How can I find the equation of current in the following circuit, when switch is closed at t=0. Capacitor is initially not charged.
simulate this circuit – Schematic created using CircuitLab
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\$\begingroup\$ If there is not resistance (also no internal resistane in the voltage source, switch or capacitor) there is not much of an equation.... i.e. the equation is \$V_C(t)=0\$ for \$t<0\$ and \$V_C(t)=V_s\$ for \$t>0\$ \$\endgroup\$– CurdCommented Jan 20, 2017 at 10:32
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\$\begingroup\$ What will be the nature of current ? \$\endgroup\$– VedanshuCommented Jan 20, 2017 at 10:41
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\$\begingroup\$ In ideal case (R=0) current will be infinite for an infinitely short time after switching on. In practice there will be some resistance R and you'll get a \$I(t)=I_0e^{-\frac{t}{RC}}\$ where \$I_0=\frac{Vs}{R}\$. \$\endgroup\$– CurdCommented Jan 20, 2017 at 10:44
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\$\begingroup\$ That means capacitor will be charged within a very short time, to the source voltage; doesn't that changes the property of capacitor of sudden change in voltage, "capacitor does not allow sudden change in voltage across it" ? \$\endgroup\$– VedanshuCommented Jan 20, 2017 at 10:47
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1\$\begingroup\$ You could also ask what would be the equation of a switch shortcutting a power source. V/0 -> infinite current \$\endgroup\$– Claudio Avi ChamiCommented Jan 20, 2017 at 10:51
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2 Answers
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There is no equation.
With an ideal voltage source and ideal switch, wires and capacitor, infinite current is taken for an infinitesimal length of time to charge the capacitor to Vs instantly.
answered Jan 20, 2017 at 15:18
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Your schematic makes only sense if theres a series resistor between the capacitor and the power source.
In this case the voltage and current waveforms for charging the capacitor are:
\$ V_{cap}(t) = V_s \cdot \left( 1 - e^{-t/\tau}\right) \$ and
\$ I_{cap}(t) = C \cdot \frac{dV_{cap}(t)}{dt} = C \cdot V_s \cdot \frac{1}{\tau}\cdot e^{-t/\tau} = V_s \cdot \frac{1}{R}\cdot e^{-t/\tau}\$
where \$ \tau = R \cdot C\$
It all results from the equation: \$ I(t) = C\cdot \frac{dV(t)}{dt}\$. This is the characteristic eqaution for a capacitor.
If you have a series resistance in your schematic and use the KVL You get:
\$ V_s = V_R(t) + V_{cap}(t)\$.
With
\$ U_R(t) = R \cdot I_{cap}(t) = R \cdot C \cdot \frac{dV_{cap}(t)}{dt}\$
you get:
\$ V_s = R \cdot C \cdot \frac{dV_{cap}(t)}{dt} + V_{cap}(t) \$
you can solve this differential equation using different methods, for example using Laplace Transform.
The solution is the equation above.
The circuit without resistance doesn't produce a waveform. If you close the switch, the voltage change on the cap would be infinitely high which results in an infinite current. The capacitor is charged to its full voltage instantly.
