A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of r

Question

A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances $\left( R_{1} = R_{2} = R_{3} = R_{4} \right)$. When $R_{3}$ resistance is heated to some temperature, its resistance value has gone up by $10\%$. The potential difference $\left( V_{a} - V_{b} \right)$ (after $R_{3}$ is heated) is $\_\_\_\_$ V.

Accepted Answer

$${V_{A} = \frac{V}{2} }{V_{B} = \frac{V}{2.1R} \times R = \frac{V}{2.1} }{\therefore V_{A} - V_{B} = V\left\lbrack \frac{1}{2} - \frac{1}{2.1} \right\rbrack }{V_{A} - V_{B} = \frac{0.1}{2 \times 2.1} \times 40 }{V_{A} - V_{B} = \frac{4}{4.2} = 0.95}$$

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