A Level Physics 10.3 Potential Dividers MCQs – 30 Practice Questions with Answers (Cambridge 9702)

These 30 original practice MCQs cover Cambridge AS & A Level Physics 9702, Section 10.3, Potential Dividers, for AS Level students. They build from the basic divider rule through numerical output calculations to sensor circuits using thermistors and light-dependent resistors (LDRs).

Formula recap: For two resistors in series, Vout = Vin x R2 / (R1 + R2), where R2 is the resistor across which the output is taken. See the full A Level Physics Formula Sheet for all the equations used on this page.

Potential Divider Basics

Q1. A potential divider circuit consists of

  • (A) two resistors in parallel across a supply, with the output taken across the supply
  • (B) two or more resistors in series across a supply, with the output taken across one of them
  • (C) a single resistor connected directly across a supply
  • (D) two cells connected in opposition across one resistor
Show Answer

Answer: B – A potential divider uses resistors in series to split the supply potential difference (p.d.) into smaller parts, and the output p.d. is taken across one of the resistors.

Q2. Resistors R1 and R2 are connected in series across a supply of p.d. Vin. The output p.d. Vout is taken across R2. Which expression gives Vout?

  • (A) Vout = Vin x R1 / (R1 + R2)
  • (B) Vout = Vin x (R1 + R2) / R2
  • (C) Vout = Vin x R2 / (R1 + R2)
  • (D) Vout = Vin x R2 / R1
Show Answer

Answer: C – The output resistor R2 takes the fraction R2 / (R1 + R2) of the supply p.d., so Vout = Vin x R2 / (R1 + R2).

Q3. Two equal resistors are connected in series across a 12 V supply. The output p.d. taken across one of the resistors is

  • (A) 6.0 V
  • (B) 12 V
  • (C) 3.0 V
  • (D) 24 V
Show Answer

Answer: A – Equal resistors share the p.d. equally: Vout = 12 x R / (R + R) = 12 x 1/2 = 6.0 V.

Q4. In a potential divider, the resistor with the larger resistance takes the larger share of the supply p.d. because

  • (A) a larger current flows through the larger resistor
  • (B) charge builds up in the larger resistor
  • (C) the larger resistor stores more energy per second at any current
  • (D) the same current flows through both resistors, and V = IR, so the larger R has the larger V
Show Answer

Answer: D – In series the current is the same everywhere. Since V = IR, the p.d. across each resistor is proportional to its resistance, so the larger resistance gets the larger share.

Q5. In a divider made from R1 (top) and R2 (bottom), the output terminals for Vout = Vin x R2 / (R1 + R2) are connected

  • (A) across the whole series combination
  • (B) in parallel with R2 only
  • (C) in series with R2
  • (D) in parallel with R1 only
Show Answer

Answer: B – The formula gives the p.d. across R2, so the output terminals must be connected across (in parallel with) R2.

Q6. The simple potential divider formula assumes that the output circuit connected across R2

  • (A) draws negligible current, so the same current flows through R1 and R2
  • (B) draws a very large current
  • (C) has zero resistance
  • (D) supplies extra energy to the divider
Show Answer

Answer: A – The formula is derived assuming the current through R1 also passes through R2. If the output device draws significant current (loading), the output p.d. falls below the formula value.

Q7. A divider has R2 equal to three times R1. The fraction of the supply p.d. appearing across R2 is

  • (A) 1/4
  • (B) 1/3
  • (C) 3/4
  • (D) 3/1
Show Answer

Answer: C – Fraction = R2 / (R1 + R2) = 3R1 / (R1 + 3R1) = 3/4.

Q8. Which statement about a two-resistor potential divider is correct?

  • (A) The p.d. across each resistor is the full supply p.d.
  • (B) The current through the larger resistor is smaller than through the smaller resistor
  • (C) The p.d. across R1 plus the p.d. across R2 is less than the supply p.d.
  • (D) The p.d. across R1 plus the p.d. across R2 equals the supply p.d.
Show Answer

Answer: D – By Kirchhoff’s second law, the p.d.s around the series loop add up to the supply p.d., and the series current is the same through both resistors.

