The value of DC voltage in a half-wave rectifier can be determined by analyzing the process of rectification. Let's break down the explanation into the following sections:

1. Half-wave rectification:
A half-wave rectifier converts an AC voltage into a DC voltage by allowing only one half of the input waveform to pass through while blocking the other half. In this case, the input AC voltage is given as V = 100sin(314t), where t represents time.

2. Rectification process:
When the AC voltage is applied to a half-wave rectifier, it passes through a diode that conducts current only in one direction. During the positive half-cycle of the AC voltage, the diode conducts and allows the positive part of the waveform to pass through. However, during the negative half-cycle, the diode blocks the current flow, resulting in no output.

3. Calculation of DC voltage:
To determine the value of DC voltage, we need to find the average value of the rectified waveform. The average value of a sinusoidal waveform over a complete cycle is given by the formula:

V_avg = (2/T) ∫[0 to T/2] V sin(ωt) dt

Where T is the time period and ω is the angular frequency (2πf).

In this case, the time period (T) can be calculated as T = 2π/ω = 2π/314 ≈ 0.0201 seconds.

Now, we can calculate the average value of the rectified waveform:

V_avg = (2/0.0201) ∫[0 to 0.0201/2] 100sin(314t) dt

V_avg = (100/0.0201) ∫[0 to 0.01005] sin(314t) dt

V_avg = (100/0.0201) [-cos(314t)/314] [0 to 0.01005]

V_avg = (100/0.0201) [-cos(314(0.01005))/314 + cos(314(0))/314]

V_avg ≈ 30.3 volts

Therefore, the value of DC voltage obtained from the half-wave rectifier is approximately 30.3 volts (option c).

To summarize, the half-wave rectifier converts the AC voltage into DC by allowing only the positive half of the waveform to pass through. The average value of the rectified waveform is calculated to determine the DC voltage, which in this case is approximately 30.3 volts.