Q. 53.9( 21 Votes )

# In Exercises 3 and 4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Answer :

In inductive circuit

Root mean square value of current, I = 15.92 A

Root mean square value of voltage, V = 220 V

The net power absorbed by each circuit over a complete cycle is given by the solution:

Power = VI cosФ

Where Ф is the phase difference between the voltage and current

For a pure inductive circuit, the phase difference between voltage and current in 90^{0}.

Substituting the above values, we get

⇒ P = 0

In capacitor circuit

Root mean square value of current, I = 2.49 A

Root mean square value of voltage, V = 110 V

The net power absorbed by each circuit over a complete cycle is given by the solution:

Power = VI cos Ф

For a pure capacitor circuit, the phase difference between voltage and current in 90^{0}.

Substituting the above values, we get

⇒ P = 0

∴ Zero power is absorbed by each circuit.

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Figure shows a typical circuit for low-pass filter. An AC input V_{i} = 10 mV is applied at the left end and the output V0 is received at the right end. Find the output voltages for v = 10 kHz, 100 k Hz, 1.0 MHz and 10.0 MHz Note that as the frequency is increased the output decreases and hence the name low-pass filter.

HC Verma - Concepts of Physics Part 2