Coil 1 has L_1 = 39 mH and N_1 = 130 turns. Coil 2 has L_2 = 49 mH and N_2 = 232

Question

Coil 1 has L_1 = 39 mH and N_1 = 130 turns. Coil 2 has L_2 = 49 mH and N_2 = 232 turns. The coils are fixed in place; their mutual inductance M is 4.1 mH. A 7.5 mA current in coil 1 is changing at the rate of 6.5 A/s. (a) What magnetic flux _12 links coil 1, and (b) what self-induced emf appears in that coil? (c) What magnetic flux _21 links coil 2, and (d) what mutually induced emf appears in that coil? (a) Number 2.25 Units mu Wb (b) Number 253.5 Units mV (c) Number 0.0003075 Units nWb (d) Number 26.65 Units mV

Explanation / Answer

L(inductance) = (Number of turns * flux)/current

by using this we get flux as Li/N

a) L = 39mH , i = 7.5 mA , N = 130 turns

flux = 39*7.5/130 = 2.25 micro Weber

b) self induced emf = Ldi/dt = 39 mH * 6.5 = 253.5 mV

c) here flux in coil 2 is not due to current in coil 2 but due to current in coil 1

instead of L2 we use M(mutual inductance)

flux = Mi/N = 4.1 mH * 7.5 mA / 232 = 0.1325 micro Weber = 132.5 nano Weber

d) mutual induced emf = M di/dt = 4.1*6.5 = 26.65 mV

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