# 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? Apr 4, 2019 at 14:59
• The voltage on capacitor cannot change instantaneously. Apr 4, 2019 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. Apr 4, 2019 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? Apr 4, 2019 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. Apr 5, 2019 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 Apr 4, 2019 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?). Apr 4, 2019 at 15:43
• 1kohm is R1, 3.3kohm is R2 and 5kohm is R3 and thanks for your confirmation Apr 4, 2019 at 15:47
• Also how am i suppose to find the current for each resistor and the capacitor at t long time or infinite Apr 4, 2019 at 15:50
• "At t=infinity, the capacitor is fully charged." Apr 4, 2019 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. Apr 4, 2019 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. Apr 4, 2019 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 $\mathrm{t}=0 \implies \displaystyle \frac{d\mathrm{V_c(0)}}{dt} = \frac{\mathrm{V_{th}}}{\mathrm{R_{th}C}} \implies \mathrm{I_c(0)} = \frac{\mathrm{V_{th}}}{\mathrm{R_{th}}}$ Apr 4, 2019 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. Apr 4, 2019 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. Apr 5, 2019 at 10:19