Q9. The output of a divider is taken across the fixed resistor R2. R1 is increased while R2 and the supply p.d. stay the same. The output p.d.

  • (A) decreases
  • (B) increases
  • (C) stays the same
  • (D) becomes equal to the supply p.d.
Show Answer

Answer: A – Vout = Vin x R2 / (R1 + R2). Increasing R1 makes the denominator larger while the numerator is unchanged, so R2 takes a smaller fraction and Vout decreases.

Q10. The output of a divider is taken across R2. R2 is increased while R1 and the supply p.d. stay the same. The output p.d.

  • (A) decreases
  • (B) increases
  • (C) stays the same
  • (D) falls to zero
Show Answer

Answer: B – Increasing R2 increases its fraction R2 / (R1 + R2) of the supply p.d., so Vout increases (approaching, but never exceeding, Vin).

Divider Calculations

Q11. A divider has R1 = 2.0 kΩ and R2 = 4.0 kΩ across a 12 V supply. The output taken across R2 is

  • (A) 4.0 V
  • (B) 2.0 V
  • (C) 6.0 V
  • (D) 8.0 V
Show Answer

Answer: D – Vout = 12 x 4.0 / (2.0 + 4.0) = 12 x 4/6 = 8.0 V.

Q12. A divider has R1 = 3.0 kΩ and R2 = 1.0 kΩ across a 20 V supply. The output taken across R2 is

  • (A) 5.0 V
  • (B) 15 V
  • (C) 10 V
  • (D) 6.7 V
Show Answer

Answer: A – Vout = 20 x 1.0 / (3.0 + 1.0) = 20 x 1/4 = 5.0 V.

Q13. A divider has R1 = 4.0 kΩ and R2 = 12 kΩ across an 8.0 V supply. The output taken across R2 is

  • (A) 2.0 V
  • (B) 4.0 V
  • (C) 6.0 V
  • (D) 7.5 V
Show Answer

Answer: C – Vout = 8.0 x 12 / (4.0 + 12) = 8.0 x 12/16 = 6.0 V.

Q14. A divider has a 6.0 kΩ resistor on top and a 2.0 kΩ resistor at the bottom. The output across the 2.0 kΩ resistor is measured as 3.0 V. The supply p.d. is

  • (A) 4.0 V
  • (B) 12 V
  • (C) 9.0 V
  • (D) 24 V
Show Answer

Answer: B – The 2.0 kΩ resistor takes the fraction 2.0 / (6.0 + 2.0) = 1/4 of the supply, so Vin = 3.0 / (1/4) = 12 V.

Q15. A designer needs an output of 4.0 V from a 12 V supply, taken across the bottom resistor R2. Which pair of resistors gives this output?

  • (A) R1 = 5.0 kΩ, R2 = 10 kΩ
  • (B) R1 = 10 kΩ, R2 = 10 kΩ
  • (C) R1 = 15 kΩ, R2 = 5.0 kΩ
  • (D) R1 = 10 kΩ, R2 = 5.0 kΩ
Show Answer

Answer: D – The required fraction is 4.0 / 12 = 1/3. For option D, R2 / (R1 + R2) = 5.0 / (10 + 5.0) = 1/3, giving Vout = 12 x 1/3 = 4.0 V.

Q16. A potentiometer (a variable resistor used as a divider) is connected across a 12 V supply, and the output is taken between one end of the track and the sliding contact. As the slider is moved from that end to the far end, the output p.d. varies over the range

  • (A) 0 to 12 V
  • (B) 6.0 V to 12 V
  • (C) 0 to 6.0 V
  • (D) 12 V throughout
Show Answer

Answer: A – At one end the output is taken across zero resistance, giving 0 V; at the far end it is taken across the whole track, giving the full 12 V. The output varies smoothly between these limits.

