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Asked by Prashant DIGHE | 28 Dec, 2019, 21:41: PM
answered-by-expert Expert Answer
current i through inductance as a function of time is given by
 
i(t) = im [1 - e-t(R/L) ].................(1)
 
where im is maximum current that flows when steady state is reached , i.e. long time after switch closed.
R is the resistance in series with inductance L.
 
when steady state is reached , inductance becomes bare conducting wire of negligible resistance.
Hence at steady state, two resistors in the circuit becomes parallel combination and gives equivalenet resistance of 1Ω.
 
Hence at steady state, current drawn from the battery of emf 12 V is 12 A .
 
Equal current flows through Resistance R1 and series combination of inductance and R2 .
 
Hence, at steady state,  current that flows through inductance is 6 A.
 
Hence eqn.(1) becomes,  i(t) = 6 [ 1 - e-t(2/0.4) ] = 6 [ 1 - e-5t ]  ....................(2)
 
Voltage across inductance = L (di/dt) = 0.4 × 6 × 5 × e-5t = 12 e-5t
Answered by Thiyagarajan K | 29 Dec, 2019, 09:04: AM

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