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I was preparing for my exam by solving questions and there is this one question I was stuck on but when I saw its solution provided by book, I think KVL applied is not right or I am not able to understand it properly. Can some please help.

Books says - (Xc + 6K + 4K)I(s) = - 160/9s

But I think it should be (Xc + 6K + 4K) I(s) = 160/9s

Solution in book please help.

Here is the question.enter image description here

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It all depends on the direction you choose for the current. Book's answer is using \$I_c\$, that is, counter-clockwise direction, so the voltage across the source will be seen like a voltage drop. You are trying to use \$i_r\$, which is clockwise direction, so the source produces a voltage rise, opposite to the drop over the impedances. That is why the book's answer has the negative sign on the \$160/9s\$ term.

In your case, you will obviously arrive to \$i_r=\frac{16}{9}e^{-100t/3}\$, and when you calculate \$V_c(t)\$, you will have to use this equation: $$ V_c(t)=V_{C10}-\frac{1}{c}\int_0^t{i_r(t)dt} $$ with the minus sign, because the voltage source and the impedance have opposite sign for the current you chose.

If you replace the value for \$i_r\$, you will get the same result as the book.

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  • \$\begingroup\$ Thank you for help mate \$\endgroup\$ Apr 29 '21 at 14:09

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