance 2.0 W are connected as shown in the diagram below.

What is the equivalent resistance between points P and Q?

Suggested solution

Assumption: no current flows through middle resistor.

Thus the potential at X can be assumed to be only affected by the two resistors at the top while the potential at Y can be assumed to be only affected by the two resistors at the bottom. Hence VX and VY are the same value (using potential divider rule on each branch).

Therefore, the p.d. across the middle resistor is zero and the assumption of no current flowing through the middle resistor is proven. Hence the circuit diagram can be considered without the middle resistor.

R = 2.0 Ω

The equivalent resistance is 2.0 Ω.

3. Given that each resistor has a resistance of R, determine the total resistance between the points PQ in the circuits shown in Fig. 3(a), 3(b), and 3(c).

Suggested solution

Take note of how each resistor are being connected to another, i.e. terminals of each resistor.

(a)

(b) Note: Point X has the same potential as Point Q while Point Y has the same potential as Point P.

RPQ =

(c)

R1 and R2 are connected in series in which they are connected in parallel to R3. These resistors R1, R2 and R3 are then connected in series to R4. And the whole R1, R2, R3 and R4 are connected in parallel to R5.

Effective Resistance

RPQ =

Electric circuits with one emf

4. (J2006/II/2)

The figure below shows an electrical circuit in which the internal resistance of the battery is negligible.

Complete the table below by giving the electrical quantities for each of the components in the above circuit. You are advised to start by completing the column for component A.

Suggested solution

circuit component / A / B / C / whole circuit
potential difference / V / 12 / 10 / 2 / 12
current / A / 3.0 / 2.0 / 2.0 / 5.0
power / W / 36 / 20 / 4 / 60
resistance / W / 4.0 / 5.0 / 1.0 / 2.4

B and C are connected in series è IB = IC

VXY = VB + VC

(B+C) is in turn connected in parallel to A è VA = VXy = VB + VC

IT = IA + IB or IT = IA + IC

circuit component / A / B / C / whole circuit
potential difference / V / (1) VA = Vxy = E / (4) VB = IB × RB / (5) VA = VB + VC / 12
current / A / (2) IA = VA / RA / (3) IT = IA + IB / (5) IB = IC / 5.0
power / W / P = V2 / R = I2 R = IV
resistance / W / 4.0 / 5.0 / (6) RC = VC / IC / (7) RT = E / IT

5. N03/II/2

The variation with current of the potential difference (p.d.) across a component X is shown in Fig. 2.1.

(a) (i) State how the resistance of component X varies, if at all, with increase of current.

(ii) On Fig. 2.1, draw a line to show the variation with current of the pd across a resistor R of constant resistance 3.0 W.

(b) The component X and the resistor R of resistance 3.0 W are connected in series with a battery of negligible internal resistance, as shown in Fig. 2.2.

The current in the circuit is found to be 2.0 A.

(i) Use Fig. 2.1 to determine the p.d. across component X.

(ii) Determine

1. the p.d. across R,

2. the emf of the battery.

Use your answer to (b)(ii) and the graph of Fig. 2.1, determine the current from the battery.

Suggested solution

(a) (i) Resistance of X increases with an increase of current. (The best way is to physically calculate the resistance of two different scenario with difference current)

(ii) For a resistance with 3.0 Ω, when the current is 2.0 A, the p.d. across the resistor of 6.0 V is required. Thus to show an ohmic resistor, draw a line passing through origin and coordinates (2,6) on the graph.

(b) (i) When the current in X is 2.0 V, p.d. across X is 5.0 V [Read off from Fig 2.1(green line)]

(ii) 1. Since X and resistor R is connected in series, the current flowing through both component is the same. Thus, with a current of 2.0 A, a p.d. of 6.0 V is required. [Read off from Fig 2.1 (blue line)]

2. Since X and resistor R is connected in series, E = VX + VR = 5.0 + 6.0 =11 V

è E = VX = VR

è IT = IX + IR

Since the same battery is used, p.d. across X and R is 11 V (e.m.f. of battery).

Current through X, IX = 2.9 A (from Fig 2.1. (orange line)

Current through R, = 3.7 A

Total current drawn from battery, IT = 3.7 + 2.9 = 6.6 A

Potential divider circuits

6. In the figure below, the two lamps L1 and L2 labelled 3.0 V, 0.25 A and 6.0 V, 0.05 A, respectively, light up with normal brightness.

(a) Calculate the values of the resistance R1 and R2.

(b) What happens to L1 if L2 burns out first?

Suggested solution

L1 and R2 are connected in series. Thus the current through L1 and R2 is the same while the sum of voltage across L1 and R2 is the voltage across XY.

è IL1 = IR1

è VXY = VL1 + VR2

IL1 = IR1

IR1 = 0.25 A

Since L2 lights up normally, the p.d. across it is 6.0 V which is equivalent to VXY.

VXY = VL1 + VR2

6.0 = 3.0 + VR2

VR2 = 3.0 V

R1 is connected in series with (L2 and L1 + R2 which are connected in parallel).

