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Difficulty: HardElectric Circuits and Measuring Instruments

A driver cell of electromotive force E0=4.0 VE_0 = 4.0\text{ V} and internal resistance r=1.0 Ωr = 1.0\ \Omega is connected across a uniform potentiometer wire of length 100 cm100\text{ cm} and total resistance Rw=9.0 ΩR_w = 9.0\ \Omega. A test cell of unknown electromotive force ExE_x and internal resistance rx=0.5 Ωr_x = 0.5\ \Omega is connected in series with a sensitive galvanometer in the secondary circuit. When the jockey touches the wire at a distance of 60 cm60\text{ cm} from the high-potential end, the galvanometer indicates zero deflection. What is the electromotive force ExE_x of the test cell?

  1. 2.16 V2.16\text{ V}Answer
  2. B
    2.40 V2.40\text{ V}
  3. C
    1.96 V1.96\text{ V}
  4. D
    3.60 V3.60\text{ V}

Answer

The electromotive force of the test cell is 2.16 V2.16\text{ V}.
The correct answer is obtained by first including the driver cell's internal resistance to find the true driver current of 0.40 A. The total voltage across the 100 cm wire is therefore 3.60 V, yielding a potential gradient of 0.036 V/cm. Multiplying by the 60 cm balance length gives an EMF of 2.16 V for the test cell.

Step-by-Step Solution

1
Calculate total resistance in the primary driver circuit.
Rtotal=Rw+r=9.0 Ω+1.0 Ω=10.0 ΩR_{\text{total}} = R_w + r = 9.0\ \Omega + 1.0\ \Omega = 10.0\ \Omega
The driver cell's internal resistance is in series with the potentiometer wire resistance.
2
Calculate the current flowing through the driver circuit.
I=E0Rtotal=4.0 V10.0 Ω=0.40 AI = \frac{E_0}{R_{\text{total}}} = \frac{4.0\text{ V}}{10.0\ \Omega} = 0.40\text{ A}
Ohm's law applied to the complete primary circuit gives the steady driver current.
3
Determine the potential drop across the entire 100 cm potentiometer wire.
Vw=I×Rw=0.40 A×9.0 Ω=3.60 VV_w = I \times R_w = 0.40\text{ A} \times 9.0\ \Omega = 3.60\text{ V}
The potential drop across the wire depends on driver current and wire resistance.
4
Calculate potential gradient and solve for the unknown EMF ExE_x at the 60 cm balance point.
k=VwLtotal=3.60 V100 cm=0.036 V/cmk = \frac{V_w}{L_{\text{total}}} = \frac{3.60\text{ V}}{100\text{ cm}} = 0.036\text{ V/cm}, so Ex=k×L=0.036 V/cm×60 cm=2.16 VE_x = k \times L = 0.036\text{ V/cm} \times 60\text{ cm} = 2.16\text{ V}
At zero galvanometer deflection, no current flows through the test cell, so its terminal voltage equals its EMF ExE_x.

Key Concept

Potentiometer balance condition and primary circuit internal resistance considerations
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Topics

Electric Circuits and Measuring Instruments

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