CAIE IGCSE Physics 4.3.3 Action & use of circuit components Question Bank
Practise interpreting potential dividers, sensor components and p.d. changes using circuit diagrams, tables and calculations.
- Syllabus
- 2026–2028
- Course
- Physics 0625
Practise interpreting potential dividers, sensor components and p.d. changes using circuit diagrams, tables and calculations.
Fig. 6.1 shows a circuit which includes an LDR (light-dependent resistor).

Fig. 6.1
The LDR is in a brightly lit room. The voltmeter reads 1.8 V .
The light level in the room changes from bright to dark.
State and explain the effect on the voltmeter reading.
statement
explanation
increases
B1
resistance of LDR increases
B1
ratio of potential differences (across each component) equals ratio of the two resistances ORR1/R2=V1/V2
OR
LDR has greater share of the total / supply voltage
OR
potential difference of the source is divided between the resistor ( R ) and LDR
B1
Fig. 8.2 shows a circuit.

Fig. 8.2
The three cells are identical and have zero resistance.
The resistors R1,R2 and R3 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.
Determine the ratio of the p.d. across R2 to the p.d. across R3.
(ratio of p.d. across R2:R3= ) 1: 2
B1
All the cells are reversed.
1. State and explain the change in current in R1.
2. Determine the new value of the ratio of the p.d. across R2 to the p.d. across R3.
1.
current is zero in R1 AND diode is in wrong direction (to allow current) owtte
B1
8(c)(iii)2.
(ratio of p.d. across R2:R3= ) 1: 1
B1