Another RC circuit problem, don't understand why my answer is wrong

In summary, the current through resistor 2 at t = 4.00 ms can be calculated using the potential difference across the capacitor, which is equal to the potential difference across R2. The correct equation for discharging a capacitor is V = V0e^(-t/TAU). Plugging in the given values, the current through R2 is 0.00044535866 A.
  • #1
iiiiaann
22
0

Homework Statement


In the figure below, R1 = 13.0 kΩ, R2 = 18.0 kΩ, C = 0.600 µF, and the ideal battery has emf = 20.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 4.00 ms?

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Homework Equations



V = IR

[itex]V_{Capacitor}(t) = V_{0}(1-e^{-t/\tau})[/itex]

The Attempt at a Solution



Okay so i put a decent amount into this one but the answer I'm coming up with is wrong and I can't figure out why. Here's what I've done:

When the capacitor is fully charged, the current through the battery is [itex]\frac{\xi}{R_{1} + R_{2}}[/itex] which is [itex]\frac{20}{13 * 10^{3} + 18 * 10^{3}}[/itex] = 6.45e-4

The current through R2 would be the same as this, 6.45e-4, so the potential difference over R2 = IR = 6.45e-4 * 18e3 = 11.61 V

When the switch is first opened, the potential difference across the capacitor would be the same as this, 11.61.

The time constant for the capacitor would be RC = 18e3 * .6e-6 = 0.0108

so to calculate the potential difference at t = 4 ms, you would use the following equation:
[itex]V_{capacitor}(0.004) = 11.61 * (1 - e^{-0.004 / 0.0108})[/itex] which comes out to 3.59

This is the same as the potential difference in R2, and so current through R2 should be 3.59 / 18e3 = 1.99e-4, but that is not right. Can someone let me know where I went wrong?
 
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  • #2
The potential across the capacitor will be falling, not rising. Check your equation for the potential on the capacitor!
 
  • #3
This equation

[itex]\displaystyle V_\text{Capacitor}(t) = V_{0}(1-e^{-t/\tau})[/itex]

is for the case of charging a capacitor which initially has a charge (and thus potential) of zero. How can I tell ? Look at V when t = 0 .

What's the correct equation for discharging a capacitor which initially has a potential of V0 across its plates?
 
  • #4
is it just the same equation except without the "1 -" part? so V0(e^-t/TAU)?
 
  • #5
got it, 0.000445358666. Thanks for putting me on the right track guys
 

FAQ: Another RC circuit problem, don't understand why my answer is wrong

Why is my calculated voltage different from the given voltage?

There are a few possible reasons for this discrepancy. One possibility is that there was an error in your calculation. Double check your math and make sure you used the correct values for the resistance and capacitance. Another possibility is that there was an error in the given voltage value. Check to see if the voltage was measured accurately. Additionally, the circuit may have other components that are affecting the voltage, such as inductors or resistors in series with the capacitor.

How do I know which direction the current is flowing in the circuit?

The direction of current flow in a circuit can be determined by using the passive sign convention. This convention states that current flows from positive to negative terminals in a voltage source, and from positive to negative terminals in a resistor. In a capacitor, current flows from positive to negative plates. By following the direction of current flow through each component in the circuit, you can determine the overall direction of current flow.

Can I use a different value for the resistance or capacitance and still get the same answer?

Yes, the values of resistance and capacitance can be changed as long as the overall circuit remains the same. However, changing these values will result in different time constants and therefore affect the behavior of the circuit. It is important to use the correct values in order to accurately analyze the circuit.

How can I tell if the circuit is in a steady state?

A circuit is considered to be in a steady state when the voltage and current values have stabilized and are not changing over time. In an RC circuit, the steady state occurs when the capacitor is fully charged and there is no more current flowing through the circuit. You can also determine the steady state by looking at the voltage across the capacitor, which should be equal to the voltage of the voltage source.

Why is the voltage across the capacitor decreasing instead of increasing?

In an RC circuit, the voltage across the capacitor will initially increase as the capacitor charges. However, as the capacitor reaches its maximum charge, the voltage will start to decrease. This is because the capacitor is acting as a voltage source in the opposite direction, causing a decrease in the overall voltage. This behavior is known as capacitor discharge and is a key concept in understanding RC circuits.

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