An AC source operating at 56 Hz with a maximum voltage of 170 V is connected in

Question

An AC source operating at 56 Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 k) and an inductor (L = 2.0 H).
(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

Given Frequency f = 60 Hz Maximum voltage Vmax = 170 V Resistance of the resistor R = 1.2 k                                          = ( 1.2 k ) ( 1000 / 1 k )                                          = 1200 Inductance of the inductor L = 2.2 H a) Impedance of the circuit is              Z = ( R 2 + (2fL )2 )1/2                  = [ ( 1200 )2 + ( 2*60*2.2 )2 ] 1/2                 = 1458.633 Maximum current             I max = V max / Z                      = ( 170 V ) / 1458.633                      = 0.1165 A _________________________________________________________________ b) Maximum values of the potential difference across the resistor and the inductor          Vmax,R = I max R                       = ( 0.1165 A ) ( 1200 )                       = 139.8565 V          Vmax,L = I max ( 2 f L )                       = ( 0.1165 A ) ( 2*3.141*60* 2.2 )                       = 96.604 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 = I max R                       = ( 0.1165 A ) ( 1200 )                       = 139.8565 V          VL = I max ( 2 f L )                       = ( 0.1165 A ) ( 2*3.141*60* 2.2 )                       = 96.604 V          VSource = Vmax sin t                          = Vmax ( IR / I max )                        = Z I R                          = (  1458.633 ) ( 0.1165 A )                        = 169.93 V ________________________________________________________________________________                       = ( 0.1165 A ) ( 1200 )                       = 139.8565 V          VL = I max ( 2 f L )                       = ( 0.1165 A ) ( 2*3.141*60* 2.2 )                       = 96.604 V          VSource = Vmax sin t                          = Vmax ( IR / I max )                        = Z I R                          = (  1458.633 ) ( 0.1165 A )                        = 169.93 V ________________________________________________________________________________

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