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An L-R-C series circuit consists of a source with voltage amplitude 120 V and angular frequency 50.0 rad/s, a resistor with R = 400 Ω, an inductor with L = 3.00 H, and a capacitor with capacitance C. (a) For what value of C will the current amplitude in the circuit be a maximum? (b) When C has the value calculated in part (a), what is the amplitude of the voltage across the inductor?

Short Answer

Expert verified

The current amplitude in the circuit will be maximum at and the corresponding voltage amplitude is 45V.

Step by step solution

01

Step-1: Formulas used  

Ohm’s law is given by I=VZ, which is used to get the amplitude of current.

Impedance of the circuit is given by Z=R2+(Ó¬³¢-1Ó¬°ä)2.

Voltage amplitude across inductor VL=IÓ¬³¢.

02

Step-2: Calculation of C when current in the circuit will be maximum

Current will be maximum when impedance is maximum and that will happen when

Ó¬³¢=1Ó¬°äC=1Ó¬2L=150rad/s23H=133.3μ¹ó

03

Step-3: Calculation of amplitude of the voltage across inductor

The maximum current I represents the amplitude current and could be calculated by

I=120V400Ω=0.3A

Now, the voltage across inductor is

VL=0.3A50rad/s3H=45V

Hence, (a)the value of C for maximum current is C=133.3μ¹óand (b)the voltage amplitude is 45V.

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