Correct Answer - Option 4 :
\( \frac{{100}}{\pi } ~A\)
Concept:
- The root mean square current /voltage (rms) of an AC circuit is the effective current/voltage of that circuit.
- The maximum value of the potential in an AC circuit is called the peak value of voltage.
- The average value of the potential and current of an AC circuit for a cycle is called average potential and average current.
\(Average\;current\;over\;half\;cycle\;\left( {{I_{avg}}} \right) = \frac{{2\;I}}{\pi }\)
Average current over complete cycle = 0
\(rms\;current\;\left( {{I_{rms}}} \right) = \frac{I}{{\sqrt 2 }}\)
Where I is maximum/peak current in the circuit.
Calculation:
Given that: i = 50 sin 100 t A
∴ The peak value of current, I = 50 A
Average current over half cycle = Average value of sinusoidal current:
\(\Rightarrow \frac{{2\;I}}{\pi } = \frac{{2\times 50 }}{\pi }\;A\)
∴ Average current over half cycle = \( \frac{{100}}{\pi } ~A\)
Average current over complete cycle = 0 A
In the given question AC (Alternating Current) wave is mentioned, which consists of peak values 'I' in the positive half cycle and '- I' in the negative half cycle.
∴ Average current over the complete cycle (positive + negative half cycle) = 0 ampere
That's why the average value of AC wave is calculated over half cycle, which is equivalent to the full-wave rectifier's average value = \( \frac{{2\;I}}{\pi } \)
The average value for a half and full-wave rectifier are:
\({I_{avg}} = \frac{{{I}}}{\pi }\) for half-wave rectifier
\({I_{avg}} = \frac{{2{I}}}{\pi }\) for full-wave rectifier
Note that, half wave rectifier consists of the one-half cycle and inactive for other half cycle. So that its average value is half of the full-wave rectifier's average value.
Where the average value of AC wave is calculated over half cycle is equal to full-wave rectifier's average value.
