Question

A sinusoidal voltage of peak value $200\text{V}$ is connected to a diode and resistor $R$ in the circuit shown so that half-wave rectification occurs. If the forward resistance of the diode is negligible compared to $R$, the rms value of voltage across $R$ in volts is approximately -

• A. $40$
• B. $60$
• C. $100$
• D. $20$

Answer 1

The correct option is C $100$

For given half-wave rectifier circuit $V_{rms}^2=\frac{1}{T}{\int }_0^T{V}^{2}dt$

For half cycle, the diode will be in forward biased condition, so $V=V_0\sin \omega t$ across the resistor and for next half cycle, the diode will be reverse biased (non-conducting), hence $V=0$ across the resistor.
Therefore,
$V_{rms}^2=\frac{1}{T}\left[{\int }_0^{T/2}{V}_0^2{\sin }^2\omega t\right(dt)+{\int }_{T/2}^{T}0dt]$
$\Rightarrow V_{rms}^2=\frac{1}{T}{\int }_0^{T/2}{V}_0^2{\sin }^2\omega t\left(dt\right)$
$\Rightarrow V_{rms}^2=\frac{V_0^2}{4}$
or $V_{rms}=\frac{V_0}{2}$

Given, $V_0=200V$
$\Rightarrow V_{rms}=\frac{V_0}{2}=\frac{200}{2}$
$\Rightarrow V_{rms}=100\text{volts}$

Hence, option $\left(D\right)$ is the correct answer.