Solution:
Using the formula for the current through the conductor, which relates the current at different temperatures:
\[
i_0 R_0 = i_{100} R_{100} \quad \text{[For the same source]}
\]
This gives:
\[
2 R_0 = 1.2 R_0 [1 + 100\alpha] \quad \Rightarrow \quad 1 + 100\alpha = \frac{5}{3} \quad \Rightarrow \quad 100\alpha = \frac{2}{3}
\]
Now, calculate \( \alpha \):
\[
50 \alpha = \frac{1}{3}
\]
Thus, the current at 50°C will be:
\[
i_{50} R_{50} = i_0 R_0
\]
Substituting values:
\[
i_{50} = \frac{i_0 R_0}{R_{50}} = \frac{2 R_0}{R_0 (1 + 50\alpha)} = \frac{2}{1 + \frac{1}{3}} = \frac{2}{\frac{4}{3}} = 1.5 \, \text{A}
\]
Thus, the current at 50°C is \( 1.5 \, \text{A} = 1500 \, \text{mA} \).
Therefore, the correct answer is \( \boxed{1500} \, \text{mA} \).
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.
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: