Question

Cells $A$ and $B$ and a galvanometer $G$ are connected to a slide wire $OS$ by two sliding contacts $C$ and $D$ as shown in figure. The slide wire is $100\text{cm}$ long and has a resistance of $12\Omega$. With $OD=75\text{cm}$, the galvanometer gives no deflection when $OC$ is $50\text{cm}$. If $D$ is moved to touch the end of wire $S$, the value of $OC$ for which the galvanometer shows no deflection is $62.5\text{cm}$. The e.m.f. of cell $B$ is $1.0\text{V}$. Calculate the internal resistance & e.m.f. of cell $A$.

• A. $3\Omega$ and $2\text{V}$
• B. $2\Omega$ and $2\text{V}$
• C. $3\Omega$ and $3\text{V}$
• D. $4\Omega$ and $3\text{V}$

Answer 1

The correct option is A $3\Omega$ and $2\text{V}$

Given:
$E_B=1\text{V}$
Let $E_A$ and $r$ be the e.m.f. and internal resistance of cell $A$, respectively.

When $OD=75\text{cm}$ and $OC=50\text{cm}$

Resistance of wire $OC$ and $OD$ will be

$R_{OC}=(\frac{12}{100})\times 50=6\Omega$

$R_{OD}=(\frac{12}{100})\times 75=9\Omega$

Since, $E_B=1\text{V}$ is balanced across $50\text{cm}$, so potential gradient of wire will be

$K=\frac{1}{50}\text{V/cm}$

Therefore, voltage drop across the wire $OD$ of length $75\text{cm}$ will be

$V_{OD}=\left(1/50\right)\times 75=1.5\text{V}$

So,

$\left(\frac{E_A}{9+r}\right)\times 9=1.5........\left(1\right)$

Similarly,
When $OD=100\text{cm}$ and $OC=62.5\text{cm}$

Resistance of wire $OD$ will be

$R_{OD}=12\Omega$

Since, $E_B=1\text{V}$ is balanced across $62.5\text{cm}$, so potential gradient of wire will be

$K=\frac{1}{62.5}\text{V/cm}$

Therefore, voltage drop across the wire $OD$ of length $75\text{cm}$ will be

$V_{OD}=\left(1/62.5\right)\times 100=1.6\text{V}$

So,
$\left(\frac{E_A}{12+r}\right)\times 12=1.6........\left(2\right)$

On solving eqs. $\left(1\right)\&\left(2\right)$, we get

$r=3\Omega$ and $E_A=2\text{V}$

Hence, option $\left(a\right)$ is correct.

Key concept: Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.