Question

A cell ${E}_{1}$ of $emf6V$ and internal resistance $2\Omega$ is connected with another cell ${E}_{2}$ of $emf4V$ and internal resistance $8\Omega$ (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 $\left(emf\right)$ of cell ${E}_{1}=6V$Electromotive force $\left(emf\right)$ of cell ${E}_{1}=4V$Resistance, ${R}_{1}=2\Omega$Resistance, ${R}_{1}=8\Omega$Step2: Finding the equivalent current in the circuit.We know that:${I}_{eq}=\frac{{E}_{eff}}{{R}_{eq}}$Where, ${I}_{eq}=$equivalent current, ${E}_{eff}=$effective $emf$,${R}_{eq}=$equivalent resistance.According to Kirchhoff Voltage Law: ${E}_{eff}={E}_{1}-{E}_{2}=6V-4V=2V$In Given figure Resistance In Series: ${R}_{eq}=2\Omega +8\Omega =10\Omega$Therefore,${I}_{eq}=\frac{{E}_{eff}}{{R}_{eq}}=\frac{2V}{10\Omega }=0.2A$Step3: Finding potential difference across points $X$ and $Y$Applying Kirchhoff Voltage Law between points and .${V}_{x}-4-0.2×8+{V}_{y}=0$$⇒$ ${V}_{x}-{V}_{y}=4+1.6$$⇒$ $\left|{V}_{x}-{V}_{y}\right|=5.6V$Hence, Option C is correct option. The potential difference across points $X$ and $Y$ is $5.6V$

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