Question

Figure shows two resistors $R_1$ and $R_2$ connected to a battery having an emf of $40$V and negligible internal resistance. A voltmeter having a resistance of $300\Omega$ is used to measure potential difference across $R_1$. Find the reading of the voltmeter?

Answer 1

as given in the circuit, current is given by $I$

$\Rightarrow I=\frac{V}{R_{eq}}$

$\Rightarrow R_{eq}=R_2+\frac{R_1{R}_v}{R_1{+R}_v}$ where $R_v$ is the resistance of the given voltmeter

$\Rightarrow R_{eq}=880+\frac{200\times 300}{200+300}=1000\Omega$

$\Rightarrow I=\frac{V}{R_{eq}}=\frac{40}{1000}=0.04volt$

Reading of volmeter $=$ potential difference =

$\Rightarrow I\times \left(\frac{R_1\times R_v}{R_1+R_v}\right)\Rightarrow I\times \frac{200\times 300}{200+300})=0.04\times 120=4.8volt$