MHT CET · Physics · Alternating Current
In an AC circuit the emf \((e)\) and the current (i) at any instant are given respectively by
\(
\begin{array}{l}
e=E_{0} \sin \omega t \
i=I_{0} \sin (\omega t-\phi)
\end{array}
\)
The average power in the circuit over one cycle of AC is
- A \(\frac{E_{0} I_{0}}{2}\)
- B \(\frac{E_{0} I_{0}}{2} \sin \phi\)
- C \(\frac{E_{0} I_{0}}{2} \cos \phi\)
- D \(E_{0} I_{0}\)
Answer & Solution
Correct Answer
(C) \(\frac{E_{0} I_{0}}{2} \cos \phi\)
Step-by-step Solution
Detailed explanation
Power \(=\) rate of work done in one complete cycle.
Or
\(
P_{\mathrm{av}}=\frac{W}{T}
\)
Or
\(
P_{\mathrm{av}}=\frac{\left(E_{0} I_{0} \cos \phi\right) T / 2}{T}
\)
or
\(
P_{\mathrm{av}}=\frac{E_{0} I_{0} \cos \phi}{2}
\)
where \(\cos \phi\) is called the power factor of an \(\mathrm{AC}\) circuit.
Or
\(
P_{\mathrm{av}}=\frac{W}{T}
\)
Or
\(
P_{\mathrm{av}}=\frac{\left(E_{0} I_{0} \cos \phi\right) T / 2}{T}
\)
or
\(
P_{\mathrm{av}}=\frac{E_{0} I_{0} \cos \phi}{2}
\)
where \(\cos \phi\) is called the power factor of an \(\mathrm{AC}\) circuit.
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