- #1
Cyrus
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I have to solve this problem for a circuit. Its a capacitor, a resistor and another capacitor all in series.
Both capacitances are the same, 1uF, and the resistor is 100k-Ohm.
I have to find the current function of time.
What I did was use KVL to get:
[tex] - \frac{1}{c_1} \int i(t)dt + v_1 (t_0) + \frac{1}{c_2} \int i(t)dt + v_2 (t_0) + R i(t) = 0 [/tex]
Then I took the derivative to get:
[tex] - \frac{1}{c_1} i(t) + R \frac {di}{dt} + \frac{1}{c_2}i(t) =0[/tex]
Which simplifies to:
[tex] R( \frac{1}{c_2} - \frac {1}{c_1} )^{-1} \frac {di}{dt} +i(t) = 0 [/tex]
But the two capacitances have the same value, which means that RC-eq is zero. That's wrong...hmmmmmmmmm
Both capacitances are the same, 1uF, and the resistor is 100k-Ohm.
I have to find the current function of time.
What I did was use KVL to get:
[tex] - \frac{1}{c_1} \int i(t)dt + v_1 (t_0) + \frac{1}{c_2} \int i(t)dt + v_2 (t_0) + R i(t) = 0 [/tex]
Then I took the derivative to get:
[tex] - \frac{1}{c_1} i(t) + R \frac {di}{dt} + \frac{1}{c_2}i(t) =0[/tex]
Which simplifies to:
[tex] R( \frac{1}{c_2} - \frac {1}{c_1} )^{-1} \frac {di}{dt} +i(t) = 0 [/tex]
But the two capacitances have the same value, which means that RC-eq is zero. That's wrong...hmmmmmmmmm
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