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. $$v_o(t) = 0 \quad \text{for } T/2 < t \le T$$

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:

$$V_{avg} = \frac{1}{T} \left[ \int_0^{T/2} V_m \sin(\omega t)\,dt + \int_{T/2}^T 0\,dt \right]$$ $$V_{avg} = \frac{1}{T} \int_0^{T/2} V_m \sin(\omega t)\,dt$$

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 $\omega T = 2\pi$:

$$V_{avg} = -\frac{V_m}{2\pi} \left[ \cos\left(\frac{2\pi}{T} \frac{T}{2}\right) - \cos(0) \right]$$ $$V_{avg} = -\frac{V_m}{2\pi} \left[ \cos(\pi) - \cos(0) \right]$$ $$V_{avg} = -\frac{V_m}{2\pi} \left[ -1 - 1 \right]$$ $$V_{avg} = -\frac{V_m}{2\pi} (-2)$$ $$V_{avg} = \frac{2V_m}{2\pi}$$ $$V_{avg} = \frac{V_m}{\pi}$$

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 $\boxed{\text{C}}$.