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Question

A cell E1 of emf6V and internal resistance 2Ω is connected with another cell E2 of emf4V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is


A

3.6V

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B

10.0V

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C

5.6V

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D

2.0V

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Solution

The correct option is C

5.6V


Step1: Given data.

Electromotive force emf of cell E1=6V

Electromotive force emf of cell E1=4V

Resistance, R1=2Ω

Resistance, R1=8Ω

Step2: Finding the equivalent current in the circuit.

We know that:

Ieq=EeffReq

Where, Ieq=equivalent current, Eeff=effective emf,Req=equivalent resistance.

According to Kirchhoff Voltage Law:

Eeff=E1-E2=6V-4V=2V

In Given figure Resistance In Series:

Req=2Ω+8Ω=10Ω

Therefore,

Ieq=EeffReq=2V10Ω=0.2A

Step3: Finding potential difference across points X and Y

Applying Kirchhoff Voltage Law between points X and Y.

Vx-4-0.2×8+Vy=0

Vx-Vy=4+1.6

Vx-Vy=5.6V

Hence, Option C is correct option. The potential difference across points X and Y is 5.6V


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