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@shreyasm89-jGobVm • Jul 20, 2010
The given graph appears wrong.On plain logic,the voltage initially across the capacitor is zero.(unless some steady state condition already exists).As the capacitor charges through the resistor its impedance increases(it charges),the current reduces.
In a RC ckt. the expression for current is V/Re^(-t/RC).So the current should go on decreasing exponentialy with time. -
@e2saleh-qaLX6c • Jul 20, 2010
You are right, I just copied the wrong labels from matlab after sticking the equations for voltage and current, here is the actual matlab graph and the correct figure:
![[IMG]](proxy.php?image=http%3A%2F%2Fsur3ah.com%2Fimages%2FmatlabXc.png&hash=0668904f9d554b3a57ccba75c67bf21b)
![[IMG]](proxy.php?image=http%3A%2F%2Fsur3ah.com%2Fimages%2FCurrent-Voltage-Xc6.jpg&hash=0a046711f242b2c331d5797ddf58a3ae)
I believe it is very obvious that the impedance of the capacitor changes with time if we are applying DC!! right? -
@shreyasm89-jGobVm • Jul 20, 2010
😀...Now that's a more accurate graph.I'm not able to understand your question.???If you are performing transient analysis then the eqn. which I mentioned above holds. But the transient time is actually negligible to term the "change in impedance" of the capacitor in case of a DC ckt. Remember a capacitor always resists change in voltage across it.So when the supply is switched ON the capacitor tends to maintain zero voltage drop across it..hence it acts as short when the supply is increasing towards its final value.As the value of the DC supply is continously changing in the time interval till it reaches maximum,it is offered minimum impedance to it.(Xc=1/wC).So in this infintesimal period of time as you stated..the Capacitance impedance is dynamic.
![[IMG]](proxy.php?image=http%3A%2F%2Fsur3ah.com%2Fimages%2FCurrent-Voltage-Xc5.jpg&hash=c4093f4911ce88dbeb2cce08e167f08c)