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Problem: An RC circuit has a time constant of 68 micro-seconds. The capacitor is charged to 14 V. How much is the voltage across a resistor after 136 micro-seconds of discharge?

Attempt: What I know is that the time the capacitor fully discharge is at 5 Time constant. Then the formulas involved are: Time Constant = R*C and Voltage at the capacitor (Vc) = Vm (e^-t/time constant) but no resistor value were given so:

Time Constant = R*C ; R = (68 x 10^-6) / 14 = 4.86 micro-Ohms. I assumed that Vc = Vm at this time constant. And now I don't know what to do next. I think we have to get the Vc after 136 micro seconds then subtract it from the total voltage.

Question: How to answer this kind of problems? By the way, the correct answer is 1.96 V from the key to correction. Any help would be appreciated.

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  • \$\begingroup\$ You don't need to know the individual R and C values - you are given \$\tau\$ (not 'micro ohms', btw) \$\endgroup\$ – Chu May 25 '18 at 10:19
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What I know is that the time the capacitor fully discharge is at 5 Time constant.

Unfortunately, that's only roughly correct. A simplified model for what's happening, when you don't need precision. Rather like saying \$\pi\$ is \$\frac{22}{7}\$.

The voltage across the resistor (which is the same as the voltage across the capacitor) drops by a factor of \$e\$ every time constant.

You have been asked what the voltage is after two time constants, so the answer is it's fallen by two factors of \$e\$, or a factor of \$e^2\$. You do the rest of the sums.

So what's that about 5 time constants? The voltage has fallen by a factor of \$e^5\$, which is roughly 148, so there's less than 1% of the original voltage across it after that time. Engineers often use 1% as an approximation to zero, when doing back-of-the-envelope analyses.

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So it will be V=14*e(-t/time constant) , you have time constant and the t, solved

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  • \$\begingroup\$ 14*(e^-[(136 x 10^-6)/(68 x 10^-6)]) = 1.89 V. But isn't that the voltage of the capacitor at that time constant and not the resistor? \$\endgroup\$ – Jayce May 25 '18 at 5:15
  • \$\begingroup\$ The R and C are in parallel. \$ V_R = V_C \$. \$\endgroup\$ – Transistor May 25 '18 at 5:55

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