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9702 · 10.3

Potential dividers — practice questions

Practice and worked examples for 9702 Potential dividers. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A 12V power supply is connected across two series resistors: R1 = 200Ω and R2 = 400Ω. Calculate the output voltage taken across R2.

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  1. Identify the input voltage: Vin=12VV_{in} = 12V.
  2. Identify the resistances: R1=200ΩR_1 = 200\Omega and R2=400ΩR_2 = 400\Omega.
  3. Apply the potential divider formula: Vout=Vin×R2R1+R2V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}.
  4. Substitute the values: Vout=12V×400Ω200Ω+400ΩV_{out} = 12V \times \frac{400\Omega}{200\Omega + 400\Omega}.
  5. Calculate the sum of resistances: Vout=12V×400600V_{out} = 12V \times \frac{400}{600}.
  6. Simplify the fraction and calculate: Vout=12V×23=8VV_{out} = 12V \times \frac{2}{3} = 8V.
  7. The output voltage across R2 is 8V.

Worked example 2

A light-sensing circuit uses a 9.0 V supply, a fixed 10 kΩ resistor, and a Light Dependent Resistor (LDR) in series. The output voltage is taken across the LDR. A switch connected to the output is triggered when the voltage reaches 6.0 V, indicating it is dark. What is the resistance of the LDR at this point?

Show solution outline
  1. Identify known values:
    • Input Voltage, Vin=9.0 VV_{in} = 9.0 \text{ V}
    • Fixed Resistor, R1=10 kΩ=10,000 ΩR_1 = 10 \text{ k}\Omega = 10,000 \text{ } \Omega
    • Output Voltage, Vout=6.0 VV_{out} = 6.0 \text{ V}
    • The output is across the LDR, so R2=RLDRR_2 = R_{LDR}.
  2. State the potential divider formula: Vout=Vin×RLDRR1+RLDRV_{out} = V_{in} \times \frac{R_{LDR}}{R_1 + R_{LDR}}
  3. Rearrange the formula to solve for RLDRR_{LDR}:
    • Divide by VinV_{in}: VoutVin=RLDRR1+RLDR\frac{V_{out}}{V_{in}} = \frac{R_{LDR}}{R_1 + R_{LDR}}
    • Cross-multiply: Vout(R1+RLDR)=VinRLDRV_{out}(R_1 + R_{LDR}) = V_{in}R_{LDR}
    • Expand the bracket: VoutR1+VoutRLDR=VinRLDRV_{out}R_1 + V_{out}R_{LDR} = V_{in}R_{LDR}
    • Group RLDRR_{LDR} terms: VoutR1=VinRLDRVoutRLDRV_{out}R_1 = V_{in}R_{LDR} - V_{out}R_{LDR}
    • Factor out RLDRR_{LDR}: VoutR1=RLDR(VinVout)V_{out}R_1 = R_{LDR}(V_{in} - V_{out})
    • Isolate RLDRR_{LDR}: RLDR=VoutR1VinVoutR_{LDR} = \frac{V_{out}R_1}{V_{in} - V_{out}}
  4. Substitute the values and calculate: RLDR=6.0 V×10,000 Ω9.0 V6.0 VR_{LDR} = \frac{6.0 \text{ V} \times 10,000 \text{ } \Omega}{9.0 \text{ V} - 6.0 \text{ V}} RLDR=60,0003.0 ΩR_{LDR} = \frac{60,000}{3.0} \text{ } \Omega RLDR=20,000 ΩR_{LDR} = 20,000 \text{ } \Omega
  5. Final Answer: The resistance of the LDR is 20 kΩ when the switch is triggered.