Question

A cell with unknown emf $E$ is connected in a circuit shown in figure. The value of $E$ for which the potential at $\text{A}$ is equal to the potential at $\text{B}$ will be

• A. $8\text{V}$
• B. $5\text{V}$
• C. $2.5\text{V}$
• D. $3.5\text{V}$

Answer 1

The correct option is B $5\text{V}$

Two cells connected in the given closed circuit are in series with the same polarity. So

Net emf and resistance of circuit,

$E_{net}=15+E$
$R_{eq}=1+2+5=8\Omega$


Current in circuit will be

$i=\frac{E_{net}}{R_{eq}}$

$\Rightarrow i=\frac{15+E}{8}.........\left(1\right)$

For potential at $\text{A}$ and $\text{B}$, we can write the expression by using $KVL$ is,

$V_A-E+(i\times R)=V_B$

$\because R=2\Omega$

$V_A-E+[\left(\frac{15+E}{8}\right)\times 2]=V_B$

Given that $V_A=V_B$, hence $V_A-V_B=0$

Thus, $V_A-V_B=E-\frac{2(15+E)}{8}$

$\Rightarrow 0=E-\frac{(15+E)}{4}$

$\Rightarrow 4E-E=15$

$\therefore E=\frac{15}{3}=5\text{V}$

Hence, option $\left(b\right)$ is correct.
$$\begin{array}{c}\text{Why this question?}\\ \text{Tip: In case of closed circuit involved in a}\\ \text{problem always try to find the net emf \&}\\ \text{equivalent resistance of circuit. This will}\\ \text{help in finding the current in circuit.}\\ \\ \text{Concept: Two or more cells are said}\\ \text{to be connected in same polarity, if they}\\ \text{support the direction of current flow}\\ \text{in circuit.}\end{array}$$