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. $3.6V$
• B. $10.0V$
• C. $5.6V$
• D. $2.0V$

Answer 1

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 .

$V_x-4-0.2\times 8+V_y=0$

$\Rightarrow$ $V_x-V_y=4+1.6$

$\Rightarrow$ $\left|V_x-V_y\right|=5.6V$

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