The question

The time constant in the solution is 0.5. (-2t implies tau = 0.5) I am wondering how they got this.

tau = RC

So I am wondering how to calculate the Rth. The capacitor is inbetween the two resistors in series, so would the Rth be

Rth = 6/(6+2)

But then tau = (6/8)*(1/3) = 0.25

if Rth is not calculated like this and is just = (6+2)

tau = 8*(1/3) which is not equal to 0.5

if Rth is also not calculated like this and is 2/(6+2)

tau = (2/8)*(1/3) which is not equal to 0.5

I know that I can get the Rth from the solution ie. since the solution says tau = 0.5, then the Rth must be 1.5ohms (since 1.5 * (1/3) = 0.5)

But I cannot see how the resistance at the terminals of the resistor is equivalent to 1.5ohms.

So my question is how to calculate Rth and how to calculate tau.

  • 1
    \$\begingroup\$ Do you know the Thevenin's theory? What is the resistance seen from capacitor terminal ? \$\endgroup\$ – G36 Nov 29 '16 at 17:03

2 ohm and 6 ohm are in parallel. So the net resistance as seen by the capacitor(or thevenin's resistance) is 12/8 Ohm

  • \$\begingroup\$ Thanks yes this is where I went wrong, the arrangement of the resistors didn't look like they were in parallel but I can see why they are now. \$\endgroup\$ – lgdl.y Nov 29 '16 at 17:14

One way to get Rth is to find V open circuit and I short circuit at the capacitor terminals. Voc/Isc = Rth. After you have done this a few times, Rth will be obvious to you.


Combine the two sources (and their respective resistors) into one source. Easy way is to convert to two current sources: -

  1. 10 volt and 2 ohms converts to 5 amp in parallel with 2 ohms
  2. 5 volt and 6 ohms converts to 0.83333 amps in parallel with 6 ohms

Next combine the two current sources into one at 5.83333 amps with a single resistor of 1.5 ohms in parallel. If you need to convert back to a single voltage source this becomes 8.75 volts sourced from 1.5 ohms.

Hence the time constant is 1.5 x 1/3 = 0.5.


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