If $\mathrm{E}_1=4 \mathrm{~V}$ and $\mathrm{E}_2=12 \mathrm{~V}$, the current in the circuit and potential…
If $\mathrm{E}_1=4 \mathrm{~V}$ and $\mathrm{E}_2=12 \mathrm{~V}$, the current in the circuit and potential difference between the points P and Q respectively are
$1 \mathrm{~A}, 8 \mathrm{~V}$
$1 \mathrm{~A}, 6 \mathrm{~V}$
$0.8 \mathrm{~A}, 6.4 \mathrm{~V}$
$0.8 \mathrm{~A}, 8 \mathrm{~V}$
Solution
The current in the circuit is
$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{12-4}{8+1+1}=\frac{8}{10}=0.8 \mathrm{~A}$
$\therefore \quad$ The potential difference between points $P$ and $Q$ is
$V_{P Q}=I R=0.8 \times 8=6.4 \mathrm{~V}$
