Rankers Physics

Power in AC Circuits: Practice Problem & Solution

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

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Power in AC Circuits. 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, $P = \frac{\varepsilon^2 R}{[ R^2 + (L\omega - \frac{1}{C\omega})^2 ]}$.

Detailed Solution (Best of Understanding Complex Concepts)

Concepts Involved:
  • Average Power in AC Circuits: The average power $P$ dissipated is calculated as $P = V_{rms} I_{rms} \cos\phi$, where the source emf $\varepsilon$ represents the RMS voltage ($V_{rms}$).
  • Impedance ($Z$): For a series LCR circuit, the total opposition to the current is $Z = \sqrt{R^2 + (X_L - X_C)^2}$, with inductive reactance $X_L = L\omega$ and capacitive reactance $X_C = \frac{1}{C\omega}$.
  • Power Factor ($\cos\phi$): The ratio of real power to apparent power in the circuit, given by $\cos\phi = \frac{R}{Z}$.
  • RMS Current ($I_{rms}$): Using Ohm's law for AC, the current is $I_{rms} = \frac{\varepsilon}{Z}$.
Step-by-Step Derivation:
  1. State the general power formula:
    $$P = V_{rms} I_{rms} \cos\phi$$
  2. Substitute the specific circuit parameters: Replace $V_{rms}$ with $\varepsilon$, $I_{rms}$ with $\frac{\varepsilon}{Z}$, and $\cos\phi$ with $\frac{R}{Z}$:
    $$P = \varepsilon \left(\frac{\varepsilon}{Z}\right) \left(\frac{R}{Z}\right)$$
  3. Simplify the expression:
    $$P = \frac{\varepsilon^2 R}{Z^2}$$
  4. Expand the squared impedance ($Z^2$): Squaring the base impedance equation gives:
    $$Z^2 = R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2$$
  5. Construct the final equation: Substitute the expanded $Z^2$ back into the simplified power expression:
    $$P = \frac{\varepsilon^2 R}{\left[ R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2 \right]}$$

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): For an electric lamp connected in series with a variable capacitor and \( \text{ac} \) source, its brightness increases with increase of capacitance. Reason (R): Capacitive reactance decreases with increase in capacitance of capacitor.  
2. 15. A series RC circuit is connected to an alternating voltage source. Consider two situations A. When capacitor is air filled. B. When capacitor is mica filled. Current through resistor is i and voltage across capacitor is V then: (2015 Re)
3. Assertion (A): Average power consumed in an AC circuit is equal to average power consumed by resistors in the circuit. Reason (R): Average power consumed by capacitor and inductor is zero.
4. 20. In a circuit L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $45^\circ$. The value of C is: (2005)
5. For a series LCR circuit the power loss at resonance is: (2002)