Question

An ammeter $A$ of finite resistance, and a resistor $R$ are joined in series to an ideal cell $C$. A potentiometer $P$ is joined in parallel to $R$. The ammeter reading is $I_0$ and the potentiometer reading is $V_0$. $P$ is now replaced by a voltmeter of finite resistance. The ammeter reading now is $I$ and the voltmeter reading is $V$ –

• A. $I>I_0,V<V_0$
• B. $I>I_0,V=V_0$
• C. $I=I_0,V<V_0$
• D. $I<I_0,V=V_0$

Answer 1

The correct option is A $I>I_0,V<V_0$

Reading of potentiometer is reading of ideal voltmeter = $V_0$

$V_0=\frac{R\times ϵ}{R+R_A}.....\left(1\right)$
Ammeter reading $=I_0$
$I_0=\frac{ϵ}{R+R_A}....\left(2\right)$
When potentiometer is replaced by voltmeter of $R_V$

$I=\frac{ϵ}{R_A+\frac{R\times R_V}{R+R_V}}$
$I=\frac{ϵ}{R_A+\frac{R}{1+\frac{R}{R_V}}}.....\left(3\right)$
from (2) & (3)
$I>I_0$
Now, $V=\frac{ϵ(R+R_V)}{R_A\times R+R_A{R}_V+R\times R_V}\times \frac{R{R}_V}{(R+R_V)}$
$V=\frac{ϵR}{R+R_A+\frac{R_{A}R}{R_V}}.....\left(4\right)$
from (1) & (4)
$V<V_0$