Q17. A 10 V supply feeds a divider in which the top resistor is a variable resistor set to 6.0 kΩ and the bottom resistor is a fixed 4.0 kΩ resistor. The output across the fixed resistor is

  • (A) 6.0 V
  • (B) 4.0 V
  • (C) 5.0 V
  • (D) 2.5 V
Show Answer

Answer: B – Vout = 10 x 4.0 / (6.0 + 4.0) = 10 x 4/10 = 4.0 V.

Q18. A divider on a 10 V supply has a 2.0 kΩ resistor on top and an 8.0 kΩ resistor at the bottom, with the output across the bottom resistor. The two resistors are then swapped over. The new output p.d. is

  • (A) 8.0 V
  • (B) 5.0 V
  • (C) 2.0 V
  • (D) 10 V
Show Answer

Answer: C – Originally Vout = 10 x 8.0 / 10 = 8.0 V. After swapping, the bottom resistor is 2.0 kΩ, so Vout = 10 x 2.0 / (2.0 + 8.0) = 2.0 V.

Q19. An output of 9.0 V is required from a 12 V supply, taken across the bottom resistor R2. The top resistor is 1.0 kΩ. The value of R2 must be

  • (A) 0.33 kΩ
  • (B) 1.0 kΩ
  • (C) 4.0 kΩ
  • (D) 3.0 kΩ
Show Answer

Answer: D – The fraction needed is 9.0 / 12 = 3/4, so R2 / (1.0 + R2) = 3/4. This gives 4R2 = 3(1.0 + R2), so R2 = 3.0 kΩ.

Q20. A 15 V supply feeds a divider with a 20 kΩ resistor on top and a 10 kΩ resistor at the bottom. This time the output is taken across the TOP resistor. The output p.d. is

  • (A) 10 V
  • (B) 5.0 V
  • (C) 7.5 V
  • (D) 15 V
Show Answer

Answer: A – The output is the fraction for the resistor it is taken across: Vout = 15 x 20 / (20 + 10) = 15 x 2/3 = 10 V. (The remaining 5.0 V appears across the bottom resistor.)


Sensor Circuits: Thermistors & LDRs

Q21. An NTC thermistor is used in a potential divider. As the temperature of the thermistor rises, its resistance

  • (A) increases
  • (B) stays the same
  • (C) decreases
  • (D) first increases, then decreases
Show Answer

Answer: C – An NTC (negative temperature coefficient) thermistor releases more charge carriers as it warms up, so its resistance decreases as temperature rises.

Q22. A thermistor is the bottom resistor of a divider, with a fixed resistor on top, and the output is taken across the thermistor. The temperature rises. The output p.d.

  • (A) increases, because the thermistor resistance increases
  • (B) stays the same
  • (C) increases, because the total resistance increases
  • (D) decreases, because the thermistor resistance falls and it takes a smaller share of the supply p.d.
Show Answer

Answer: D – Temperature up means Rthermistor down. The output fraction Rthermistor / (Rfixed + Rthermistor) falls, so the output across the thermistor decreases. (At the same time, the p.d. across the fixed resistor rises.)

Q23. A thermistor is the bottom resistor of a divider, with a fixed resistor on top, and the output is taken across the thermistor. The temperature falls. The output p.d.

  • (A) decreases
  • (B) increases
  • (C) stays the same
  • (D) falls to zero
Show Answer

Answer: B – Temperature down means Rthermistor up. The thermistor then takes a larger fraction of the supply p.d., so the output across it increases.

Q24. A thermistor is the TOP resistor of a divider, and the output is taken across the fixed resistor at the bottom. The temperature rises. The output p.d.

  • (A) increases
  • (B) decreases
  • (C) stays the same
  • (D) becomes equal to the supply p.d. immediately
Show Answer

Answer: A – Temperature up means Rthermistor (top) down. The fixed bottom resistor then takes a larger fraction, Rfixed / (Rthermistor + Rfixed), of the supply, so the output across it increases.

