In parallel RC circuit, what is the behavior of current of capacitor C1? What is the time constant?
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5\$\begingroup\$ Was this assigned as homework to you? If it was, we won't ruin your learning experience by doing it for you. If you expect help from us, please edit your question to tell us what you have tried and what specifically are you having trouble with. \$\endgroup\$– RicardoNov 29, 2014 at 18:48
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\$\begingroup\$ It is not a hw question. Is the current of capacitor goes to infinity at the beginning? \$\endgroup\$– tntNov 29, 2014 at 19:11
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\$\begingroup\$ infinity is a mathematical construct, doesn't exist in the real world. \$\endgroup\$– vicatcuNov 29, 2014 at 19:29
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5\$\begingroup\$ We don't just answer homework questions here, especially when no effort is shown on your part. \$\endgroup\$– Olin LathropNov 29, 2014 at 19:39
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\$\begingroup\$ -1 for horrible schematic! You should read this: Rules and guidelines for drawing good schematics \$\endgroup\$– KamilNov 29, 2014 at 20:36
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2 Answers
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In the circuit as drawn, I am assuming the voltage source and capacitor are ideal and the voltage source is producing an instantaneous step from 0 to 5 V. In this case, the model gives an infinite current (or a current impulse) into the capacitor at the time of the voltage step.
But that does not tell you much about what a real circuit would do. It just tells you you have created an unrealistic model.
A more realistic model would include series resistance in the voltage source and capacitor models, and a nonzero rise time for the voltage source's output step. Any one of these improvements in the model would result in a finite current into the capacitor.
answered Nov 29, 2014 at 20:56
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The governing equation for a capacitor: \begin{equation} I = C \frac{dV}{dt} \end{equation}
If the circuit has been at DC "forever", then no current flows through the capacitor because V never changes.
However, forever is a long time and you can't actually have this in the real world. So replace the source with some V(t) which has V(0) = 0. Now you have an equation which describes what I is vs. t.
answered Nov 29, 2014 at 19:36
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\$\begingroup\$ ...also which doesn't transition "instantaneously" \$\endgroup\$– vicatcuNov 29, 2014 at 19:46
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\$\begingroup\$ My question is: It is initially 0 Volt, when 5 volts is applied, what is the current passes throu the capacitor? Is it infinity therotically? \$\endgroup\$– tntNov 29, 2014 at 19:49
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\$\begingroup\$ @tnt to understand what "really" happens (in the brief period of time before "steady state") you have to model the circuit as a transmission line - look up "bounce diagrams" \$\endgroup\$– vicatcuNov 29, 2014 at 19:56
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\$\begingroup\$ electronics.stackexchange.com/questions/844/… \$\endgroup\$– vicatcuNov 29, 2014 at 20:00