A Wheatstone bridge has four resistors: \( R_1 = R_2 = R_3 = 1\,\Omega \) and \( R_4 = 2\,\Omega \). 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) → \( 2.0\,\Omega \)
A Wheatstone bridge is balanced when the ratios of the resistances in its two arms are equal.
\( \displaystyle \frac{R_1}{R_2} = \frac{R_3}{R_4'} \;\Rightarrow\; \frac{1}{1} = \frac{1}{R_4'} \;\Rightarrow\; R_4' = 1\,\Omega \)
So the \( 2\,\Omega \) resistor must be reduced to \( 1\,\Omega \) by a shunt \( x \) connected in parallel with it.
\( \displaystyle \frac{2x}{2 + x} = 1 \;\Rightarrow\; 2x = 2 + x \;\Rightarrow\; x = 2\,\Omega \)
Two equal \( 2\,\Omega \) resistors in parallel give \( 1\,\Omega \), which confirms the result.