Question

Difficulty: HardElectric Circuits and Measuring Instruments

A battery with an electromotive force (e.m.f.) of 12.0 V12.0\text{ V} and an internal resistance of 1.0 Ω1.0\text{ }\Omega is connected to a circuit containing two resistors, 6.0 Ω6.0\text{ }\Omega and 3.0 Ω3.0\text{ }\Omega, connected in parallel with each other. What is the potential difference across the 3.0 Ω3.0\text{ }\Omega resistor?

  1. 8.0 V8.0\text{ V}Answer
  2. B
    12.0 V12.0\text{ V}
  3. C
    4.0 V4.0\text{ V}
  4. D
    3.6 V3.6\text{ V}

Answer

The potential difference across the 3.0 Ω3.0\text{ }\Omega resistor is 8.0 V8.0\text{ V}.
The parallel external resistors of 6.0 Ω6.0\text{ }\Omega and 3.0 Ω3.0\text{ }\Omega combine to give an equivalent resistance of 2.0 Ω2.0\text{ }\Omega. Including the internal resistance of 1.0 Ω1.0\text{ }\Omega, the total circuit resistance is 3.0 Ω3.0\text{ }\Omega. This leads to a total circuit current of 4.0 A4.0\text{ A}. Multiplying this current by the external equivalent resistance of 2.0 Ω2.0\text{ }\Omega gives the terminal voltage of 8.0 V8.0\text{ V}, which is identical across both parallel branches.

Step-by-Step Solution

1
Calculate the equivalent resistance (RpR_p) of the parallel combination.
Rp=6.0×3.06.0+3.0=18.09.0=2.0 ΩR_p = \frac{6.0 \times 3.0}{6.0 + 3.0} = \frac{18.0}{9.0} = 2.0\text{ }\Omega
Resistors in parallel combine according to reciprocal addition.
2
Find the total circuit resistance (RtotalR_{\text{total}}) including the internal resistance (rr).
Rtotal=Rp+r=2.0 Ω+1.0 Ω=3.0 ΩR_{\text{total}} = R_p + r = 2.0\text{ }\Omega + 1.0\text{ }\Omega = 3.0\text{ }\Omega
The internal resistance of the power supply acts in series with the external equivalent load resistance.
3
Calculate the total current (II) drawn from the battery using Ohm's law for a complete circuit.
I=ERtotal=12.0 V3.0 Ω=4.0 AI = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{3.0\text{ }\Omega} = 4.0\text{ A}
Current is determined by the total electromotive force divided by the total resistance of the circuit.
4
Determine the terminal potential difference (VV) across the parallel network.
V=I×Rp=4.0 A×2.0 Ω=8.0 VV = I \times R_p = 4.0\text{ A} \times 2.0\text{ }\Omega = 8.0\text{ V}
Components connected in parallel experience the same potential difference, which equals the terminal voltage of the battery.

Key Concept

Terminal potential difference in circuits with internal resistance
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