Q. Two batteries of e.m.f. $4\ V$ and $8\ V$ with internal resistances $1\ \Omega$ and $2\ \Omega$ are connected in a circuit with a resistance of $9\ \Omega$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are
A) $\frac{1}{3}\ A$ and $3V$
B) $\frac{1}{6}\ A$ and $4V$
C) $\frac{1}{9}\ A$ and $9V$
D) $\frac{1}{2}\ A$ and $12V$
Explanation
(a) Applying Kirchhoff’s voltage law in the given loop
$- 2i + 8 - 4 - 1 \times i - 9i = 0 \Rightarrow i = \frac{1}{3}A$
Potential difference across $PQ = \frac{1}{3} \times 9 = 3V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.