Question

The potential difference $V_A-V_B$ in the circuit shown is :
Given:
$E_1=1.5\text{V},E_2=2.0\text{V},E_3=2.5\text{V}$
$R_1=10\Omega ,R_2=20\Omega ,R_3=30\Omega$

• A. $\frac{11}{20}\text{V}$
• B. $\frac{10}{11}\text{V}$
• C. Zero
• D. $\frac{8}{3}\text{V}$

Answer 1

The correct option is B $\frac{10}{11}\text{V}$

The given circuit can be visualized as a combination of three non-ideal batteries in parallel connection across points $\text{A}$ and $\text{B}$.

So, net emf of circuit is given by

$E_{net}=\frac{\frac{E_1}{R_1}+\frac{E_2}{R_2}-\frac{E_3}{R_3}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

$\Rightarrow E_{net}=\frac{\frac{1.5}{10}+\frac{2}{20}-\frac{2.5}{30}}{\frac{1}{10}+\frac{1}{20}+\frac{1}{30}}$

$\Rightarrow E_{net}=\left(\frac{\frac{9+6-5}{60}}{\frac{6+3+2}{60}}\right)$

$\therefore E_{net}=\frac{1}{6}\times \frac{60}{11}=\frac{10}{11}\text{V}$

Thus given circuit can be simplified to a single cell whose potential difference of both end will be equal to potential difference across points $\text{A}$ and $\text{B}$.

$\therefore V_A-V_B=\frac{10}{11}\text{V}$

Hence, option $\left(b\right)$ is correct.
$$\begin{array}{c}\text{Why this question?}\\ \text{On simplifying the given circuit we obtain}\\ \text{a single cell with net emf and an equivalent}\\ \text{resistance, this constitutes an open circuit.}\\ \text{Hence, potential difference between points}\\ \text{A}\text{and}\text{B}\text{will be equal to}\\ \text{net emf of cell.}\end{array}$$