Question

A potential difference of $220V$ is maintained across a 12000 ohm rheostat $AB$ as shown in figure. The voltmeter $V$ has a resistance of $6000ohm$ and point $C$ is at one fourth of the distance from $A$ to $B$. What is the reading in the voltmeter ?

• A. 32 V
• B. 36 V
• C. 40 V
• D. 42 V

Answer 1

Answer: (C)

As in case of linear rheostat, $R\propto$ length $(L)$
$\frac{R_{AC}}{R_{AB}}=\frac{AC}{AB}$
Here, $R_{AB}=12000\Omega$ and $AC=\frac{1}{4}AB$
$\therefore R_{AC}=12000\times \frac{1}{4}=3000\Omega$
As the resistance $R_{AC}(=3000\Omega )$ is in parallel with voltmeter of resistance of $6000\Omega$.
Therefore, the effective resistance between points $A$ and $C$ will be
$R_{AC}^{\prime }=\frac{3000\times 6000}{(3000+6000)}=2000\Omega$
Resistance between points $B$ and $C$
$R_{BC}=R_{AB}-R_{AC}=12000\Omega -3000\Omega =9000\Omega$
$R_{BC}$ and $R_{AC}^{\prime }$ are in series.
Therefore, the voltmeter reading will be
$V_{AC}=\frac{R_{AC}^{\prime }{V}_{AB}}{\left(R_{BC}+R_{AC}^{\prime }\right)}$
$=\frac{2000}{(9000+2000)}\times 220=40V$