To find the average value of the output voltage, we first need to analyze the circuit. Step 1: Identify the circuit type and output waveform. The circuit shown is a half-wave rectifier. The input voltage is $v_i = V_m \sin(\omega t)$. During the positive half-cycle of the input voltage ($0 \le t \le T/2$), the diode is forward-biased (assuming an ideal diode with 0V drop). Thus, the output voltage $v_o$ is equal to the input voltage $v_i$. $$v_o(t) = V_m \sin(\omega t) \quad \text{for } 0 \le t \le T/2$$ During the negative half-cycle of the input voltage ($T/2 < t \le T$), the diode is reverse-biased. It acts as an open circuit, so no current flows through the resistor $R$. Therefore, the output voltage $v_o$ is 0. $$v_o(t) = 0 \quad \text{for } T/2 < t \le T$$ Step 2: Calculate the average value of the output voltage. The average value of a periodic waveform $v_o(t)$ over one period $T$ is given by the formula: $$V_{avg} = \frac{1}{T} \int_0^T v_o(t)\,dt$$ Substitute the output voltage waveform into the integral: $$V_{avg} = \frac{1}{T} \left[ \int_0^{T/2} V_m \sin(\omega t)\,dt + \int_{T/2}^T 0\,dt \right]$$ $$V_{avg} = \frac{1}{T} \int_0^{T/2} V_m \sin(\omega t)\,dt$$ Step 3: Perform the integration. We know that $\omega = \frac{2\pi}{T}$. $$V_{avg} = \frac{V_m}{T} \left[ -\frac{\cos(\omega t)}{\omega} \right]_0^{T/2}$$ $$V_{avg} = -\frac{V_m}{\omega T} \left[ \cos\left(\omega \frac{T}{2}\right) - \cos(0) \right]$$ Substitute $\omega T = 2\pi$: $$V_{avg} = -\frac{V_m}{2\pi} \left[ \cos\left(\frac{2\pi}{T} \frac{T}{2}\right) - \cos(0) \right]$$ $$V_{avg} = -\frac{V_m}{2\pi} \left[ \cos(\pi) - \cos(0) \right]$$ $$V_{avg} = -\frac{V_m}{2\pi} \left[ -1 - 1 \right]$$ $$V_{avg} = -\frac{V_m}{2\pi} (-2)$$ $$V_{avg} = \frac{2V_m}{2\pi}$$ $$V_{avg} = \frac{V_m}{\pi}$$ Step 4: Convert to decimal value. Using the approximation $\pi \approx 3.14159$: $$V_{avg} \approx \frac{1}{3.14159} V_m \approx 0.3183 V_m$$ Comparing this result with the given options: A) $V_m$ B) $0.5V_m$ C) $0.318V_m$ D) $0.636V_m$ The calculated average value matches option C. The final answer is $\boxed{\text{C}}$.