To find the voltage across the capacitor (Vc) at time t = 200 ms, we first need to understand the behavior of the RC circuit when the switch is moved from position A to position B.
When the switch has been in position A for a long time, the capacitor is fully charged to the voltage of the source, which is given as Vb = 3.35 V. At t = 0 ms, when the switch is moved to position B, the capacitor will start to discharge through the resistor.
The formula to find the voltage across the capacitor during discharge is:
Vc(t) = Vb * e^(-t/(R*C))
Where:
- Vc(t) is the voltage across the capacitor at time t.
- Vb is the initial voltage across the capacitor, 3.35 V.
- R is the resistance, 495.5 ohms.
- C is the capacitance, 4.00 µF (which is 4.00 x 10^-6 F).
- t is the time in seconds.
Substituting the values:
- R = 495.5 ohms
- C = 4.00 x 10^-6 F
- t = 200 ms = 0.200 seconds
First, calculate the time constant (τ):
τ = R * C = 495.5 * (4.00 x 10^-6) = 0.001982 seconds (or 1.982 ms).
Now, substituting values into the formula:
Vc(0.200) = 3.35 * e^(-0.200 / 0.001982)
Calculating the exponent:
-0.200 / 0.001982 = -100.908
Now, using a calculator to find e^(-100.908), we get a very small number, approximately 1.7 x 10^-44.
So we can now calculate Vc:
Vc(0.200) ≈ 3.35 * 1.7 x 10^-44 ≈ approximately 0 V.
Therefore, at time t = 200 ms, the voltage across the capacitor (Vc) is effectively 0 V because it has discharged nearly completely.