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:


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 $$ 2\frac{di_1}{dt}+i_1+2(i_1-i_2)+2(i_1)=0$$ $$ 2\frac{di_1}{dt}+i_1 +2i_1-2i_2+2i_1=0$$ $$2\frac{di_1}{dt}+5i_1-2i_2=0 \tag1$$

Applying KVL in loop 2 $$ 6i_2-2i_1+(i_2-i_1)2=0 $$ $$2i_2=i_1$$ $$i_2=\frac{i_1}{2} \tag2$$

Putting \$i_2\$ in (1)

$$2\frac{di_1}{dt}+5i_1-2\frac{i_1}{2}=0$$ $$2\frac{di_1}{dt}+4i_1=0$$ $$\frac{di_1}{dt}+2i_1=0$$ $$\frac{di_1}{dt}=-2i_1$$ $$\int \frac{di_1}{i_1}=\int -2 dt$$ $$ln i_1=-2t+lnC$$ $$ln\frac{i_1}{C}=-2t$$ $$\frac{i_1}{C}=e^{-2t} \tag 3$$ $$i_1=i(0)e^{-2t}$$ $$i_1=12 e^{-2t}$$

  • 1
    \$\begingroup\$ Ah, I found the problem. Loop 2: It should be \$(i_2-i_1)2\$ in your first equation. \$\endgroup\$ – KingDuken Jul 27 '17 at 20:53
  • \$\begingroup\$ @KingDuken Sir, can you please help me with this question of mine: electronics.stackexchange.com/questions/316246/… \$\endgroup\$ – Soumee Jul 28 '17 at 8:43

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