The modified Kirchoff loop rule for the primary circuit is as follows: V_1 - L_1
ID: 1515195 • Letter: T
Question
The modified Kirchoff loop rule for the primary circuit is as follows: V_1 - L_1 dI_1/dt - M dI_2/dt = 0. There are two source of back-emf entering this equation, one from the changing current in solenoid 1, having inductance L_1, and the other from the changing current in solenoid 2, which affects the primary through their mutual inductance M. (I have dropped the subscripts on M, following your textbook.) The modified Kirchoff loop rule for the secondary circuit has the same form, but includes the wit age drop across the resistor rather than the battery: -L_2 dI_2/dt - M dI_1/dt - I_2 R = 0. Combine these two equations in a manner such that dl_1/dt is eliminated, and write down the resulting closed equation for I_2. You should find that I_2 obeys the following equation:) Using the equation immediately above, show that, when steady-state is reached, the current in the secondary circuit is given by I_2 = V_2/R.Explanation / Answer
d)
a)
given that, v2(transformed voltage)=13800 V
v1=345000 V
v2=-v1(N2/N1)
N2/N1=-v2/v1
=-(13800 V/345000 V)
=-0.04
b)
v2(transformed voltage)=110 V
v1=345000 V
v2=-v1(N2/N1)
N2/N1=-v2/v1
=-(110 V/345000 V)
=-0.000318
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