In Exercises 7.3 and 7.4, what is the net power absorbed by each circu

1.	In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Answer» In the inductive circuit, Rms value of current, I = 15.92 A Rms value of voltage, V = 220 V Hence, the net power absorbed can be obtained by the relation, P=VI cosϕ Where, ϕ= Phase difference between V and I. For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., ϕ= 90°. Hence, P = 0 i.e., the net power is zero. In the capacitive circuit, rms value of current, I = 2.49 A rms value of voltage, V = 110 V Hence, the net power absorbed can be obtained as: P=VI cosϕ For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., ϕ= 90°. Hence, P = 0 i.e., the net power is zero.

In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Answer»

In the inductive circuit,
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,

P=VI cosϕ

Where,
ϕ= Phase difference between V and I.

For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., ϕ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit,
rms value of current, I = 2.49 A
rms value of voltage, V = 110 V
Hence, the net power absorbed can be obtained as:

P=VI cosϕ

For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., ϕ= 90°.

Hence, P = 0 i.e., the net power is zero.

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