Q. A resistance of $4\Omega$ and a wire of length $5\ metres$ and resistance $5\Omega$ are joined in series and connected to a cell of e.m.f. $10\ V$ and internal resistance $1\Omega.$ A parallel combination of two identical cells is balanced across $300\ cm$ of the wire. The e.m.f. $E$ of each cell is
A) $1.5\ V$
B) $3.0\ V$
C) $0.67\ V$
D) $1.33\ V$
Explanation
(b) $E = xl = \frac{V}{l} = \frac{iR}{L} \times l \Rightarrow E = \frac{r}{(R + R_{h} + r)} \times \frac{R}{L} \times l$
$\Rightarrow E = \frac{10}{(5 + 4 + 1)} \times \frac{5}{5} \times 3 = 3V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.