Q. In the circuit shown in the figure below the cells $E_{1}$ and $E_{2}$ have e.m.f. $4 \, V$ and $8 \, V$ and internal resistance $0.5 \, Ω$ and $1.0 \, Ω$ respectively. Then the potential difference across the cell $E_{1}$ and $E_{2}$ will be-
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Solution:

Equivalent resistance of circuit is :
$R_{e q}=8 \, \Omega$
So, $I=\frac{8 - 4}{8}=0.5 \, A$
$V_{8 V}=E- \, Ir=8-\left(0.5\right)\left(1\right)=7.5 \, V$
$V_{4 V}=E+Ir=4+\left(0.5\right)\left(0.5\right)=4.25 \, V$