Question

A resistor (R = 9.00 ✕ 102 Ω), a capacitor (C = 0.250 μF), and an inductor (L = 2.40 H) are connected in series across a 2.40 ✕ 102-Hz AC source for which ΔVmax = 1.05 ✕ 102 V.

(a) Calculate the impedance of the circuit. _____kΩ

(b) Calculate the maximum current delivered by the source. ____A

(c) Calculate the phase angle between the current and voltage. _____°

(d) Is the current leading or lagging behind the voltage?

1)The current leads the voltage.

2)The current lags behind the voltage.

3)There is no phase difference between the current and voltage.

Answer 1

given that

R = 9*10^2 ohm

C = 0.250*10^(-6) F

L = 2.40 H

f = 2.40*10^2 Hz

delta Vmax = 1.05*10^2 V

we know that

w = 2*pi*f

w = (2*3.14*(2.40*10^2)) = 15.07*10^2 rad/s^2

we know that

X_L = w*L = 15.07*10^2*2.40 = 36.17*10^2 rad*H/s^2

X_C = 1/(w*c) = 1/(15.07*10^2*0.250*10^(-6)) = 26.59*10^2 s^2/F*rad

part(a)

impedence Z = sqrt (R^2 + (X_L-X_C)^2)

Z = sqrt ((9*10^2)^2 + (36.17*10^2 - 26.59*10^2)^2)

Z = sqrt (10^4*(81+ 601.24)

Z = 26.11*10^2 ohm

Z = 2.61*10^3 ohm

Z = 2.61 kohm

part(b)

Imax = Vmax/Z

Imax = 1.05*10^2 / 2.61*10^3

Imax = 0.04 A

part(c)

Let phase angle b/w current voltage is theta

So

cos(theta) = R/Z

cos(theta) = 9*10^2/2.61*10^3

theta = cos^-1(0.34)

theta = 19.87 degree

part(d)

in given condition we found that X_L > X_C

so voltage is leadind

so answer is (2) current lags behind the voltage