You have a resistor of resistance 180? , an inductor of inductance 0.400H , a ca

Question

You have a resistor of resistance 180? , an inductor of inductance 0.400H , a capacitor of capacitance 6.10?F and a voltage source that has a voltage amplitude of 31.0V and an angular frequency of 300rad/s . The resistor, inductor, capacitor, and voltage source are connected to form an L-R-C series circuit.

Part A

What is the impedance of the circuit?

Part B

What is the current amplitude?

Part C

What is the phase angle of the source voltage with respect to the current?

Part D

Does the source voltage lag or lead the current?

Part E

What is the voltage amplitude across the resistor?

Part F

What is the voltage amplitude across the inductor?

Part G

What is the voltage amplitudes across the capacitor?

Explanation / Answer

The impedance of the LCR circuit is given by

Z =Sqrt(R2+(wL-1/wC)2=

R =180ohm then R2 =32400ohm

wL =(300rad/s)(0.400) =72

1/wC =1/(300rad/s)(6.10*10-6F) =546.44

Then impedance is given by Z =Sqrt(1802+(72-546.44)2 =507.43ohms

The amplitude of current is given by

io =Eo/Z =31V/507.43 =0.0610A

The phase angle is given by tan@ =((wL-1/wC)/R) =(72-546.44)/180 =-2.635

@ = tan-1(-2.635) =-1.208degrees

When wL<1/wC , then @ will be negative i.e current leads the e.m.f

The voltage amplitude across the resistor is

   Vo =IoR =(0.0610A)(180ohms) =10.98V =11V

The voltage amplitude across the inductor is given by

V0 =I0*XL =Io*wL =(0.0610)(72) =4.392V

The voltage amplitudes across the capacitor isgiven by

vo =Io(Xc) =Io*(1/wC) =(0.0610)(546.44)=33.33V

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