(a) What is the maximum value of the current in the circuit? I max = _______ A (

Question

(a) What is the maximum value of the current in the circuit?
Imax = _______ A

(b) What are the maximum values of the potential difference across the resistor and the inductor?
VR,max = ________ V
VL,max =________ V

(c) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the inductor, and the AC source?
VR =______ V
VL =______ V
Vsource = _______V

(d) When the current is zero, what are the magnitudes of the potential difference across the resistor, the inductor, and the AC source?
VR = ______ V
VL = _______V
Vsource = _______V

Explanation / Answer

formuala for impedance is

Z = sqrt R^2 + ( XL)^2

= sqrt ( 1.2 * 10^3)^2 + ( 2pi ( 59) ( 2.4))^2

=1493.57 ohm

(a)

I max = V max/ Z = 170/1493.5 = 0.113 A

(b)

V amx R = I max R = 0.113 ( 1200) = 136.58 V

Vmax L = I max XL = 0.113 (  2pi ( 59) ( 2.4)= 100.48 V

(c)

When the instantaneous current is a maximum i = Imax ( ), the instantaneous voltage across
the resistor is

del VR = i R = I max R = V max R = 136.58 V

. The instantaneous voltage across
an inductor is always 90° or a quarter cycle out of phase with the instantaneous current.
Thus, when

i = I max

del vL = 0

Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path.
Thus, for this series circuit

del V source = Vmax R + del V L and at this instant

when i = I max

del v source = I max R + 0 = 136.58 V

(d)

When the instantaneous current i is zero, the instantaneous voltage across the resistor is
vR = iR = 0 . Again, the instantaneous voltage across an inductor is a quarter cycle out
of phase with the current. Thus, when i = 0, we must have vL = VL, max = 100.48 V

Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit
at the instant when del v source = del vR del v L = 0 + del vL max = 100.48 V

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