Question

As given in the figure, a series circuit connected across a $200V,60Hz$ line consists of a capacitor of capacitive reactance $30\mathrm{\Omega }$, a non-inductive resistor of $44\mathrm{\Omega }$, and a coil of inductive resistance $90\mathrm{\Omega }$ and resistance $36\mathrm{\Omega }$. The power dissipated in the coil is?

Answer 1

Correct Answer: 144

The total impedance \(Z\) of the series circuit can be calculated as follows:

• Inductive reactance \(X_L = 90 \, \Omega\)
• Capacitive reactance \(X_C = 30 \, \Omega\)
• Resistances \(R_1 = 44 \, \Omega\) and \(R_2 = 36 \, \Omega\) (for the coil)

First, calculate the overall resistance \(R\) in series:

\(R = R_1 + R_2 = 44 + 36 = 80 \, \Omega\)

The net reactance \(X\) is:

\(X = X_L - X_C = 90 - 30 = 60 \, \Omega\)

The impedance \(Z\) is:

\(Z = \sqrt{R^2 + X^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, \Omega\)

The current \(I\) in the circuit is given by:

\(I = \frac{V}{Z} = \frac{200}{100} = 2 \, \text{A}\)

The power dissipated in the coil (only resistive part) is:

\(P = I^2 \times R_2 = 2^2 \times 36 = 4 \times 36 = 144 \, \text{W}\)

The calculated power is \(144 \, \text{W}\), which matches the range of 144,144 as expected.