Alternating Current: Practice Problem & Solution
37. In an AC circuit the e.m.f (e) and the current (i) at any instant are given respectively by $e = E_0 sinomega t$, $i = I_0 sin(omega t - phi)$. The average power in the circuit over one cycle of a.c. is: (2008)
Solution Explained:
To solve this problem, we apply the core principles of Alternating Current. Understanding the underlying formula is key to arriving at the correct answer below:
Average power is given by $P_{avg} = E_{rms} I_{rms} \cos\phi$.
Substituting RMS values: $E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.
$P_{avg} = \left(\frac{E_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos\phi$.
$P_{avg} = \frac{E_0 I_0}{2} \cos\phi$.
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