Voltage across capacitor

Thread Starter


Joined Jan 23, 2009
The current through an initially uncharged 4uF capacitor is as attached. Find the voltage across the capacitor for 0<t<3.

I have following calculation:

i = 40mA 0 <t < 1
i = 0 1 < t < 2
i = -40mA 2 < t < 3

since the voltage across capacitor is given by:

v = (1/c) ∫ i dt + v(t0)

v (@40mA) = (1/c) ∫40m dt + 0 = 10t KV
v(@0mA) = (1/c) ∫0m dt + 10KV = 10KV
v(@ -40mA) = (1/c) ∫-40m + 10KV = -10t + 10 KV

But it seems v(@-40mA) is wrong and also understand I am missing something but do not know where exactly and how.

I will appreciate the pointers.



Joined Feb 7, 2012
It may help to know that if you charge it and discharge it by the same amount the final voltage will be the initial voltage = 0. Between 1<t<2 the voltage doesnt vary and as it is a constant current, the voltage graph will be all straight lines. So all you realy have to know is how much charge is contained in 40mA x 1 sec. Which by definition is 40mC. Then voltage on a cap V=q/c=0.04/0.000004=10,000V=10kV. so it rises linearly to 10kV stays there for 1 second and linearly decays to 0v.

v=10000t V 0<t<1
V=10000 V 1<t<2
V=10000-10000(t-2) V 2<t<3

Note the minus t-2 on time because you discharge starting at 2 seconds.

Hope thats ok.

Thread Starter


Joined Jan 23, 2009
Thanks for the reply...

The thing I did not get is the last part where voltage decays linearly to 0V.

What I understood is the voltage in capacitor is

∫ idt + v(t0), where i is current and v(t0) is initial voltage on the capacitor.

In this case between time span 2 to 3 sec the current is constant -40mA and hence the voltage will linearly increase. Looking back at t=2 sec the initial voltage is 10KV due to steady capacitor voltage and which was not discharge. So the real answer is -10t+ 30 KV. I am unable to understand where 30 KV come from? I assume the v(t0) is 10KV.

I will appreciate your clarification.


Joined Feb 11, 2012
It would help to understand the answer if you were to plot the capacitor voltage.

For 1<t<2, there is no capacitor current, so the voltage remains the same as it was when t=1.

At t=2, the voltage is still 10 kV, and it begins its downward slope. Note that the downward slope is exactly the negative of the upward slope. Furthermore, since the up integration time is the same as the down integration time, the capacitor voltage ends up where it began (zero). The answer for 2<t<3 is v = 10000 * (3-t).

If you plot the capacitor voltage, you'll find it forms an isosceles (symmetrical) trapezoid.
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