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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.

enter image description here


schematic

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} \$

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    \$\begingroup\$ Can you transcribe your work and rotate the image to be the right way up? \$\endgroup\$ – Hearth Apr 4 at 14:59
  • \$\begingroup\$ The voltage on capacitor cannot change instantaneously. \$\endgroup\$ – Dirceu Rodrigues Jr Apr 4 at 15:08
  • \$\begingroup\$ 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. \$\endgroup\$ – Warren Hill Apr 4 at 15:13
  • \$\begingroup\$ 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? \$\endgroup\$ – Warren Hill Apr 4 at 15:16
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    \$\begingroup\$ @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. \$\endgroup\$ – JimmyB Apr 5 at 10:15
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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:

enter image description 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.

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  • \$\begingroup\$ So at t=0 are my calculations correct? I highly doubt that tho. I really hope u can understand the hand writing \$\endgroup\$ – Abdullah Apr 4 at 15:40
  • \$\begingroup\$ 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?). \$\endgroup\$ – JimmyB Apr 4 at 15:43
  • \$\begingroup\$ 1kohm is R1, 3.3kohm is R2 and 5kohm is R3 and thanks for your confirmation \$\endgroup\$ – Abdullah Apr 4 at 15:47
  • \$\begingroup\$ Also how am i suppose to find the current for each resistor and the capacitor at t long time or infinite \$\endgroup\$ – Abdullah Apr 4 at 15:50
  • \$\begingroup\$ "At t=infinity, the capacitor is fully charged." \$\endgroup\$ – JimmyB Apr 4 at 15:51
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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.

schematic

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.

schematic

simulate this circuit

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  • \$\begingroup\$ 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. \$\endgroup\$ – Elliot Alderson Apr 4 at 18:53
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    \$\begingroup\$ @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. \$\endgroup\$ – Elliot Alderson Apr 4 at 20:08
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    \$\begingroup\$ @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}}}\$ \$\endgroup\$ – Unknown123 Apr 4 at 21:26
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    \$\begingroup\$ @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. \$\endgroup\$ – Elliot Alderson Apr 4 at 21:59
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    \$\begingroup\$ 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. \$\endgroup\$ – JimmyB Apr 5 at 10:19

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