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For a general RC circuit that looks like the following:

enter image description here

(Sorry for the rough sketch of the circuit, I couldn't find a picture online)

What would the time constant or RC equations be?

My thinking: The time constant would be R2C.

I drew a large loop around the circuit (contains battery, R2 and C): So from KVL I have : V - R2 I2 - Q(t)/C = 0 and solving the differential equation, I get that Q(t) = VC(1-e(-t/R2C)).

I'm not sure I'm right so please correct me if I'm wrong.

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4 Answers 4

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Correct. The parallel resistor R1 has no effect if the components are ideal. If you are using a more realistic model of a battery as an ideal voltage source with some finite internal resistance then it would come into play.

Edit: however if you were to remove the battery and observe the time constant as the capacitor discharged through R1 and R2 the time constant would be different.

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  • \$\begingroup\$ Thank you. I'd also like to confirm the max current in the current equation would thus be V/R_2 right? \$\endgroup\$ Commented Apr 5, 2022 at 22:15
  • \$\begingroup\$ Maximum current out of the battery when charging is V/R1 + V/R2 which decays to V/R1 with the time constant previously mentioned. \$\endgroup\$
    – vir
    Commented Apr 5, 2022 at 22:17
  • \$\begingroup\$ Yea. I was referring to the current in the branch that contains the capacitor though. Also how come the max current is V/R1+ V/R2? Should the max current occur when the capacitor has 0 charge? So I have 2 resistors in parallel which means my current is VR1R2/(R1+R2)? \$\endgroup\$ Commented Apr 5, 2022 at 22:20
  • \$\begingroup\$ Yes. The current through R1 is always V/R1. The current through R2 starts at V/R2 and drops to 0 as the capacitor charges up to V. V/R1 + V/R2 is equal to V[R1R2/(R1+R2)]. \$\endgroup\$
    – vir
    Commented Apr 5, 2022 at 22:29
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The time constant of a circuit is obtained by reducing the excitation or the stimulus to zero. Zeroing your voltage source is similar to replacing it by a short circuit in the network under study. Once done, temporarily disconnect the capacitor and "look" through its connecting terminals to determine the resistance R:

enter image description here

The pole, for a 1st-order circuit is the inverse of the time constant. For more information on time constants and poles, you can have a look at a seminar I taught in 2016 and available to download from my webpage.

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  • \$\begingroup\$ How can this R = R1 + R2 be true? \$\endgroup\$
    – G36
    Commented Apr 6, 2022 at 14:10
  • \$\begingroup\$ Oh, it is not and needs a clean up! : ) Thanks for pointing out this miserable typo. \$\endgroup\$ Commented Apr 6, 2022 at 14:32
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Yes, you're right.

For charging, it's R2 and C. For discharging, it would be R1+R2 and C.

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R1 is in parallel with an ideal voltage source. It has no effect, since ideal voltage sources have zero impedance. Paralleling a zero impedance with any other impedance doesn't change anything. The equivalent circuit is:

schematic

simulate this circuit – Schematic created using CircuitLab

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