A real inductor can be modeled as an ideal inductor in series withan internal re

Question

A real inductor can be modeled as an ideal inductor in series withan internal resistance as illustrated in the figure above.

A time-varying current I is passed through a realinductor. At a time when I = +7 A anddI/dt = +14 A/s, the voltage drop across theinductor is observed to be VA -VB = +190 V. At another time when I =+7 A and dI/dt = -14 A/s, the observed voltagedrop is V'A - V'B= +60 V.

(a) Find the internal resistance.

R =    

(b) Find the inductance.

A real inductor can be modeled as an ideal inductor in series withan internal resistance as illustrated in the figure above. A time-varying current I is passed through a realinductor. At a time when I = +7 A anddI/dt = +14 A/s, the voltage drop across theinductor is observed to be VA -VB = +190 V. At another time when I =+7 A and dI/dt = -14 A/s, the observed voltagedrop is V'A - V'B= +60 V. (a) Find the internal resistance. R = Omega (b) Find the inductance. L = H

Explanation / Answer

We know , Potential drop across a resistor = IR Also Potential drop across a inductor = L*dI/dt Let here resistance be R and inductance be L Here in initial case , Potential drop = 190 V IR + L dI/dt = 190 7R + 14L = 190 V   .............(i) Similarly in second case Potential drop = 60 V IR + L dI/dt = 60 7R - 14L = 60 ..........(ii) Adding (i) and (ii) 14R = 250 R = 17.86 Putting in equation (i) 7*17.86 + L*14 = 190 L = 4.64 H

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