Question

A circuit contains a resistor R = 40 $\Omega$, an inductor L = 0.5 H and a capacitor C = 0.5 mF connected in series with a voltage source $V(t) = 200sin(100t)$. The real power dissipated in the circuit is

Answer 1

320 W

Answer 2

To determine the real power dissipated in the circuit, we follow these steps:

1. Identify the given parameters and extract angular frequency and peak voltage:

   • Resistance, $R = 40 \, \Omega$
   • Inductance, $L = 0.5 \, \text{H}$
   • Capacitance, $C = 0.5 \, \text{mF} = 0.5 \times 10^{-3} \, \text{F}$
   • Voltage source, $V(t) = 200\sin(100t)$

   From $V(t) = V_0\sin(\omega t)$, we have:

   • Peak voltage, $V_0 = 200 \, \text{V}$
   • Angular frequency, $\omega = 100 \, \text{rad/s}$

2. Calculate the inductive reactance ($X_L$): $X_L = \omega L = 100 \, \text{rad/s} \times 0.5 \, \text{H} = 50 \, \Omega$

3. Calculate the capacitive reactance ($X_C$): $X_C = \frac{1}{\omega C} = \frac{1}{100 \, \text{rad/s} \times 0.5 \times 10^{-3} \, \text{F}} = \frac{1}{0.05} = 20 \, \Omega$

4. Calculate the impedance ($Z$) of the series RLC circuit: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

   $Z = \sqrt{40^2 + (50 - 20)^2}$

   $Z = \sqrt{40^2 + 30^2}$

   $Z = \sqrt{1600 + 900}$

   $Z = \sqrt{2500} = 50 \, \Omega$

5. Calculate the RMS voltage ($V_{rms}$): $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} \, \text{V}$

6. Calculate the RMS current ($I_{rms}$): $I_{rms} = \frac{V_{rms}}{Z} = \frac{200/\sqrt{2}}{50} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{A}$

7. Calculate the real power ($P$) dissipated in the circuit:

   Real power is dissipated only in the resistive component of the circuit.

   $P = I_{rms}^2 R$

   $P = (2\sqrt{2})^2 \times 40$

   $P = (4 \times 2) \times 40$

   $P = 8 \times 40$

   $P = 320 \, \text{W}$