Question

An LRC series circuit consists of a 3.15 H inductor, a 6.36 ohm resistor, and a 5.48 microFarand capacitor. The combination is connected to an AC votage source that has a peak voltage of 147 V and an angular frequency of 441 rad/s. What is the peak voltage measured across the inductor?

Answer 1

Please check Underline part

In LRC circuit:

X_L = w*L

Using given values:

w = 441 rad/s

L = 3.15 H

X_L = 441*3.15 = 1389.15 ohm

X_C = 1/(w*C)

C = 5.48*10^-6 F

So,

X_C = 1/(441*5.48*10^-6)

X_C = 413.79 Ohm

Now Impedance of circuit will be:

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

Given, R = 6.36 Ohm (This value is very low please check given value in your question once again and if different let me know)

Z = sqrt (6.36^2 + (1389.15 - 413.79)^2)

Z = 975.38 Ohm

Using ohm's law:

I_max = V_max/Z

I_max = 147/975.38 = 0.1507 Amp

Since LRC are in series, So current in each of them will be equal to max current, So

Now peak voltage across the inductor will be given by:

V_{L_max} = X_L*I_max

V_{L_max} = 1389.15*0.1507 = 209.34 V

V_{L_max} = 209.34 V (If your given values are true)

2nd solution: If R = 636 ohm, then

Z = sqrt (636^2 + (1389.15 - 413.79)^2)

Z = 1164.40 Ohm

Using ohm's law:

I_max = V_max/Z

I_max = 147/1164.40 = 0.1262 Amp

Since LRC are in series, So current in each of them will be equal to max current, So

Now peak voltage across the inductor will be given by:

V_{L_max} = X_L*I_max

V_{L_max} = 1389.15*0.1262 = 175.0 V

V_{L_max} = 175.0 V (If R = 636 ohm)