Question 187: AC fundamentals

Question

An inductor of reactance $100 \Omega$, a capacitor of reactance $50 \Omega$, and a resistor of resistance $50 \Omega$ are connected in series with an AC source of $10 \mathrm{~V}, 50 \mathrm{~Hz}$. Average power dissipated by the circuit is ___________ W.

Accepted Answer

To calculate the average power dissipated in the circuit, we follow these steps:
Formula for Power
:
$    P = V_{	ext{rms}} 	imes I_{	ext{rms}} 	imes \cos \phi    $
Expressing Power in Terms of Voltage and Impedance
:
$    P = V_{	ext{rms}} 	imes \frac{V_{	ext{rms}}}{Z} 	imes \frac{R}{Z}    $
Simplifying the Expression
:
$    P = \frac{V_{	ext{rms}}^2 	imes R}{Z^2}    $
Calculate Impedance (Z)
:
The impedance of the circuit (Z) is calculated as:
$    Z = \sqrt{R^2 + (X_L - X_C)^2}    $
Given:
- $ X_L = 100 \, \Omega $
- $ X_C = 50 \, \Omega $
- $ R = 50 \, \Omega $
$    X_L - X_C = 100 \, \Omega - 50 \, \Omega = 50 \, \Omega    $
Thus:
$    Z = \sqrt{50^2 + 50^2} = 50\sqrt{2} \, \Omega    $
Substitute Values to Find Power (P)
:
Given $ V_{	ext{rms}} = 10 \, 	ext{V} $, we calculate $ P $ as follows:
$    P = \frac{10^2 	imes 50}{(50\sqrt{2})^2}    $
Simplifying further:
$    P = \frac{100 	imes 50}{2500 	imes 2} = 1 \, 	ext{W}    $
Therefore, the average power dissipated by the circuit is $ 1 \, 	ext{W} $.

AC fundamentals

An inductor of reactance 100 \Omega, a capacitor of reactance 50 \Omega, and a resistor of resistance 50 \Omega are connected in series with an AC source of 10 \mathrm{~V}, 50 \mathrm{~Hz}. Average power dissipated by the circuit is ___________ W.

Integer / Numeric Answer Type