The balance condition for a Wheatstone bridge with arms $P, Q, R, S$ (galvanometer between B and D) is:
A$P + Q = R + S$
B$P/Q = R/S$
C$P/Q = S/R$
D$PQ = RS$
Answer & Solution
Correct answer: C. $P/Q = S/R$
At balance the galvanometer current $I_g = 0$. From the loop equations, $I_1P = I_2S$ and $I_1Q = I_2R$, dividing gives **P/Q = S/R**.
Related questions
The reactance of a capacitor $C$ at frequency $f$ in an a.c. circuit is:A bird perched on a single high voltage wire is unharmed because:The two conservation laws embodied in Kirchhoff's rules are conservation of:The resistance offered by the cell itself is called:Which device detects the current in a Wheatstone bridge?In a Wheatstone bridge, the source is connected across the:The Wheatstone bridge is made up of how many resistors?At a junction, the current entering equals the current: