In the circuit shown, the average power dissipated in the resistor is (assume diode to be ideal)

Options

A. `E_(0)^(2)/2R`
B. `E_(0)^(2)/4R`
C. `E_(0)^(2)/R`
D. Zero

Answer

To find the average power dissipated in the resistor in the given circuit with an ideal diode, we can follow these steps:

### Step 1: Understand the Circuit
The circuit consists of an ideal diode (D1) in series with a resistor (R), connected to an AC voltage source \( V(t) = V_0 \sin(\omega t) \). The diode will only allow current to flow in one direction, effectively clipping the negative half of the AC waveform.

### Step 2: Determine the Current through the Resistor
For an ideal diode, the current through the resistor when the diode is forward-biased (i.e., during the positive half-cycle of the AC signal) can be expressed as:
\[ I(t) = \frac{V(t)}{R} = \frac{V_0 \sin(\omega t)}{R} \]
for \( 0 \leq \omega t \leq \pi \) (the positive half-cycle).

### Step 3: Calculate the Average Power without the Diode
The average power \( P_{\text{avg}} \) dissipated in the resistor without the diode can be calculated using the formula:
\[ P_{\text{avg}} = \frac{1}{T} \int_0^T V^2(t) dt \]
where \( T \) is the period of the AC signal. The average power without the diode (considering the full cycle) is:
\[ P_{\text{avg, without diode}} = \frac{1}{2} \cdot \frac{V_0^2}{R} \]
This is because the average of \( \sin^2(\omega t) \) over one complete cycle is \( \frac{1}{2} \).

### Step 4: Calculate the Average Power with the Diode
Since the diode only allows current during the positive half-cycle, the average power dissipated in the resistor with the diode can be given by:
\[ P_{\text{avg, with diode}} = \frac{1}{2} \cdot P_{\text{avg, without diode}} = \frac{1}{2} \cdot \left( \frac{1}{2} \cdot \frac{V_0^2}{R} \right) = \frac{1}{4} \cdot \frac{V_0^2}{R} \]

### Step 5: Final Expression for Average Power
Thus, the average power dissipated in the resistor when the diode is ideal is:
\[ P = \frac{V_0^2}{4R} \]

### Conclusion
The average power dissipated in the resistor is \( \frac{V_0^2}{4R} \).