Q187 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics

In a Wheatstone's bridge, three resistance 'Q', 'R' and 'S' are connected in three arms and first arm is formed by three resistances '$P_1$', '$P_2$' and '$P_3$' in parallel. The condition for the bridge to be balanced is
Figure

  • A
    $\dfrac{P_1 P_2 P_3 / (P_1 P_2 + P_2 P_3)}{Q} = \dfrac{S}{R}$
  • Correct answer
    B
    $\dfrac{P_1 P_2 P_3 / (P_2 P_3 + P_1 P_3 + P_1 P_2)}{Q} = \dfrac{S}{R}$
  • C
    $\dfrac{P_1 P_2 P_3 / (P_2 P_3 + P_1 P_3 + P_1 P_2)}{Q} = \dfrac{R}{S}$
  • D
    $\dfrac{P_1 P_3 + P_1 P_2}{P_1 P_2 P_3 Q} = \dfrac{S}{R}$

Solution

For a balanced Wheatstone bridge, the ratio of resistances in the adjacent arms must be equal.

From the given circuit diagram, the resistances in the four arms are:
Resistance of arm AB $= P$
Resistance of arm BC $= Q$
Resistance of arm AD $= S$
Resistance of arm DC $= R$

The condition for the bridge to be balanced is:
$\dfrac{P}{Q} = \dfrac{S}{R}$

The resistance $P$ is formed by three resistors $P_1$, $P_2$, and $P_3$ connected in parallel. The equivalent resistance $P$ is given by:
$\dfrac{1}{P} = \dfrac{1}{P_1} + \dfrac{1}{P_2} + \dfrac{1}{P_3}$

$\dfrac{1}{P} = \dfrac{P_2 P_3 + P_1 P_3 + P_1 P_2}{P_1 P_2 P_3}$

$P = \dfrac{P_1 P_2 P_3}{P_1 P_2 + P_2 P_3 + P_1 P_3}$

Substituting the value of $P$ in the balance condition, we get:
$\dfrac{P_1 P_2 P_3 / (P_1 P_2 + P_2 P_3 + P_1 P_3)}{Q} = \dfrac{S}{R}$

Answer: $\dfrac{P_1 P_2 P_3 / (P_2 P_3 + P_1 P_3 + P_1 P_2)}{Q} = \dfrac{S}{R}$

Hence, the correct answer is option: 'B'.

Relevant Questions