è E = VR1 + VL2 or E = VR1 + VL1 + VR2

è IT = IR1 = IL1 + IL2 or IT = IR1 = IR2 + IL2

E = VR1 + VL2

12 = VR1 + 6.0

VR1 = 6.0 V

IT = IR1 = IL1 + IL2

IR1 = 0.25 + 0.05

= 0.30 A

(b) If L2 burns out first, no current flows through L2, Thus only R1, L1 and R2 are connected in series.

Total resistance in circuit, RT

= R1 + R2 + Resistance of L1

= 20 + 12 +

= 44 W

E = ITRT

12 = IT × 44

IT = 0.27 A (> 0.25 A)

Hence, the lamp L1 shines more brightly

(c) If L1 burns out first, no current can flow through L1 thus no current will flow through R2 as well. Thus the only R1 and L2 are connected in series.

Total resistance in circuit

= R1 + Resistance of L2

= 20 +

= 140 W

E = ITRT

12 = IT × 140

IT = 0.086 A (> 0.05 A)

Hence, the lamp L2 shines more brightly.

7. A thermistor has a resistance 3900 W at 0 °C and resistance 1250 W at 30 °C. The thermistor is connected into the circuit below in order to monitor temperature changes.

The battery of e.m.f 1.50 V has negligible internal resistance and the voltmeter has infinite resistance.

(a) The voltmeter is to read 1.00 V at 0 °C. Show that the resistance of resistor R is 7800 W.

(b) The temperature of the thermistor is increased to 30 °C . Determine the reading on the voltmeter.

(c) The voltmeter in the figure is replaced with one having a resistance of 7800 W. Calculate the reading on this voltmeter for the thermistor at a temperature of 0 °C.

Suggested solution

(a) Since the resistance of voltmeter is infinite, there is negligible current through the voltmeter. Thus, the thermistor and resistor are connected in series.

Thus, using the potential divider principle, having the same current

At 0 °C:

R = 7800 W

(b) When the temperature is 30 °C , the resistance of the thermistor is 1250 W.

Using the potential divider principle, still having the same current

VR = 1.29 V

Effective resistance of voltmeter and R

R ‘ = 3900 Ω

This effective resistance R ‘ is still connected in series to the thermistor. Thus potential divider rule is still applicable.

VR ’ = 0.75 V

8. The galvanometer reading is zero in the circuit.

What is the value of resistor X?

Suggested solution

When there is no current flowing through the galvanometer, VA = VC (i.e. VAC = 0)

And VB = VD

Therefore VAB = VCD

The circuit can thus seen as the circuit diagram below:

Looking at the right circuit, it is an open circuit. Thus VCD = E = 2.0 V

With VAB = VCD

VX = 2.0 V (Since VAB = VX)

Looking at the left circuit, the two resistors of 75 Ω and X are connected in series. Thus using potential divider

RX = 50 Ω

9. Fig. 9 shows a 120 V supply that is connected to two loads A and B through distribution wires of resistance per unit length 1.00 x 10-4 Ω m-1. Load A is 200 m away from the supply, while load B is a further 100 m away from load A.

(a) Calculate current I.

(b) Determine the potential difference across load A and load B.

Suggested solution

The distribution wire can be considered as resistor that is connected along the path where the current flow in the circuit.

(Look at the circuit diagram and try to imagine how each resistor is connected to each other)

(a) I = 50 + 30 = 80 A.

(b) For load A:

E = I(0.020) + IA RA + I(0.020)

120 = (80 × 0.020) + VA + (80 × 0.020)

VA = 117 V

For load B:

E = I(0.020) + IB(0.01) + IB RB + IB(0.01) + I(0.020)

120 = (80 × 0.020) + (30 × 0.010) + VB + (30 × 0.010) + (80 × 0.020)

VB = 116 V

Alternative: VA = IB(0.01) + IB RB + IB(0.01)

Potentiometer

10 (a) A uniform wire AB of resistance 1.5 W is connected in series with a resistor of 5.0 W, a cell of e.m.f. of 9.0 V and internal resistance 1.0 W, as shown in Fig. 2.1. Calculate the potential difference (p.d.) across the wire AB.

Suggested solution

(a) There are a total of 3 resistors (i.e. 1.0 W, 5.0 W and 1.5 W) connected in series.

By using potential divider rule:

VAB = 1.8 V

(b) A cell C of e.m.f. 1.5 V and internal resistance 0.80 W is connected to the circuit of Fig. 2.1, as shown in Fig. 2.2.

The moveable contact can be connected to any point along the wire AB. At point D, there is zero current in the galvanometer.

(i) Calculate the length of AD.

(ii) State and explain one advantage of using Fig. 2.2 to measure the e.m.f. of C compared to using a laboratory voltmeter.

Suggested solution

(b) At point D, there is zero current in the galvanometer, there is no p.d. between D and F,

i.e. VD = VF.

and since point A and point E are essentially the same ‘point’ in the wire, VA = VE

Thus VAD = VEF.

The circuit can thus seen as the circuit diagram below:

Looking at the lower circuit, it is an open circuit.

Thus the p.d. of EF is the same as that of the e.m.f. of cell C, i.e. VEF = EC = 1.5 V

Thus with VAD = VEF

VAD = 1.5 V

V = IR and

Therefore

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