Question

A sinusoidal voltage Δv = 45.0 sin(100t), where Δv is in volts and t is in seconds, is applied to a series RLC circuit with L = 140 mH, C = 99.0 µF, and R = 61.0 Ω.

(a) What is the impedance (in Ω) of the circuit?

________ Ω

(b) What is the maximum current (in A)?

_________A

(c) Determine the numerical value for ω (in rad/s) in the equation i = I_max sin(ωt − ϕ).

_________rad/s

(d) Determine the numerical value for ϕ (in rad) in the equation i = I_max sin(ωt − ϕ).

__________rad

(e)What If? For what value of the inductance (in H) in the circuit would the current lag the voltage by the same angle ϕ as that found in part (d)?

__________ H

(f) What would be the maximum current (in A) in the circuit in this case?

__________ A

Answer 1

(a) The voltage Vis given as follows: V = V, sin ot = (45.0) sin (100t) Therefore, peak voltage and angular frequency is given as follows: V = 45.0 V o =100 rad/s The impedance in RLC circuit is given as follows: X = (X, -X +R? The impedance by inductance in the circuit is given as follows: X, = @L =(100 rad/s) (140 mH) = 14 2 The impedance by capacitance in the circuit is given as follows: х. (100 rad/s)(99.0 µF) = 101 Q The impedance in RLC circuit is given as follows: X = (X, - X.)* + R? -ν(14.0 Ω-101 Ω) -(61.0 ) =106.25 O

Xc (100 rad/s)(99.0 µF) =101 2 The impedance in RLC circuit is given as follows: X = (x,- X.)+R² (14.0 0-101 2) +(61.0 2) = 106.25 O Therefore, the impedance of the circuit is 106.25 (b) The maximum current in the circuit is given as follows: Imax %3D 45.0 V 106.25 O = 0.42 A Therefore, the maximum current in circuit is 0.42 A

(c) The numerical value of @ for current would be same as that of value of @ for voltage. Therefore, the value of @ for current is 100. (d) The phase angle o for RLC circuit is given as follows: -1 X,-X. 0 = tan 14.0 Ω-101 Ω = tan 61.0 O =-0.959 rad Therefore, the value of phase angle is -0.959 rad.

(e) The phase angle will be positive and implies current will lag the voltage by same angle if (x, - Xc) is positive and have same magnitude as of earlier case. Therefore, calculate new impedance of inductor X, as follows: Px-'x|= °x-'x °x+l°x- x|= "x = |14.0 2-101 2|+101 N =188 N Calculate the value inductance L' as follows: X = @L %3D L' 188 Q 100 rad/s =1.88 H Therefore, the value of new inductance is 1.88 H (f) The value of maximum current for the inductance in which current lags behind voltage is given as follows: V. (x -x.) +R° 45.0 V (188 Ω-101 Ω) + (61.0 Ω) = 0.42 A Therefore, current has same value and is equal to 0.42 A