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In Qs. $7.3$ and $7.4$, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
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To find: net power in Q. $7.3$ and $7.4$
Formula: power $P_{a v}=E_{\mathrm{v}} I_{\mathrm{v}} \cos \phi$
For Q7.3 (inductive circuit) current lags behind emf by $\pi$ / 2
Hence, $\mathrm{P}_{\mathrm{av}}=\mathrm{E}_{\mathrm{v}} \mathrm{I}_{\mathrm{v}} \cos \phi=220 \times 15.9 \times \cos 90^{\circ}=0$
In Q. 7.4 (pure capacitative circuit) current leads emf by phase $\pi / 2$
hence $P_{a v}=E_v I_v \cos \phi=E_v I_v \cos \pi / 2=E_v I_v \cos 90^{\circ}=0$
Formula: power $P_{a v}=E_{\mathrm{v}} I_{\mathrm{v}} \cos \phi$
For Q7.3 (inductive circuit) current lags behind emf by $\pi$ / 2
Hence, $\mathrm{P}_{\mathrm{av}}=\mathrm{E}_{\mathrm{v}} \mathrm{I}_{\mathrm{v}} \cos \phi=220 \times 15.9 \times \cos 90^{\circ}=0$
In Q. 7.4 (pure capacitative circuit) current leads emf by phase $\pi / 2$
hence $P_{a v}=E_v I_v \cos \phi=E_v I_v \cos \pi / 2=E_v I_v \cos 90^{\circ}=0$
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