Looking for classes? Ksquare Career Institute, Bengaluru →
The reading of an ideal voltmeter in the circuit shown is:
Detailed Solution
Each branch has 50 $\Omega$, so the total current is $\frac{2}{25}$ A and each branch carries $\frac{1}{25}$ A. Voltmeter reading $=V_1-V_2=\frac{1}{25}\times30-\frac{1}{25}\times20=\frac{10}{25}=0.4$ V.