An AC source operating at 60 Hz with a maximum voltage of 170 V is connected in

Question

An AC source operating at 60 Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kW) and an inductor (L = 2.8 H). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the inductor? (c) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the inductor, and the AC source? (d) When the current is zero, what are the magnitudes of the potential difference across the resistor, the inductor, and the AC source?

Explanation / Answer

here f = 60 Hz so w = 2 x pi x f

w = 2 x 3.14 x 60

= 376.8 rad/s

Vmax = 170 V L = 2.8 H R = 1.2 ohm

so the maximum value of current

Imax = Vmax / Z

= Vmax / root of ( R2 + xL2)

here XL = wL = 376.8 x 2.8

= 1055.04 ohm

so Imax = 170 / root of (1.22 + 1055.042)

= 0.1611 A Ans

the current is same in series

so voltage across resistor VR = Imax R

= 0.1611 x 1.2 = 0.1933 V Ans

voltage across inductor VL = Imax XL

= 0.1611 x 1055.04

= 169.966 V Ans

the voltage of AC will be same of 170 V

if the current is zero

there will not drop any voltage across resistor an inductor

the total voltage stored on AC source of 170 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.