Q. If in the circuit shown below, the internal resistance of the battery is $1.5\ \Omega$ and $V_{P}\ $and$\ V_{Q}$ are the potentials at $P$ and $Q$ respectively, what is the potential difference between the points $P$ and $Q$
A) Zero
B) $4\ volts\ (V_{P} > V_{Q})$
C) $4\ volts\ (V_{Q} > V_{P})$
D) $2.5\ volts\ (V_{Q} > V_{P})$
Explanation
(d) $R_{eq} = \frac{5}{2}\Omega$
$i = \frac{20}{\frac{5}{2} + 1.5} = 5A$
Potential difference between $X$ and $P,$
$V_{X} - V_{P} = \left( \frac{5}{2} \right) \times 3 = 7.5V$ ….(i)
$V_{X} - V_{Q} = \frac{5}{2} \times 2 = 5V$ ….(ii)
On solving (i) and (ii) $V_{P} - V_{Q} = - 2.5\ volt;V_{Q} > V_{P}$
Short Trick : $\left( V_{P} - V_{Q} \right) = \frac{i}{2}\left( R_{2} - R_{1} \right) = \frac{5}{2}(2 - 3) = - 2.5$
$\Rightarrow V_{Q} > V_{P}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.