Q. In a Wheatstone bridge, $P = 90\Omega,Q = 110\Omega,\ R = 40\Omega\ and\ S = 60\Omega$ and a cell of 4 V emf. Then the potential difference between the diagonal along which a galvanometer is connected is
A) $- 0.2\ V$
B) $+ 0.2\ V$
C) $- 1\ V$
D) $+ 1\ V$
Explanation
(a) Current through resistance $P$ and $Q$,
$i_{1} = \frac{4}{90 + 110} = \frac{1}{50}A$
$V_{A} - V_{B} = Pi_{1} = 90 \times \frac{1}{50} = 1.8\ V$
Current through resistance $R$ and $S$,
$i_{2} = \frac{4}{40 + 60} = \frac{1}{25}A$
$V_{A} - V_{D} = Ri_{2} = 40 \times \frac{1}{25} = 1.6\ V$
$V_{B} - V_{D} = \left( V_{A} - V_{D} \right) - (V_{A} - V_{B})$
$= 1.6 - 1.8 = - 0.2V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.