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I've connected a 6V DC voltage source via a 1Ω resistor to a 1F capacitor. I was expecting the simulation to reveal a less constant t-vs-v graph: it would start at 0 and after a while stabilize at 6 volts. But it seems that the voltage across the capacitor begins at the source level and remains so. Two questions:

  1. Why is the voltage across the capacitor constant for the 10 seconds?
  2. (Secondary) Why isn't the current through the resistor I = V/R = 6/1 = 6 A (at least while the capacitor is being charged)?

enter image description here

I'm making my first steps in electronics. Any feedback would be very helpful to me.

Solution. Thanks to everyone for the feedback. Lack of reputation doesn't allow me to upvote them, but all were very helpful to me. Turned out that I didn't check "use initial voltage", so while the initial voltage of the cap was set to 0, it wasn't being used:

enter image description here

Circuit designed and simulated using SystemVision.

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    \$\begingroup\$ Try starting the simulation from real t=0. It could be that under advanced settings the transient analysis isn't beginning at t= 0. \$\endgroup\$ – Andy aka Jun 22 '16 at 17:46
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When you first apply a voltage across a capacitor, assuming the capacitor is discharged, it acts as a short, and thus will show 0 volts across it. However, depending on the value of R and C, the capacitor will eventually charge, and when it is "full" it will not allow any current to pass. It acts as an open circuit, meaning no current will flow (and thus, no current will flow through the resistor). Your simulation is showing the steady-state (after the capacitor has charged), so it shows the full 6V across it and no current flowing through the circuit. It seems that your simulation thinks your capacitor is fully charged when it begins. This is definitely not what I would have expected to see. I would have expected a voltage curve on the capacitor increasing up to around 6 volts in around 4.5 seconds, and then level off (and the current with an inverted version of the graph, eventually dropping to 0). Check the initial conditions of your simulation to make sure it doesn't treat your capacitor as charged on startup.

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    \$\begingroup\$ That did cross my mind, so I clicked on the cap to make sure there is no "initial voltage" it begins with. But I'll tinker with some of the other settings to see what's going on. \$\endgroup\$ – Readingtao Jun 22 '16 at 17:48
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    \$\begingroup\$ Thanks for your help. Turned out I hadn't "forced" the simulation to use the initial voltage of 0. \$\endgroup\$ – Readingtao Jun 22 '16 at 18:23
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1) The simulator starts by calculating a DC solution of the circuit. In the DC solution capacitors are charged to the voltage that would be on a node after a very long time. The transient (time) simulation you performed starts calculating at time = 0 with this DC solution. So that is the solution where the capacitor is already charged.

If you want to see the charging behaviour of the capacitor, you must convince the simulator to calculate a DC solution where the voltage across the capacitor is zero. In some simulators this can be done with an "initial condition".

However it is much simpler to replace your DC voltage source with a pulse-source. Let it make a pulse that starts at 1 second and ends at 100 seconds. Then you will see the charging curve.

2) You're not charging the capacitor so that is why. If you were you would see that predicted current briefly when the voltage across the capacitor is zero.

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The simulator tries to find a reasonable initial condition for your circuit. In this case it assumes that the capacitor is already charged.

Either use a pulse (square wave) as an input source or specify an initial condition directly.

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    \$\begingroup\$ With pulses it had worked; I just wanted to understand why with constant voltage it wasn't working. Thanks for your help. \$\endgroup\$ – Readingtao Jun 22 '16 at 18:22

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