ElectroMagnetic InductionHard
Question
The battery shown in the figure is ideal. The values are e = 10 V, R = 5Ω, L = 2H . Initially the current in the inductor is zero. The current through the battery at t = 2s is
Options
A.12 A
B.7 A
C.3 A
D.3 A
Solution
I + I1 + I2
I1 = E/R
L
= E
I2 =
I = E/R +
I = 12A.
I1 = E/R
L
= EI2 =
I = E/R +
I = 12A.
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