Rankers Physics

Alternating Current: Practice Problem & Solution

36. Power dissipated in an LCR series circuit connected to an AC source of emf $varepsilon$ is: (2009)
$\frac{\varepsilon^2 \sqrt{R^2 + (L\omega - \frac{1}{C\omega})^2}}{R}$
$\frac{\varepsilon^2 [ R^2 + (L\omega - \frac{1}{C\omega})^2 ]}{R}$
$\frac{\varepsilon^2 R}{\sqrt{R^2 + (L\omega - \frac{1}{C\omega})^2}}$
$\frac{\varepsilon^2 R}{[ R^2 + (L\omega - \frac{1}{C\omega})^2 ]}$

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 dissipated is $P = V_{rms} I_{rms} \cos\phi$.
$P = \varepsilon \left(\frac{\varepsilon}{Z}\right) \left(\frac{R}{Z}\right) = \frac{\varepsilon^2 R}{Z^2}$.
In a series LCR circuit, $Z^2 = R^2 + (L\omega - \frac{1}{C\omega})^2$.
Thus, $P = \frac{\varepsilon^2 R}{[ R^2 + (L\omega - \frac{1}{C\omega})^2 ]}$.

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