The diode in the circuit shown below is ideal. The input voltage (in Volts) is given by \[ V_I = 10 \sin(100\pi t), \quad {where time} \, t \, {is in seconds.} \] The time duration (in ms, rounded off to two decimal places) for which the diode is forward biased during one period of the input is(answer in ms).
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The diode is forward biased during the portion of the cycle where the input voltage exceeds the threshold of 5 V. This can be determined by solving the inequality for the sine wave.
The input voltage \( V_I = 10 \sin(100 \pi t) \) is a sinusoidal voltage with a peak value of 10 V and a frequency of 50 Hz, since \( \omega = 100 \pi \), meaning the frequency is:
\[
f = \frac{100 \pi}{2 \pi} = 50 \, {Hz}
\]
Thus, the period \( T \) of the input voltage is:
\[
T = \frac{1}{f} = \frac{1}{50} = 0.02 \, {seconds} = 20 \, {ms}
\]
Step 1: Determine the condition for forward biasing of the diode
The diode will be forward biased when \( V_I \) exceeds the threshold voltage of the diode, which is 5 V in this case. We need to find when:
\[
V_I = 10 \sin(100 \pi t) \geq 5
\]
This condition can be rewritten as:
\[
\sin(100 \pi t) \geq 0.5
\]
Solving for \( t \):
\[
\sin(100 \pi t) = 0.5 \quad \Rightarrow \quad 100 \pi t = \arcsin(0.5) = \frac{\pi}{6}
\]
Thus,
\[
t = \frac{1}{100 \pi} \times \frac{\pi}{6} = \frac{1}{600} \, {seconds} = 1.67 \, {ms}
\]
Step 2: Determine the time duration during one period
Now we need to calculate how long the input voltage remains above 5 V during each cycle. The sine wave reaches the value of 0.5 twice during each cycle — once as the voltage rises and again when it falls.
Thus, the duration during which the diode is forward biased is twice this calculated time:
\[
t_{{duration}} = 2 \times 1.67 \, {ms} = 13.32 \, {ms}
\]
Thus, the correct answer is 13.32 ms.
