A 1.63-k resistor and a 539-mH inductor are connected in series to a 1050 Hz gen

Question

A 1.63-k resistor and a 539-mH inductor are connected in series to a 1050 Hz generator with an rms voltage of 12.0 V.

(a) What is the rms current in the circuit?
3.07 mA

(b) What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part (a)

i need help to solve for part b

i got my answer was c=21.32 nf but it was wrong

please show me steps to get to the answer.

Especially when we use this z^2=r^2+(xL-xC)^2 equation to solve for xc..

i used the quadradic formula, found xc=7111

(i plug in the number z=3911 , r=1603 , XL=3555.97)

i dont know what i did wrong..thank you

Explanation / Answer

We have,

z^2=r^2+(xL-xC)^2

r=1603 ohms

xL = 2*pi*f*L

xL = 6.28*1050*0.539

xL = 3554.16

xC = 1/ 2*pi*f*C

xC = 1/(6.28*1050*C)

Given I = 3.07/2 mA = 1..535 mA

I = Vs/Z

Z = Vs/I

Z^2 = Vs^2/I^2

z^2=r^2+(xL-xC)^2 = Vs^2/I^2

2569609 + ( 3554.16 -0.00015165/C )^2 = (12000/1.535)^2

( 3554.16 -0.00015165/C )^2 = 76442736

3554.16 -0.00015165/C = 8743.15

0.00015165/C = 5188.99

C = 29.22 nF

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