Problem 1 In an RLC series circuit, the voltage amplitude and frequency of the s

Question

Problem 1 In an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively, and R 600 ?, L :-0.29 H, and C 21 ?F (a) What is the impedance of the circuit (in ?)? (b) What is the amplitude of the current from the source (in A)? (c) If the emf of the source is given by v(t) (100 V) sin (1000rt), how does the current vary with time? (i.e., write an expression for I(t), using the variable t as necessary.) (d) Repeat the above calculations with C changed to 0.21 HF

Explanation / Answer

a)

indtuctive reactance, XL = 2*pi*f*L

= 2*pi*500*0.29

= 911.1 ohms

capacitive recatnce, XC = 1/(2*pi*f*C)

= 1/(2*pi*500*2.1*10^-6)

= 151.6 ohms

impedance if the circuit, z = sqrt(R^2 + (XL - Xc)^2)

= sqrt(600^2 + (911.1 - 151.6)^2)

= 968 ohms

b) Imax = Vmax/z

= 100/968

= 0.103 A

c) here Xc < XL . so current lags voltage.

phase angle, theta = tan^-1( (XL - Xc)/R)

= tan^-1( (911.1 - 151.6)/600)

= 51.7 degrees

= 0.902 radians

so,

I(t) = Imax*sin(w*t - phi)

= Imax*sin(1000*t - 0.902)

d)

a)

indtuctive reactance, XL = 2*pi*f*L

= 2*pi*500*0.29

= 911.1 ohms

capacitive recatnce, XC = 1/(2*pi*f*C)

= 1/(2*pi*500*0.21*10^-6)

= 1516 ohms

impedance if the circuit, z = sqrt(R^2 + (XL - Xc)^2)

= sqrt(600^2 + (911.1 - 1516)^2)

= 852 ohms

b) Imax = Vmax/z

= 100/852

= 0.113 A

c) here Xc > XL . so current leads voltage.

phase angle, theta = tan^-1( (XL - Xc)/R)

= tan^-1( (911.1 - 1516)/600)

= 45.2 degrees

= 0.789 radians

so,

I(t) = Imax*sin(w*t - phi)

= 0.113*sin(1000*t - 0.789)

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