Question

A resistance of $4\Omega$ and a wire of length $5\text{m}$ and resistance $5\Omega$ are joined in series and connected to a cell of EMF $10\text{V}$ and internal resistance $1\Omega$. A parallel combination of two identical cells is balanced across $300\text{cm}$ length of the wire. The EMF $E$ of each cell is -

• A. $1.5\text{V}$
• B. $3.0\text{V}$
• C. $0.6\text{V}$
• D. $1.3\text{V}$

Answer 1

The correct option is B $3.0\text{V}$


$\Delta V_{AB}=i\times R_{AB}=\left(\frac{10}{4+5+1}\right)\times 5=5\text{V}$

Potential gradient across $AB=\frac{\Delta V_{AB}}{5}$

$=\frac{5}{5}=1\text{V/m}$

Now, the net EMF of the cell combination will be $E$ as the two cells are in parallel.

So, potential drop across balancing length of $3\text{m}=E$

$\Rightarrow E=\text{Balancing length}\times \text{Potential Gradient}$

$=3\times 1=3\text{V}$

Hence, option $\left(B\right)$ is the correct answer.