Two batteries of emf 4 V and 8 V with internal resistance $1 \Omega$ and $2 \Omega$ are connected in

Question

Two batteries of emf 4 V and 8 V with internal resistance $1 \Omega$ and $2 \Omega$ are connected in a circuit with resistance of $9 \Omega$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are\

Professor PC Thomas Classes · Answer

(a) : $I=\frac{8-4}{1+2+9}=\frac{4}{12}=\frac{1}{3} \mathrm{~A}$ $$ V_{P}-V_{Q}=4-\frac{1}{3} 	imes 3=3 \mathrm{volt} $$