Question

A resistor with R=300 ? and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 216 W. (a) What is the impedance Z of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?

Answer 1

Answer:

impedance Z = 416.66 ohm

voltage across inducance V = 346.99 V

power factor = 0.720

Explanation:

given data

resistor R = 300

voltage amplitude = 500 V

resistor = 216 W

to find out

impedance and amplitude of the voltage and power factor

solution

we apply here average power formula that is

average power = I²×R     ............1

I = $\frac{Vrms}{Z}$

so

average power =  ($\frac{Vrms}{Z}$)²×R

Vrms = $\frac{1}{\sqrt{2} }$ × Vmax

Z = V × $\sqrt{\frac{R}{2P} }$

Z = 500 × $\sqrt{\frac{300}{2*216} }$

impedance Z = 416.66 ohm

and

we know voltage across inductor is here express as

V = I × X     .............2

so here X will be by inductance

Z² = R² + (X)²  

(X)²  = 416.66² - 300²  

X = 289.15 ohm

and I = $\frac{V}{Z}$

I = $\frac{500}{416.66}$

I = 1.20 A

so from equation 2

V = 1.20 × 289.15

voltage across inducance V = 346.99 V

and

average power = Vmax × Imax  ×  cos∅

tan∅ = $\frac{289.15}{300}$

tan∅ = 43.95°

so power factor is

power factor = cos43.95°

power factor = 0.720