A Wheatstone bridge has four resistors: $R_1=R_2=R_3 = 1Ω$ and $R_4 = 2Ω$. In order to balance the bridge, resistor $R_4$ has to be shunted by resistance of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.0 Ω
Given:
$R_1 = R_2 = R_3 = 1\,Ω$, $R_4 = 2\,Ω$
Condition for balance of Wheatstone bridge:
$\frac{R_1}{R_2} = \frac{R_3}{R_4'}$
where $R_4'$ is the effective resistance after shunting.
Substitute values:
$\frac{1}{1} = \frac{1}{R_4'}$
⟹ $R_4' = 1\,Ω$
Let shunt resistance be $S$:
$\frac{1}{R_4'} = \frac{1}{R_4} + \frac{1}{S}$
$\frac{1}{1} = \frac{1}{2} + \frac{1}{S}$
$1 - \frac{1}{2} = \frac{1}{S}$
$\frac{1}{2} = \frac{1}{S}$
$S = 2\,Ω$
Final Answer:
Shunt resistance = 2 Ω