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

2.0 Ω

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