A resistor,inductor and capacitor are connected in series,each with effective (rms)voltage of 65 V,140 V and 80 V,respectively.If the resistor is rated at 24 W,what is the average power dissipated in the circuit?
A)88 W
B)176 W
C)238 W
D)323 W

Question

Quizplus · Accepted Answer

Using the formula for impedance in a series circuit:
Z = sqrt(R^2 + (XL - XC)^2)
where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

Given:
Effective voltage (rms) of resistor, V_R = 65 V
Effective voltage (rms) of inductor, V_L = 140 V
Effective voltage (rms) of capacitor, V_C = 80 V
Resistance, R = 24 ohms

We can first find the values of XL and XC using the formulas:
XL = 2πfL
where f is the frequency and L is the inductance
XC = 1/(2πfC)
where f is the frequency and C is the capacitance

Assuming a frequency of 50 Hz (standard AC frequency), we get:
XL = 2π(50)(L) = 140 ohms
XC = 1/(2π(50)(C)) = 31.8 ohms

We can then find the total impedance of the circuit:
Z = sqrt(R^2 + (XL - XC)^2) = 134.5 ohms

The total current in the circuit is given by:
I = V/Z = (65 + 140 + 80)/134.5 = 2.30 A (rms)

The average power dissipated in the circuit is given by:
P = I^2R = (2.30)^2(24) = 126.7 W

However, this only accounts for the power dissipated in the resistor. The inductor and capacitor have reactive power, which means they store and release energy without dissipating it. Therefore, we need to also consider the reactive power in the circuit.

The reactive power in the inductor is given by:
Q_L = V_L^2/XL = 140^2/140 = 140 W

The reactive power in the capacitor is given by:
Q_C = V_C^2/XC = 80^2/31.8 = 202 W

The total reactive power in the circuit is given by:
Q = Q_L - Q_C = -62 W

The negative sign indicates that the circuit is capacitive, meaning the capacitor is absorbing reactive power while the inductor is releasing it.

The total apparent power in the circuit is given by:
S = VI = (65 + 140 + 80)(2.30) = 654 VA

The total power factor of the circuit is given by:
pf = P/S = 126.7/654 = 0.193

The average power dissipated in the circuit is then given by:
P_avg = S(pf) = 654(0.193) = 126.3 W

Since the resistor is the only component dissipating real power, the average power dissipated in the circuit is approximately equal to the power dissipated in the resistor, which is 126.7 W. The closest answer choice is 176 W, which is option B.

A Resistor,inductor and Capacitor Are Connected in Series,each with Effective