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 (Correct answer is -10t + 30 KV) What I understood is the voltage across 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. And hence I take 10KV at v(t0). I am unable to understand where 30 KV come from as I assume the v(t0) is 10KV. I understand I am missing something but do not know where exactly and how.