Discussion
In the given figure, the value of \(R = 0\), the source voltage \(V_s = 300\) V (constant) and the load inductance \(L = 300\ \mu\)H. If the switch S1 is closed till time \(t = 100\ \mu\)s, what is the value of load current I at time \(t = 100\ \mu\)s?
Answer: C
With \(R = 0\) the load is a pure inductance, and when the switch S1 closes at \(t = 0\) the constant source voltage \(V_s\) is applied across \(L\) (the diode \(D_1\) conducts, and \(D_m\) is reverse biased). The current rises linearly.
\[ V_s = L\frac{di}{dt} \Rightarrow i(t) = \frac{V_s}{L}t \]
\[ \begin{aligned} i(100\ \mu\text{s}) \\ &= \frac{300}{300 \times 10^{-6}} \times 100 \times 10^{-6} \\ &= 100\ \text{A} \end{aligned} \]
The answer is 100 A, option C. The value of 50 A would be at \(t = 50\ \mu\)s and 200 A at \(t = 200\ \mu\)s.
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