As given in the figure a series circuit connected across a 200 V, 60 Hz line consists of a capacitor of capacitive reactance 30 \( \Omega \), a non-inductive resistor 44 \( \Omega \), a coil of inductive reactance 90 \( \Omega \), and another resistance of 36 \( \Omega \). The power dissipated in the circuit is
Show Hint
When dealing with AC circuits, always check the total impedance to calculate the current and then use \( P = I^2 R \) for power dissipation.
The total impedance \( Z \) of the circuit is given by:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
where \( R \) is the total resistance, \( X_L \) is the inductive reactance, and \( X_C \) is the capacitive reactance. Substituting the values:
\[
Z = \sqrt{(44 + 36)^2 + (90 - 30)^2} = \sqrt{80^2 + 60^2} = 100 \, \Omega
\]
The current \( I \) is given by:
\[
I = \frac{V}{Z} = \frac{200}{100} = 2 \, \text{A}
\]
The power dissipated is:
\[
P = I^2 R = 2^2 \times (44 + 36) = 320 \, \text{W}
\]
