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?
Answer
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) ...
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