Question

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

Answer 1

In Exercise 7.3

rms value of current, I = 15.9A

rms value of voltage , E = 220V

so, the net power absorbed by circuit can be given by, $P=VIcos\phi$

where, $\phi$ is phase difference between V and I. for a pure inductive circuit, the phase difference between voltage and current is 90° i.e., $\phi=90^{\circ}$

so, power absorbed by circuit, P = (220V)(15.9A)cos90° = 0

similarly, in Exercise 7.4

rms value of current, I = 2.5A

rms value of voltage , E = 110V

so, the net power absorbed by circuit can be given by, $P=VIcos\phi$

where, $\phi$ is phase difference between V and I. for a pure capacitive circuit, the phase difference between voltage and current is 90° i.e., $\phi=90^{\circ}$

so, power absorbed by circuit, P = (110V)(2.5A)cos90° = 0

Answer 2

Answer:the net power absorbed is zero for both the circuits

Explanation: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.