# Capacitor and resistors at t=0 and t after long time

I’m pretty sure my answer is wrong because I forgot to include the capacitor in the calculation and I don't know how I'm supposed to do that.

Find the current for each resistor and in the capacitor at t=0 and after a long time.

simulate this circuit – Schematic created using CircuitLab

$$\ \mathrm{I_1} = \mathrm{I_2} + \mathrm{I_3} \$$

\ \begin{aligned} \mathrm{R_{eq2,3}} &= {\left(\frac{1}{\mathrm{R_2}}+\frac{1}{\mathrm{R_3}}\right)}^{\displaystyle-1}\\ &= {\left(\frac{1}{3300}+\frac{1}{5000}\right)}^{\displaystyle-1}\\ &= 1987.95\Omega \end{aligned} \

\ \begin{aligned} \mathrm{R_{eq}} &= \mathrm{R_1} + \mathrm{R_{eq2,3}}\\ &= 1000 + 1987.95\\ &= 2987.95 \Omega \end{aligned} \

\ \begin{aligned} \mathrm{I_1} &= \frac{12}{2987.95}\\ &= 4\!\times\!10^{-3} \mathrm{A} \end{aligned} \

\ \begin{aligned} \Delta\mathrm{V} &= \mathrm{I_1} \cdot \mathrm{R_1}\\ &= 4\!\times\!10^{-3} \cdot 1000\\ &= 4 \mathrm{V} \end{aligned} \

\ \begin{aligned} \mathrm{I_2} &= \frac{8}{3300}\\ &= 2.42\!\times\!10^{-3}\mathrm{A} \end{aligned} \

\ \begin{aligned} \mathrm{I_3} &= \mathrm{I_1} - \mathrm{I_2}\\ &= 1.58\!\times\!10^{-3}\mathrm{A} \end{aligned} \

• Can you transcribe your work and rotate the image to be the right way up? – Hearth Apr 4 '19 at 14:59
• The voltage on capacitor cannot change instantaneously. – Dirceu Rodrigues Jr Apr 4 '19 at 15:08
• Please redraw your circuit using the schematic tool and label each component. This site also uses MathJax so you can write maths so it is easy to read. – Warren Hill Apr 4 '19 at 15:13
• Also for $t = 0$ do you mean $t = 0^-$, The switch has been open for a long time or $t=0^+$ the instant the swich has been closed? – Warren Hill Apr 4 '19 at 15:16
• @Unknown123 Yes, of course. But the question asks for voltages and currents of the R's and the C, which, with switch open and C uncharged, you don't have to "calculate" for because it's (obviously?) 0. – JimmyB Apr 5 '19 at 10:15

At t=0, the capacitor is completely discharged and has 0V across it.

At t=infinity, the capacitor is fully charged. How much current will flow into/through the capacitor then? What voltage drop will occur due to that current across the 3.3k resistor?

Look at the charge curve of a capacitor, e.g. here:

It also has the values for t=0 and t->infinity.

You can see that for long t the charge and the voltage approach 100% while the current goes towards 0. Theoretically, the capacitor will only ever be exactly 100% charged after an infinite amount of time, but it will be very close to 100% (99.9999...%) after a small multiple of the R*C time constant. This can be seen from the 1-exp(-t/...) term, which approaches 1 (100%) exponentially over t.

• So at t=0 are my calculations correct? I highly doubt that tho. I really hope u can understand the hand writing – Abdullah Apr 4 '19 at 15:40
• Not sure which is R1, R2 or R3, but yes, at t=0 you have one resistor (R1?) in series with the two other resistors in parallel (R2||R3?). – JimmyB Apr 4 '19 at 15:43
• 1kohm is R1, 3.3kohm is R2 and 5kohm is R3 and thanks for your confirmation – Abdullah Apr 4 '19 at 15:47
• Also how am i suppose to find the current for each resistor and the capacitor at t long time or infinite – Abdullah Apr 4 '19 at 15:50
• "At t=infinity, the capacitor is fully charged." – JimmyB Apr 4 '19 at 15:51

At t=0 the capacitor has 0V (zero voltage) across it. At that instant t=0 it will act like a short circuit. That allows you to solve for the currents using the circuit shown below.

simulate this circuit – Schematic created using CircuitLab

At t=inf the capacitor is fully charged has 0A (zero current) through it. So at t=inf it behaves like an open circuit. Solve for currents per circuit below. Obviously I_R3 = 0.

simulate this circuit

• I think it is very misleading to say that the capacitor is a short circuit at t=0. Yes, the voltage across the capacitor is zero but not because the capacitor is a short circuit. A capacitor is always an open circuit at dc. – Elliot Alderson Apr 4 '19 at 18:53
• @Unknown123 It really has nothing to do with ESR. Since $i_C = C\,dV/dt$ it must be true that $i_C = 0$ when $dV/dt = 0$...in other words, at dc. So, what we know for certain is that the current through a capacitor must be zero at dc, which means that a capacitor looks like an open circuit at dc. If the voltage across the capacitor happens to also be zero then that is just a coincidence, it doesn't imply a short circuit. – Elliot Alderson Apr 4 '19 at 20:08
• @ElliotAlderson Yes, I agree it doesn't imply a short circuit, it only acts or looks like short circuit at $\mathrm{t}=0$ as in this case it would be $\displaystyle \frac{d\mathrm{V_c(t)}}{dt} = \frac{\mathrm{V_{th}}}{\mathrm{R_{th}C}}\cdot e^{\displaystyle\left(-\frac{\mathrm{t}}{\mathrm{R_{th}C}}\right)}$, thus at $\mathrm{t}=0 \implies \displaystyle \frac{d\mathrm{V_c(0)}}{dt} = \frac{\mathrm{V_{th}}}{\mathrm{R_{th}C}} \implies \mathrm{I_c(0)} = \frac{\mathrm{V_{th}}}{\mathrm{R_{th}}}$ – Unknown123 Apr 4 '19 at 21:26
• @Unknown123 At t=0+ we no longer have a dc situation, we have a transient situation. The capacitor does not act like a short circuit in that instant. Because the capacitor voltage must be continuous, the capacitor acts like an ideal voltage source with a value equal to $v_C(0-)$. It is only a coincidence that in this case that voltage is zero...that does not mean that the capacitor acts like a short circuit. If you put that in your head you will struggle when $v_C(0-) \ne 0$ so you should use the correct concepts from the start. – Elliot Alderson Apr 4 '19 at 21:59
• I think it's perfectly reasonable to say that at t=0 the capacitor "looks/acts like" a short circuit in this case. This is specifically what the question aims at, the understanding that at t=0 the cap has 0V and max. current and at t=infiniy the cap has 0A. – JimmyB Apr 5 '19 at 10:19