An L - R - C series circuit consists of a 50.0 Ohm resistor, a 20.0 uF capacitor

Question

An L-R-C series circuit consists of a 50.0 Ohm resistor, a 20.0uF capacitor, a 3.00mH inductor, and an ac voltage source of voltage amplitude 65.0V operating at 1300Hz

a) Find the current amplitude across the inductor, the resistor, and the capacitor.

I=???A

b) Find the voltage amplitudes across the inductor, the resistor, and the capacitor.

Vl,Vr,Vc=???V

c) Why can the voltage amplitudes add up to more than 65.0V ?

d) If the frequency is now doubled, but nothing else is changed, which of the quantities in part A and B will change?

-only voltage amplitude across the inductor will change
-only current amplitude will change
-only voltage amplitude across the inductor and capacitor will change
-current amplitude and voltage across the any circuit element will change


e) Find new current amplitude across the inductor, the resistor, and the capacitor.
I=???A

f) Find new voltage amplitudes across the inductor, the resistor, and the capacitor.
Vl,Vr,Vc=???V

Please show all your calculations so I can see how you got the answers.

Explanation / Answer

w = 2pif = 8168rad/s

XL = wL = 24.50ohms

Xc = 1/wC = 6.12 ohms

Z = sqrt(R^2 + (XL-Xc)^2)

Z = 53.27 ohms

current = V/Z = 65/53.27 = 1.22A

B )V across R = I*R = 61 V

V across L = I*XL = 29.89 V

V across C = I*Xc = 7.4664V

C)as the circuit contains non resistive and reactive elements,it tends to be higher than 65 V

D)All options

E)w = 2*8168 = 16336 rad/s

XL = 49 ohms

Xc = 3.06 ohms

Z = 67.9 ohms

current = V/Z = 65/67.9 = 0.957A

F)V = IR = 47.86V

V = IXL = 46.893A

V=I*Xc = 2.92V

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