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: Understand when the diode conducts.
The diode only carries current in one direction. Looking at the circuit, the diode is forward biased and conducts only when the source voltage \(v_s\) exceeds the opposing battery voltage \(V_{dc}=50\ \text{V}\). When \(v_s\) is below \(50\ \text{V}\), including its entire negative half cycle, the diode is reverse biased and blocks current completely.
Step 2: Find the conduction angles.
Let \(\theta=2\pi50t=\omega t\). The diode turns on when
\[ 100\sin\theta>50 \]
\[ \sin\theta>0.5 \]
This is true between
\[ \theta_1=30^{\circ}=\frac{\pi}{6} \] and
\[ \theta_2=150^{\circ}=\frac{5\pi}{6} \]
so the diode conducts only for this window in every cycle.
Step 3: Write the current during conduction.
While the diode conducts, it behaves like a short, so the current is set entirely by the source, the battery, and the resistor \(R\):
\[ i(\theta)=\frac{v_s-V_{dc}}{R}=\frac{100\sin\theta-50}{10} \]
Step 4: Set up the average current formula.
The average of a periodic waveform is the integral over one full cycle divided by the period, and the current is zero outside the conduction window:
\[ I_{avg}=\frac{1}{2\pi}\int_{\theta_1}^{\theta_2}\frac{100\sin\theta-50}{10}\,d\theta \]
Step 5: Carry out the integration.
\[ I_{avg}=\frac{1}{20\pi}\int_{\pi/6}^{5\pi/6}\left(100\sin\theta-50\right)d\theta \]
\[ =\frac{1}{20\pi}\Big[-100\cos\theta-50\theta\Big]_{\pi/6}^{5\pi/6} \]
Step 6: Substitute the limits.
At \(\theta=5\pi/6\):
\[ -100\cos\left(\frac{5\pi}{6}\right)-50\left(\frac{5\pi}{6}\right)=86.603-130.900=-44.297 \]
At \(\theta=\pi/6\):
\[ -100\cos\left(\frac{\pi}{6}\right)-50\left(\frac{\pi}{6}\right)=-86.603-26.180=-112.782 \]
Subtracting,
\[ -44.297-(-112.782)=68.485 \]
Step 7: Compute the final value.
\[ I_{avg}=\frac{68.485}{20\pi}=\frac{68.485}{62.832}=1.09\ \text{A} \]
Final Answer:
\[ \boxed{I_{avg}=1.09\ \text{A}} \]