Question

An AC voltage of the form Δv = 100 sin 1 000t, where Δv is in volts and t is in seconds, is applied to a series RLC circuit. Assume the resistance is 415 Ω, the capacitance is 5.35 µF, and the inductance is 0.500 H. Find the average power delivered to the circuit.

Answer 1

Answer:

power delivered $P=Vi=70.72\times 0.1360=9.62W$

Explanation:

We have given $\Delta V=100\ sin(1000t)$

From the equation $V_{MAX}=100volt$ ,$\omega =1000$

We know that $V_{RMS}=\frac{V_{MAX}}{\sqrt{2}}=\frac{100}{1.414}=70.7213volt$

Resistance R = 415 OHM

Capacitance $C=5.35\mu F=5.35\times 10^{-6}F$

Capacvitive reactance $X_C=\frac{1}{\omega C}=\frac{1}{1000\times 5.35\times 10^{-6}}=186.9158ohm$

Inductance = 0.5 H

Inductive reactance $X_L=\omega L=1000\times 0.5=500ohm$

Impedance $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{415^2+(500-186.9158)^2}=519.8525ohm$

Now current $i=\frac{V}{Z}=\frac{70.72}{519.8525}=0.1360A$

So power delivered $P=Vi=70.72\times 0.1360=9.62W$