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9702 · 10.1

Practical circuits — practice questions

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

Worked example 1

A potential divider is constructed with a 12.0 V supply connected to two resistors in series, R₁ = 2.0 kΩ and R₂ = 4.0 kΩ. A high-resistance voltmeter is connected across R₂. What is the reading on the voltmeter?

Show solution outline
  1. Identify the known values: Vin=12.0 VV_{in} = 12.0 \text{ V}, R1=2.0 kΩR_1 = 2.0 \text{ k}\Omega, R2=4.0 kΩR_2 = 4.0 \text{ k}\Omega.
  2. The output voltage, VoutV_{out}, is the voltage across R2R_2. The total resistance of the series combination is RtotalR_{total}.
  3. Calculate the total resistance: Rtotal=R1+R2=2.0 kΩ+4.0 kΩ=6.0 kΩR_{total} = R_1 + R_2 = 2.0 \text{ k}\Omega + 4.0 \text{ k}\Omega = 6.0 \text{ k}\Omega.
  4. Apply the potential divider formula: Vout=Vin×R2RtotalV_{out} = V_{in} \times \frac{R_2}{R_{total}}.
  5. Substitute the values into the formula: Vout=12.0 V×4.0 kΩ6.0 kΩV_{out} = 12.0 \text{ V} \times \frac{4.0 \text{ k}\Omega}{6.0 \text{ k}\Omega}.
  6. The units of kΩ cancel out, simplifying the fraction: Vout=12.0 V×23V_{out} = 12.0 \text{ V} \times \frac{2}{3}.
  7. Calculate the final answer: Vout=8.0 VV_{out} = 8.0 \text{ V}. The voltmeter reads 8.0 V.

Worked example 2

A cell with an e.m.f. of 1.5 V is connected to an external resistor of 4.0 Ω. The current flowing through the circuit is measured as 0.30 A. Calculate the internal resistance of the cell.

Show solution outline
  1. Identify the known values: ε=1.5 V\varepsilon = 1.5 \text{ V}, R=4.0ΩR = 4.0 \Omega, I=0.30 AI = 0.30 \text{ A}.
  2. We need to find the internal resistance, rr.
  3. Use the e.m.f. equation that relates all these quantities: ε=I(R+r)\varepsilon = I(R + r).
  4. Substitute the known values into the equation: 1.5 V=0.30 A×(4.0Ω+r)1.5 \text{ V} = 0.30 \text{ A} \times (4.0 \Omega + r).
  5. Rearrange the equation to solve for the term in the brackets: 1.50.30=4.0+r\frac{1.5}{0.30} = 4.0 + r.
  6. This simplifies to: 5.0=4.0+r5.0 = 4.0 + r.
  7. Therefore, the internal resistance r=5.0Ω−4.0Ω=1.0Ωr = 5.0 \Omega - 4.0 \Omega = 1.0 \Omega.
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