You have a 200-ohm resistor, a 0.400-H inductor, and a 6.00-mu F capacitor. Supp

Question

You have a 200-ohm resistor, a 0.400-H inductor, and a 6.00-mu F capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 V and an angular frequency of 250 rad/s. What is the impedance of the circuit? What is the current amplitude? What are the voltage amplitude across the resistor and across the inductor? What is the phase angle phi of the source voltage with respect to the current? Does the source voltage lag or lead the current? Construct the phasor diagram. The resistor, inductor, capacitor, and voltage source described in Exercise 31.14 are connected to form an L-R_C series circuit. What is the impedance of the circuit? What is the current amplitude? What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? What are the voltage amplitudes across the resistor, inductor, and capacitor? Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.

Explanation / Answer

Best Answer:  1)
XL = (250)*(.4) = 100 Ohms
Z = 200 Ohms / [cos (arctan 100/200)] = 223.61 Ohms |_26.565'

2)
Peak current amplitude = 30V/223.61 Ohms = .134164 A = 134.164mA
RMS current amplitude = [(30Vp)*(.7071)] / (223.61 Ohms) = 94.87 mA
Current lags Voltage by - (arc tan 100/200) = -26.565 degrees

3. With Cap added
XC = 1/[(250)*(.000006F)] = 666.66 Ohms
X = (100 - 200) Ohms = -566.66 Ohms
1)
Z = (200 - j100)Ohms = 223.61Ohms |_-26.565'

2)
Peak current amplitude = 30V/223.61 Ohms = 134.162 mA
RMS current = 94.87mA
Current leads Voltage by - [(arc tan (-100)/(200)] = 26.565 degrees

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