Q25. In a thermostat circuit, a thermistor is the bottom resistor of a divider and the output across the thermistor drives a transistor that switches a heater ON when the output p.d. is high. The heater switches on when it is cold because

  • (A) the thermistor resistance is low when cold, giving a high output
  • (B) the supply p.d. rises when it is cold
  • (C) the thermistor resistance is high when cold, so it takes a large share of the supply p.d. and the output is high
  • (D) the fixed resistor takes the whole supply p.d. when it is cold
Show Answer

Answer: C – When cold, the NTC thermistor has a high resistance, so its fraction of the supply p.d. is large and the output across it is high, switching the transistor and heater on. As the room warms, the resistance and output fall and the heater switches off at the threshold.

Q26. An LDR is the bottom resistor of a divider, with a fixed resistor on top, and the output is taken across the LDR. The light falling on the LDR gets brighter. The output p.d.

  • (A) increases, because the LDR resistance increases in bright light
  • (B) stays the same
  • (C) increases, because the total current increases
  • (D) decreases, because the LDR resistance falls in bright light and it takes a smaller share of the supply p.d.
Show Answer

Answer: D – Brighter light means RLDR down. The fraction RLDR / (Rfixed + RLDR) falls, so the output across the LDR decreases. (The p.d. across the fixed resistor rises instead.)

Q27. An LDR is the bottom resistor of a divider, with a fixed resistor on top, and the output is taken across the LDR. The surroundings get darker. The output p.d.

  • (A) decreases
  • (B) increases
  • (C) stays the same
  • (D) falls to zero
Show Answer

Answer: B – In darkness the LDR resistance rises, so the LDR takes a larger fraction of the supply p.d. and the output across it increases. This is why this arrangement is used in automatic lighting circuits that respond to darkness.

Q28. An LDR is the TOP resistor of a divider, and the output is taken across the fixed resistor at the bottom. The light falling on the LDR gets brighter. The output p.d.

  • (A) decreases
  • (B) stays the same
  • (C) increases
  • (D) falls to zero
Show Answer

Answer: C – Brighter light means RLDR (top) down. The fixed bottom resistor then takes a larger fraction, Rfixed / (RLDR + Rfixed), of the supply, so the output across it increases.

Q29. A designer wants a divider output p.d. that RISES as the temperature rises, using an NTC thermistor and one fixed resistor. The correct arrangement is

  • (A) thermistor as the top resistor, with the output taken across the fixed resistor at the bottom
  • (B) thermistor as the bottom resistor, with the output taken across the thermistor
  • (C) thermistor and fixed resistor in parallel
  • (D) thermistor as the top resistor, with the output taken across the thermistor
Show Answer

Answer: A – As temperature rises, the thermistor resistance falls. With the thermistor on top, the fixed bottom resistor takes a growing share of the supply p.d., so the output across the fixed resistor rises with temperature. Taking the output across the thermistor itself would give an output that falls with temperature.

Q30. In an automatic street-light circuit, an LDR is the bottom resistor of a divider and the output across the LDR switches the lamp ON when the output p.d. is high. The lamp switches on

  • (A) at midday, because the LDR resistance is then at its highest
  • (B) in bright light, because the output across the LDR is then at its highest
  • (C) whenever the supply is connected, regardless of light level
  • (D) when it gets dark, because the LDR resistance rises and the output across it rises
Show Answer

Answer: D – In darkness RLDR is high, so the LDR takes a large share of the supply p.d. and the output across it is high, switching the lamp on. In bright light its resistance and output fall, and the lamp switches off.

Keep practising: work through more topic pages and full practice tests from the Physics MCQs hub.

Revise all the circuit equations, including the potential divider rule, on the A Level Physics Formula Sheet.

These are original practice questions written for revision, not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.