Question

A resistance of $4\Omega$ and a wire of length $5m$ and resistance $5\Omega$ are joined in series and connected to a cell of e.m,f, $10V$ and internal resistance $1\Omega$. A Parallel combination of two identical cells is balanced across $300cm$ of the wire. the e.m.f. $E$ each cell is then:

• A. $1.5V$
• B. $3.0V$
• C. $0.67V$
• D. $1.33V$

Answer 1

The correct option is B $3.0V$

In the potentiometer circuit, net resistance,

$R_1=5+4+1=10\Omega$ (due to wire $+4\Omega$series resistor & internal resistance of cell ).

and any provide $E_1=10V$.

For secondary circuit,

equivalent any of cells is parallel, $E^{\prime }=E$ ( as no internal resistance)

and resistance $=\frac{5}{5}\times 3=3\Omega$ (due to $3$m balance length )

At balance length, both circuits have equal currents (I)

$\Rightarrow I=\frac{E_1}{R_1}=\frac{E_2}{R_2}\Rightarrow \frac{10}{10}=\frac{E}{3}\Rightarrow E=3V.$