Johnny_YU
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hi Johnny,
This is what LTSpice shows for that circuit.
As the Vsrc impedance is zero and the two capacitors are the same value, at t= 0, Vout will be Vsrc/2
The bottom cap will then charge up to Vsrc exponentially
E
Hi Johnny, Let me try a more or less descriptive/intuitive explanation:
At t=0 both capacitors have no charge.
After closing the switch the initial current through both capacitors must be equal (both capacitors equal and "empty"). More than that, at t=0 the resistor has not yet any influence.
Because of equal currents also both initial voltages across the capacitors must be equal. This is possible only if the voltage across C2 is 0.5Vcc.
Condition: the initial energy stored in this simple RC circuit is ZERO.
AT t=0s ,a DC voltage source of 5 Volts is applied to the circuit.
Strictly speaking, your problem description is physically impossible.
A DC voltage can't be applied at a specific time, in analysis terms it's either present since infinite time, or you have a pulsed voltage. You can fool the SPICE analysator by omitting the initial transient solution, but in this case, the result doesn't fully conform with laws of physics.
A switched voltage source will however cause infinite initial current when it tries to charge the capacitors in no time, which is physically impossible as well.
In case you want a numerical calculation, just start from the solution of Vo(t) and apply, as also you said:
Ic2(t)=C2*dVo(t)/dt
remembering that Vo(t)=Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*1(t) then
Ic2(t)=Vcc*C2*C2/(C1+C2)]*1/[R1*(C1+C2)]*exp[-t/(R1*(C1+C2))]}*1(t)+
+Vcc*{1-C2/(C1+C2)*exp[-t/(R1*(C1+C2))]}*δ(t)
for t=0+:
Ic2(0)=Vcc*C2²/[R1*(C1+C2)]²*1(0)+Vcc*{1-C2/(C1+C2)}*δ(0)
this means there will be an (ideally) infinite pulse in 0, than the current will decreases starting from 125 uA
The "physically impossible" comment was related to your simulation waveform in post #10 and the question about the actual initial current.
You can of course describe the theoretical waveforms generated when switching a voltage source to a capacitor in no time with a symbolic δ operator. But you can't see the waveforms in a circuit simulator, because it's not able to handle infinity.
HI,albbg
Am I right? waiting for your reply~:???:
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