CAIE IGCSE Physics Potential-Divider Ratio R1/R2 = V1/V2

Practise using resistance ratios to divide supply voltage, calculate an unknown resistor or voltmeter reading and interpret sensor-based divider circuits.

Exam points

apply R_1/R_2 = V_1/V_2 to match each resistor with the p.d. measured across it
use V_out = V_supply R/(R_1 + R_2) to calculate an output or an unknown resistance
combine a thermistor or LDR resistance with the fixed resistor to find the changing voltmeter reading

[Maximum number: 3]

(a)

Fig. 8.2 shows a circuit.

Fig. 8.2

The three cells are identical and have zero resistance.
The resistors $R_{1}, R_{2}$ and $R_{3}$ are identical.
The reading on the voltmeter is 6.0 V .
When the diode is conducting, it has zero resistance and zero potential difference (p.d.) across it.

[ 3 ]

(i)

Determine the ratio of the p.d. across $R_{2}$ to the p.d. across $R_{3}$ .

[ 1 ]

(ratio of p.d. across $\mathrm{R}_{2}: \mathrm{R}_{3}=$ ) 1: 2

B1

(ii)

All the cells are reversed.
1. State and explain the change in current in $R_{1}$ .
2. Determine the new value of the ratio of the p.d. across $R_{2}$ to the p.d. across $R_{3}$ .

[ 2 ]

1.

current is zero in $\mathrm{R}_{1}$ AND diode is in wrong direction (to allow current) owtte

B1

8(c)(iii)2.

(ratio of p.d. across $\mathrm{R}_{2}: \mathrm{R}_{3}=$ ) 1: 1

B1

All question bank results loaded

4.3.3.3—Potential divider relation: R1/R2 = question 1

Question (a)

Question (i)

Question (ii)