Question

A battery with EMF of $9\text{V}$ is connected in series with an ammeter and a voltmeter. Resistance of these devices are unknown. If the voltmeter is connected to a resistance $($its value is also unknown$)$ in parallel, the ammeter reading will be doubled, and the voltmeter reading halved. What will be the voltmeter reading (in $\text{V}$) after connecting the resistance ?

Answer 1

Before connecting to resistor,
Let voltmeter reading be $V,$ potential across ammeter is $9-V$
from KVL,



After connecting to resistor, the voltage across the ammeter will be doubled as current is doubled $(V=IR)$ keeping resistance same.


$2(9-V)+V/2=9$

$\Rightarrow 18-2V+V/2=9$

$\Rightarrow 3V/2=9$

$\Rightarrow V/2=3\text{V}$