Question

Question 1

A series RLC circuit has a capacitor with a capacitance of 36.0 ?F , an inductor with an inductance of 0.300 H and a resistor with a resistance of 65.0 ?. The circuit is attached to a source that has a rms voltage of 52.0 V and a frequency of 85.0 Hz. What is the peak current, the phase angle and the average power loss?

The Peak Current:

The Phase Angle:

The Average Power Loss:

Answer 1

For peak current, we need to find Peak Voltage V_p

V_p = Vrms /

Vp = 52/ =36.7V

Equivalent resistance for RLC circuit ,Z=

R = resistance

XL = inductive resistance (reactance) = jwL =j*2**f*L = j*2**85*0.3 = 103.67j

XC = capacitive resistance capacitance = 36uF=1/jwC = 1/j*2**85*36*10^-6 = 52j

hence

Z = (65*65 +(103.6 - 52)²)^1/2

^{Z =} (4225 +2667.67)^1/2

|Z| = 83.02

1. Peak current = Vp/|Z| = 36.7/83.02 = 0.44A

2.Phase angle

tan = (XL -XC)/R = (103.67 -52)/65 = 0.794

= tan^-1(0.794) = 0.67

3.Average Power Loss

Pav = V_rms *I_rms = Vp*Ip /2 = 36.7*0.44/2

Pav = 8.074Watts ------average power