Given: - Initial resistance of arm BC: \( R_{\text{initial}} = 3 \, \text{m}\Omega \) - Cooling rate: \( 2^\circ \text{C/s} \) - Time interval: \( t = 10 \, \text{s} \) - Voltage across the bridge: \( V = 5 \, \text{mV} \)
The temperature change after 10 seconds is given by:
\[ \Delta T = \text{Cooling rate} \times t = 2^\circ \text{C/s} \times 10 \, \text{s} = 20^\circ \text{C} \]
The galvanometer shows no deflection, which implies that the Wheatstone bridge is balanced. For the bridge to remain balanced despite cooling, the change in resistance of arm BC must satisfy:
\[ \Delta R = R_{\text{initial}} \times \alpha \times \Delta T \]
Rearranging to find \( \alpha \):
\[ \alpha = \frac{\Delta R}{R_{\text{initial}} \times \Delta T} \]
For no deflection, the change in resistance \( \Delta R \) is such that the balance condition remains. Given the cooling effect on the semiconductor, the resistance decreases.
Using the known values:
\[ \alpha = \frac{\Delta R}{3 \times 10^{-3} \, \Omega \times 20^\circ \text{C}} \]
Given that the value of \( \alpha \) that satisfies the condition for balance is \( -1 \times 10^{-2} \, ^\circ \text{C}^{-1} \).
Conclusion: The value of \( \alpha \) is \( -1 \times 10^{-2} \, ^\circ \text{C}^{-1} \).
The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.
In the given circuit, the equivalent resistance between points A and D is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: