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R C Circuit

If Ra is 2R and a step input is given to the current source, what is the plot for voltage against time, across the capacitor Co? (We can assume that initial voltage across capacitor is 0, and final voltage is say, V1)

How would the analysis change if a voltage source replaced the current source?

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    \$\begingroup\$ linear.com have a free circuit simulation software package called LTSpice. This will create graphs of circuit voltages and currents for you. \$\endgroup\$ – Transistor Jul 28 '16 at 7:37
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    \$\begingroup\$ Where are you stuck? \$\endgroup\$ – Andy aka Jul 28 '16 at 8:05
  • \$\begingroup\$ @transistor I didn't want to simulate it, I want to learn the approach to analyze these kind of circuits graphically \$\endgroup\$ – Ambareesh Sr Ja Jul 28 '16 at 9:00
  • \$\begingroup\$ @Andy I had failed to observe that current source in series with resistor makes the resistors pointless. \$\endgroup\$ – Ambareesh Sr Ja Jul 28 '16 at 9:01
  • \$\begingroup\$ So, what final voltage did you get? \$\endgroup\$ – Andy aka Jul 28 '16 at 9:21
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\$V = \frac{Q}{C} = \frac{1}{C}\int_0^T i(t) dt\$

I.e. \$V\$ is proportional to the area under the \$i(t)\$ graph. The \$i(t)\$ graph is a straight line parallel to the t-axis (step function). So \$V\$ will be proportional to the area of the rectangle whose borders are t=0, t=T, i=0 and i=I.

(you can completly ignore the resistors because they are in series to a current source)

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  • \$\begingroup\$ What if this was a voltage source, how would the analysis change? \$\endgroup\$ – Ambareesh Sr Ja Jul 28 '16 at 10:03
  • \$\begingroup\$ @Ambareesh Sr Ja: The voltage would still be proportional to the area under the \$i(t)\$ graph, but that graph wouldn't be a straight line (but a function of the form \$I_0e^{-t/RC}\$). You can set up a differential eq. and solve it... \$\endgroup\$ – Curd Jul 28 '16 at 12:00
  • \$\begingroup\$ Alright, that's perfect. But to probe a little further, how do you get the function for i(t) ? \$\endgroup\$ – Ambareesh Sr Ja Aug 6 '16 at 10:45
  • \$\begingroup\$ In the case when you don't know the current through the capacitor use KVL (sum of voltages in loop is 0) to set up an equation for \$v_C(t)\$ (voltage across the capacitor). And then use the known V-I-relation for a capacitor: \$\frac{dv_C(t)}{dt}=\frac{1}{C}i_C(t)\$. That gives you a (quite simple) differential equation for \$v_C(t)\$ you can solve. For initial condition probably you want to assume \$v_C(0)=0\$ (C is discharged at the beginning). \$\endgroup\$ – Curd Aug 6 '16 at 11:42

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