Consider the circuit shown. Assume that the diode D is ideal. The supply voltage vs = 325sin(2π50t) V, L = 500 µH, and R = 10 Ω. The peak diode current (in amperes) is _________ (Round off to one decimal place)

The correct option is 32.5 A.

1. Identify the Circuit Parameters from the Image:
From the given circuit diagram, we can observe a sinusoidal voltage source connected in series with an ideal diode $D$, an inductor $L$, and a resistor $R$.
The parameters are:
Supply voltage: $v_s=325sin\left(2\pi 50t\right)V$
Peak voltage: $V_m=325V$
Angular frequency: $\omega =2\pi f=2\pi \times 50=100\pi rad/s\approx 314.16rad/s$
Inductance: $L=500\mu H=500\times {10}^{-6}H$
Resistance: $R=10\Omega$

2. Calculate the Inductive Reactance and Impedance:
The inductive reactance ($X_L$) is given by:
$X_L=\omega L=100\pi \times 500\times {10}^{-6}=0.05\pi \Omega \approx 0.1571\Omega$
The total impedance ($Z$) of the RL circuit is:
$Z=\sqrt{R^2+X_L^2}=\sqrt{{10}^2+{0.1571}^2}=\sqrt{100+0.0247}\approx 10.0012\Omega$

3. Analyze the Time Constant and Transient Behavior:
When the diode conducts during the positive half-cycle, the current expression contains both a steady-state component and a decaying transient component:
$i\left(t\right)=\frac{V_m}{Z}sin(\omega t-\varphi )+A{e}^{-\frac{R}{L}t}$
Let us evaluate the inductive time constant ($\tau$):
$\tau =\frac{L}{R}=\frac{500\times {10}^{-6}}{10}=5\times {10}^{-5}s=50\mu s$
The period of the AC source is $T=\frac{1}{f}=\frac{1}{\mathrm{50}}=20ms$, so the positive half-cycle lasts for $10ms=10000\mu s$.
The peak current occurs approximately at the quarter cycle, near $t\approx 5ms=5000\mu s$.
At this moment, the elapsed time is:
$\frac{t}{\tau }=\frac{5000}{50}=100$
Since $t=100\tau$, the transient term $e^{-t/\tau }=e^{-100}\approx 0$ has completely decayed to zero.

4. Calculate the Peak Current:
Because the transient component is negligible at the time of the peak, the peak diode current is purely determined by the steady-state maximum value:
$I_p\approx \frac{V_m}{Z}=\frac{325}{10.0012}\approx 32.496A$
Rounding to one decimal place, we obtain:
$I_p\approx 32.5A$