Find $i$ and in the circuit . Let $i(0)=12 A$
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(i1−i2)+2(i1)=0
2di1dt+i1+2i1−2i2+2i1=0
2di1dt+5i1−2i2=0
Applying KVL in loop 2 6i2−2i1+(i2−i1)2=0
2i2=i1
i2=i12
Putting $i_2$ in (1)
2di1dt+5i1−2i12=0
2di1dt+4i1=0
di1dt+2i1=0
di1dt=−2i1
∫di1i1=∫−2dt
lni1=−2t+lnC
lni1C=−2t
i1C=e−2t
i1=i(0)e−2t
i1=12e−2t
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