RC Circuit, finding time it takes for Voltage across capacitor to rise from 0 to 5V

For the circuit shown below what is the time it takes for Vc to rise from 0 V to 5 V in (second). Assume Vc(0+) = 0 V The answer is $t = 2\ln{2}$, but I am unsure why. What I have so far:

$$V(t) = [V(\infty) + V(0)]e^{-t/T} \qquad \text{where \;T = RC}$$ R = 2e3 and C = 1e-3 , so T = 2

$$V(t) = [V(\infty) + V(0)]e^{-t/2}$$

at t= infinite, the capacitor acts like a broken circuit so V(inf) = 0 but the problem also states that V(0) = 0 so that would set whole thing to 0

Am I doing something wrong, or am I even on the right track here?

• Just as a comment here in terms of the practical applications. You will notice that you obviously can't wait until infinity for the result to settle, so how are these circuits used? The short answer is , it depends upon the accuracy of the circuit. a general rule of thumb is that the signal has substantially settled to the asymptotic value after 6 time constants, in some case 3 time constants are used. – placeholder Oct 14 '12 at 18:02
• RC circuit, its time constant, exponential decay - hint hint ;) – Dzarda Jan 2 '15 at 12:29

No, at t = ∞, the fact that the capacitor "acts as an open circuit" means that the voltage across it is the same as the source: 10V.

To get this, apply Kirchoff's voltage law around the loop. Since the current in the loop at t = ∞ is zero, there can be no voltage across the resistor. Therefore, the source voltage must also appear across the capacitor in order to make the voltages around the loop sum to zero.

With this value plugged into your equation, you can now solve for the time at which V(t) equals 5V.

Additional: Your equation is slightly incorrect, too. The voltage across the capacitor is actually

V(t) = V(∞) - [V(∞) - V(0)]e-t/T

The first term represents the final static result, and the second term represents the transient result.

Also, keep in mind that ln(x) = -ln(1/x) ...

• so V(t) = 5,V(inf) = 10,V(0) = 0 and Time constant = 2; solved it, I got 2 ln (0.5) = t , but the answer is supposed to be 2 ln (2) – 40Plot Oct 14 '12 at 16:21
• @40Plot: See the additional information in my answer. – Dave Tweed Oct 14 '12 at 16:38

This looks like homework to me so I'll point you in the right direction and let you finish the job.

Yes there is a formula.

First you need to determine the time constant for your circuit which is:

Time constant (t) in Seconds = Capacitance in Farads (C) * Resistance in Ohms (R)

or

t = CR

Then comes the difficult bit. Charging and discharging a capacitor follows an exponential curve. I'll let you do some research to find what it is. You can find the formula by a simple Google search or just looking through some past answers on this stack (after all we do like to see folk making an effort to solve their problem.)

Another way of finding the answer is to use a graph (see below).

The X axis is in time constant and the vertical axis is in % of final voltage. So 100% = 10V (in your case). You can always use the graph to confirm your calculations. The important points to note on the graph are that after 1 time constant the voltage has reached about 63% of its final value. After 5 time constants it has reached almost 100% of the final value (it never actually gets there but its near enough).

Good luck at t= infinite, the capacitor acts like a broken circuit so V(inf) = 0

This assumption is wrong. An open circuit can support any voltage across it. A short circuit will always have 0 volts across it.

Consider the capacitor as an open circuit, that means there is no current through it. Since it's in series with the resistor, that means there's no current through the resistor. From Ohm's law, you can now find the voltage across the resistor is 0 V.

Then Kirchoff's voltage law tells you the asymptotic voltage across the capacitor must be the sum of the voltage across the source (10 V) and the voltage across the resistor (0 V). So the final voltage $V(\infty)$ across the capacitor is 10 V.

From there you should be able to work the rest of the problem.