Sunday, 21 December 2014

Doubt on source free RL circuit


Find $i$ and in the circuit . Let $i(0)=12 A$


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Ans.$12 e^{-2t}$



I have worked out the problem(with a little mistake in it)...But I have a few doubts:



Doubt:


1)Why is the constant equal to the initial current?


2)$ i=-i_1=-12 e^{-2t}$ . Bit in the answer $ i $ is positive. Where is the mistake?



Applying KVL in loop 1 2di1dt+i1+2(i1i2)+2(i1)=0

2di1dt+i1+2i12i2+2i1=0
2di1dt+5i12i2=0


Applying KVL in loop 2 6i22i1+(i2i1)2=0

2i2=i1
i2=i12


Putting $i_2$ in (1)


2di1dt+5i12i12=0

2di1dt+4i1=0
di1dt+2i1=0
di1dt=2i1
di1i1=2dt
lni1=2t+lnC
lni1C=2t
i1C=e2t
i1=i(0)e2t
i1=12e2t





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