Hi all!
Here is a circuit with two sources and a switch. The right source should be a dependent source according to the notation used in my book, although I fail to see any dependance. Could this be a typing error in the book?
The task is to find \( v=? \)
The initial conditions before closing the switch would be:
\( i_1 (0) = 0 \)
\( i_2 (0) = - \frac{k}{2} \)
\( v(0) = k \), this assumes that the right source is active starting from \( - \infty \). Could this assumption be wrong?
Starting with loop analysis:
(1): \( -2 + 2i_1 + 2 (i_1 - i_2) = 0 \)
(2): \( 2 (i_2 - i_1) + \frac{di_2}{dt} + k = 0 \)
Eliminating \( i_1 \) gives:
\( i_2 + \frac{di_2}{dt} = 1 - k \)
Solving this equation:
\( i_2^h = A e^{-t} \)
\( i_2^p = 1 - k \)
\( i_2 = i_2^h + i_2^p = Ae^{-t} + 1 - k\)
Inserting the initial condition:
\( i_2(0) = A + 1 - k = - \frac{k}{2} \)
\( A = \frac{k}{2} - 1\)
It follows:
\( i_2 = (\frac{k}{2} - 1) e^{-t} + 1 - k \)
From Eq. (1):
\( i_1 = \frac{i_2+1}{2} \)
\( i_1 = \frac{1}{2} (\frac{k}{2} - 1 ) e^{-t} + 1 - \frac{1}{2}k \)
Observe that \( i_1 \) does not satisfy the initial condition for 0 in the last equation, only for 0+, after the switch is closed. So it would probably be better to multiply the result of \( i_1\) with the unit step function? I am asking this, because we can derive the formula for \( i_1 \) without the step function, by simply inserting the initial condition in the diff. eq.
Now to calculate v:
\( v = 2 (i_1 - i_2) \)
\( v = (1 - \frac{k}{2}) e^{-t} + k\)
For example, for \( k = 0.5 \) we obtain \( v = \frac{3}{4} e^{-t} + \frac{1}{2} \). The solution according to my book should be: \( v = e^{-\frac{t}{2}} \). For \( k = 0 \) the solution is correct: \( v = e^{-t}\).
From this I surmise that the exponential part should look like \( e^{-kt} \), but I do not see how this could be obtained from the differential equation above. Any help or hints would be appreciated.
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