Question 9 (1 point) You have a 200 ? resistor, a 0.400 H inductor, and a 6.00 ?

Question

Question 9 (1 point) You have a 200 ? resistor, a 0.400 H inductor, and a 6.00 ?F capacitor. Suppose you take the three components and attach them in series with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 rad/s. a) What is the impedance of the circuit? b) What is the current amplitude? c) What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? d) What are the voltage amplitudes across the resistor, inductor, and capacitor? e) Is it possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source? a) Z-6010 a) Z-605 ? b)1-0.0499 A b)1-0.0496 A c) ?-706", lead d) VR-9.93 V, VL-4.89 V,Vc-32.3 V d) VR-9.98 V, VL-4.99 V, Vc-33.3 V e) No. e) Yes

Explanation / Answer

Part A:

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

R = 200 Ohm

XL = w*L = 250*0.4 = 100 ohm

Xc = 1/(w*C) = 1/(250*6*10^-6) = 666.67 Ohm

So impedance will be

Z = sqrt (200^2 + (100 - 666.67)^2)

Z = 600.92 = 601 Ohm

Part B

Imax = Vmax/Z

Imax = 30/601 = 0.0499

Part C

phase Angle is given by:

phi = arctan ((XL - Xc)/R)

phi = arctan ((100 - 666.67)/200)

phi = -70.56 deg, and since phase angle is negative, So voltage will lag.

Part D.

VR = Imax*R = 0.0499*200 = 9.98 V

VL = Imax*XL = 0.0499*100 = 4.99 V

VC = Imax*Xc = 0.0499*666.67 = 33.3 V

Part E.

Yes It's possible.

Please Upvote.

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