Voltage Developed Across Wheatstone Bridge Due to Resistance Temperature Change

To calculate the magnitude of the voltage developed between terminals $S$ and $T$, we first analyze the potential distribution in the circuit.

1. Circuit Topology and Initial Parameters

From the circuit diagram:

• Battery voltage applied between terminals $P$ and $Q$: $V_{PQ} = 50\text{ V}$. Let potential at terminal $Q$ be $V_Q = 0\text{ V}$, so potential at terminal $P$ is $V_P = 50\text{ V}$.
• $R_1 = 60\,\Omega$ is connected between node $P$ and node $S$.
• $R_3 = 300\,\Omega$ is connected between node $Q$ and node $S$.
• $R_2 = 100\,\Omega$ is connected between node $P$ and node $T$.
• $R_4 = 500\,\Omega$ is connected between node $Q$ and node $T$.

Initially, the bridge is balanced because: $\frac{R_1}{R_2} = \frac{60}{100} = \frac{3}{5} \quad \text{and} \quad \frac{R_3}{R_4} = \frac{300}{500} = \frac{3}{5}$

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2. Change in Resistance of $R_3$

When the temperature of resistor $R_3$ is raised by $\Delta T = 100\,^\circ\text{C}$, its resistance changes according to the temperature coefficient of resistance $\alpha = 0.0004\,^\circ\text{C}^{-1}$:

$R_3' = R_3 (1 + \alpha \Delta T)$ $R_3' = 300 \times \left(1 + 0.0004 \times 100\right)$ $R_3' = 300 \times (1 + 0.04) = 300 \times 1.04 = 312\,\Omega$

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3. Calculation of Potentials at Terminals $S$ and $T$

Potential at Terminal $T$ ($V_T$):

Since $R_2$ and $R_4$ remain unchanged, the potential at node $T$ is determined by the voltage divider rule across the branch $P-T-Q$: $V_T = V_Q + (V_P - V_Q) \cdot \frac{R_4}{R_2 + R_4}$ $V_T = 50 \times \frac{500}{100 + 500} = 50 \times \frac{500}{600} = 50 \times \frac{5}{6} = \frac{250}{6}\text{ V} = \frac{125}{3}\text{ V}$

New Potential at Terminal $S$ ($V_S'$):

Using the new resistance value $R_3' = 312\,\Omega$, the potential at node $S$ is given by: $V_S' = V_Q + (V_P - V_Q) \cdot \frac{R_3'}{R_1 + R_3'}$ $V_S' = 50 \times \frac{312}{60 + 312} = 50 \times \frac{312}{372}$

Simplifying the fraction $\frac{312}{372}$ by dividing the numerator and denominator by $12$: $\frac{312}{372} = \frac{26}{31}$

Thus, $V_S' = 50 \times \frac{26}{31} = \frac{1300}{31}\text{ V}$

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4. Voltage Developed Between Terminals $S$ and $T$

The potential difference $V_{ST}$ between terminals $S$ and $T$ is: $V_{ST} = |V_S' - V_T| = \left| \frac{1300}{31} - \frac{125}{3} \right|$

Taking the common denominator ($31 \times 3 = 93$): $V_{ST} = \left| \frac{1300 \times 3 - 125 \times 31}{93} \right|$ $V_{ST} = \left| \frac{3900 - 3875}{93} \right| = \frac{25}{93}\text{ V}$

Evaluating the numerical value: $V_{ST} \approx 0.2688\text{ V}$

Rounding off to two decimal places, the magnitude of the voltage developed is $0.27\text{ V}$ (or within the range $0.26$ to $0.28$).