(a) What is the maximum value of the current inthe circuit? I max = 1A (b) What
ID: 1674686 • Letter: #
Question
(a) What is the maximum value of the current inthe circuit?Imax = 1A
(b) What are the maximum values of the potential difference acrossthe resistor and the inductor?
VR,max = 2 V
VL,max = 3 V
(c) When the current is at a maximum, what are the magnitudes ofthe potential differences across the resistor, the inductor, andthe AC source?
VR = 4V
VL = 5V
Vsource = 6 V
(d) When the current is zero, what are the magnitudes of thepotential difference across the resistor, the inductor, and the ACsource?
VR = 7V
VL = 8V
Vsource = 9 V
Explanation / Answer
we are given with f = 56 Hz Vmax = 170 V R = 1.2 k = 1.2 x 103 L = 3.0 H the inductive reactance is given by XL = 2 f L = ........ the impedence Z is given by Z = [R2 + (XL -XC)2] as XC = 0 Z = [R2 +(XL)2] = ......... (a) the maximum current will be Imax = Vmax / Z =........ A (b) the maximum values of the potential difference acrossthe resistor will be VRmax = Imax R = ........ V the maximum values of the potential difference acrossthe capacitor will be VLmax = ImaxXL = ........ V (c) when the instantaneous current i is zero theinstantaneous voltage across the resistor is vR = i R = 0 the instantaneous voltage across the inductor isalways 90o or a quarter cycle out of phase with the instantaneous current so we get when i = Imax vL = 0 the kirchoffs rule always applies to the instantaneousvoltage around a closed loop for the series circuit vsource = vR + vL when i = Imax vsource = Imax R + 0 = ....... V (d) when the instantaneous current is zero the instantaneous voltage across the resistor willbe vR = 0 the instantaneous voltage across the inductor isa quarter cycle out of phase with the current so when i = 0 we must get vL = VL,max = ......... V now applying the kirchoofs loop rule to theinstantaneous voltage around the series circuit at the instant when i = Imax gives vsource = vR + vL = 0 + VLmax = ......... V
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