Question

Power dissipated in an $LCR$ series circuit connected to an a.c. source of emf $\varepsilon$ is :

• A. $\varepsilon^{2} R / \sqrt{ R ^{2}+\left( L \omega-\frac{1}{ C \omega}\right)^{2}}$
• B. $\varepsilon^{2} R /\left[ R ^{2}+\left( L \omega-\frac{1}{ C \omega}\right)^{2}\right]$
• C. $\varepsilon^{2} \sqrt{\left[R^{2}+\left(L \omega-\frac{1}{C \omega}\right)^{2}\right]} / R$
• D. $\frac{\varepsilon^{2}\left[ R ^{2}+\left(L \omega-\frac{1}{ C \omega}\right)^{2}\right]}{ R }$

Answer 1

The Correct Option is B

To solve the problem of determining the power dissipated in an LCR series circuit connected to an AC source, we need to understand the concept of impedance and power in AC circuits.

In an LCR series circuit, the total impedance Z is given by:

Z = \sqrt{R^{2} + (X_L - X_C)^{2}}

Where:

• R is the resistance.
• X_L = L\omega is the inductive reactance (with \omega being the angular frequency).
• X_C = \frac{1}{C\omega} is the capacitive reactance.

The formula for power dissipated in an AC circuit is represented by:

P = I^{2}R = \frac{V^{2}R}{Z^{2}}

Where:

• V is the maximum voltage (or \varepsilon in this problem).
• Z is impedance.

Substituting the expression for impedance Z and using \varepsilon as the maximum voltage, we have:

P = \frac{\varepsilon^{2}R}{R^{2} + (L\omega - \frac{1}{C\omega})^{2}}

This matches the given correct option:

$\varepsilon^{2} R /\left[ R ^{2}+\left( L \omega-\frac{1}{ C \omega}\right)^{2}\right]$

Conclusion: The power dissipated in the LCR series circuit is correctly given by the expression \frac{\varepsilon^{2} R}{R^{2} + (L\omega - \frac{1}{C\omega})^{2}}. The key to solving this problem is correctly applying the formula for power dissipated and understanding the role of impedance in the circuit.