Question

Consider the circuit shown. Assume that the diode \(D\) is ideal. The supply voltage \(v_s=325\sin(2\pi50t)\) V, \(L=500\ \mu H\), and \(R=10\ \Omega\).

The peak diode current (in amperes) is (round off to one decimal place).

Hint:

Since wL is tiny next to R here, the RL transient dies out almost instantly, so the peak current is close to Vm/Z, occurring near the middle of the positive half cycle.

Answer 1

Correct Answer: 32.5

Step 1: Compare the inductive and resistive parts.
The inductive reactance is $\omega L = 2\pi(50)(500\times10^{-6}) = 0.157\ \Omega$, tiny next to $R=10\ \Omega$. This tells us the circuit behaves almost like a pure resistor, with the inductor mattering only for a very short instant right after the diode switches on.

Step 2: Treat it as resistive first.
If the load were purely resistive, the current would simply be $i(t) = \frac{V_m}{R}\sin(\omega t)$, peaking at $\frac{325}{10} = 32.5$ A exactly at the crest of the sine wave.

Step 3: Add the small correction from L.
With the inductor included, the true peak is $\frac{V_m}{Z}$ where $Z=\sqrt{R^2+(\omega L)^2}=\sqrt{100+0.0247}=10.0012\ \Omega$, only a whisker above $10\ \Omega$. So the corrected peak is
\[ i_{peak}=\frac{325}{10.0012}=32.496\text{ A} \]

Step 4: Explain why the L barely changes anything.
The time constant $L/R = 50\ \mu s$ is minute compared to the $20$ ms period of the supply, so any starting transient dies away almost the instant the diode conducts, well before the current reaches its crest near the middle of the half cycle.

Step 5: Round off.
\[ i_{peak}\approx 32.5\text{ A} \]
\[ \boxed{32.5} \]