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V = IR in series and parallel circuits

Electricity
GCSE Physics (8463)
ORIGINAL

The same two resistors in series, then in parallel: share the p.d., add the branch currents, and explain why parallel resistance falls.

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Question

A 4.0 Ω4.0\,\Omega resistor and an 8.0 Ω8.0\,\Omega resistor are connected in series to a 12 V12\,\text{V} battery. (a) Calculate the current in the circuit and the potential difference across the 8.0 Ω8.0\,\Omega resistor. (b) The two resistors are then reconnected in parallel across the same battery. Calculate the current from the battery and explain why the total resistance is now less than 4.0 Ω4.0\,\Omega.

Every step worked, with the reasoning.

  1. 1
    Series: Rtotal=R1+R2=4.0+8.0=12 ΩR_{\text{total}} = R_1 + R_2 = 4.0 + 8.0 = 12\,\Omega

    In series, the total resistance is the sum of the resistances.

  2. 2
    I=VR=1212=1.0 AI = \dfrac{V}{R} = \dfrac{12}{12} = 1.0\,\text{A}

    Rearrange V=IRV = IR. In series the same current passes through every component.

  3. 3
    V8.0 Ω=IR=1.0×8.0=8.0 VV_{8.0\,\Omega} = IR = 1.0 \times 8.0 = 8.0\,\text{V}

    Check: 4.0 V+8.0 V=12 V4.0\,\text{V} + 8.0\,\text{V} = 12\,\text{V} — in series the battery's potential difference is shared between the components.

  4. 4
    Parallel: each resistor has the full 12 V12\,\text{V}, so I4.0 Ω=124.0=3.0 AI_{4.0\,\Omega} = \dfrac{12}{4.0} = 3.0\,\text{A} and I8.0 Ω=128.0=1.5 AI_{8.0\,\Omega} = \dfrac{12}{8.0} = 1.5\,\text{A}

    In parallel, the potential difference across each branch is the same as the supply.

  5. 5
    Ibattery=3.0+1.5=4.5 AI_{\text{battery}} = 3.0 + 1.5 = 4.5\,\text{A}, so Rtotal=124.5=2.7 ΩR_{\text{total}} = \dfrac{12}{4.5} = 2.7\,\Omega (2 s.f.)

    The total current is the sum of the branch currents. The second branch is an extra path for the charge, so more current flows for the same p.d. — the total resistance is less than the smallest single resistance.

Answer: (a) 1.0 A1.0\,\text{A} and 8.0 V8.0\,\text{V}; (b) 4.5 A4.5\,\text{A} — the second branch gives the current an extra path, so the total current rises and the total resistance (2.7 Ω2.7\,\Omega) is less than either resistor.

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