Question

In the circuit shown in figure, emf of the batteries are $E_1=3\text{V},E_2=2\text{V},E_3=1\text{V}$ and their internal resistances are $R=r_1=r_2=r_3=1\Omega$. The potential difference between the points $\text{A}$ and $\text{B}$, and the current through branch containing resistor $r_2$ will be respectively

• A. $2\text{V},1\text{A}$
• B. $3\text{V},2\text{A}$
• C. $2\text{V},0\text{A}$
• D. $4\text{V},1.5\text{A}$

Answer 1

The correct option is C $2\text{V},0\text{A}$

The given circuit can be visualized as a parallel combination of three non-ideal batteries, and equivalent battery is connected in series with resistor $R$.


So, equivalent emf and resistance of parallel combination will be

$E_{eq}=\frac{\sum \left(\frac{E}{r}\right)}{\sum \left(\frac{1}{r}\right)}$

Putting the value in expression, we get

$E_{eq}=\frac{\frac{3}{1}+\frac{2}{1}+\frac{1}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=\frac{6}{3}$

$\therefore E_{eq}=2\text{V}$

Also, equivalent internal resistance,

$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$

$\Rightarrow \frac{1}{r_{eq}}=1+1+1$

$\therefore r_{eq}=\frac{1}{3}\Omega$

From the equivalent circuit shown in figure, it can be observed that no current will be flown from cell as the circuit is not closed one.

$\therefore V_A-V_B=E_{eq}=2\text{V}$

Now in the branch containing resistance $r_2$, let the current be $i_2$.


Apply KVL between points $\text{A}$ and $\text{B}$,

$V_A+(0\times R)+i_2{r}_2-E_2=V_B$

$\Rightarrow (V_A-V_B)-E_2=-i_2{r}_2$

$\Rightarrow 2-2=-i_2{r}_2$

$\therefore i_2=0$

Current through the branch containing $r_2$ is zero.

Hence, option $\left(c\right)$ is correct.