Question

Consider the circuit shown. Assume that the diode (\(D\)) is ideal.

Given \(v_s=100\sin(2\pi50t)\ \text{V}\), \(V_{dc}=50\ \text{V}\), and \(R=10\ \Omega\), the average value of the current through the diode is A (Round off to two decimal places)

Hint:

The diode only conducts while the source voltage exceeds the battery voltage; integrate the current over that conduction window and divide by the full period to get the average.

Answer 1

Correct Answer: 1.09

Step 1: Convert the problem into the time domain.
The source is $v_s=100\sin(100\pi t)$ V, so the angular frequency is $\omega=100\pi$ rad/s and the period is $T=1/50=0.02$ s. The diode conducts only when $v_s>50$ V, which happens when $\sin(100\pi t)>0.5$.
Step 2: Find the actual turn-on and turn-off times.
$\sin(100\pi t)=0.5$ first at $100\pi t=\pi/6$, giving $t_1=1/600$ s, and again at $100\pi t=5\pi/6$, giving $t_2=1/120$ s. Current flows only between $t_1$ and $t_2$ in every cycle.
Step 3: Write the diode current as a function of time.
\[ i(t)=\frac{100\sin(100\pi t)-50}{10}=10\sin(100\pi t)-5 \]
Step 4: Average this current over the full period.
\[ I_{avg}=\frac{1}{T}\int_{t_1}^{t_2}\left[10\sin(100\pi t)-5\right]dt \]
\[ =\frac{1}{0.02}\left[-\frac{10}{100\pi}\cos(100\pi t)-5t\right]_{t_1}^{t_2} \]
Step 5: Evaluate at the two time limits.
At $t_2=1/120$ s: $100\pi t_2=5\pi/6$, so $\cos(5\pi/6)=-0.8660$, giving the bracket value $-\frac{10}{100\pi}(-0.8660)-5(1/120)=0.02757-0.04167=-0.01410$.
At $t_1=1/600$ s: $100\pi t_1=\pi/6$, so $\cos(\pi/6)=0.8660$, giving the bracket value $-\frac{10}{100\pi}(0.8660)-5(1/600)=-0.02757-0.00833=-0.03590$.
Step 6: Subtract and scale by $1/T$.
\[ -0.01410-(-0.03590)=0.02180 \]
\[ I_{avg}=\frac{0.02180}{0.02}=1.09\ \text{A} \]
\[ \boxed{I_{avg}=1.09\ \text{A}} \]