Voltage drop across capacitor in RC circuit

[Avalanche]

Homework Statement

Please look at the attachment for the circuit.

The question is this:

The switch has been closed for a long time. It is then opened. How long does it take for the potential across the 4μF capacitor to fall to half of its initial value?

Homework Equations

V = V₀e^(-t/RC)

The Attempt at a Solution

Once the switch is opened, the capacitor would begin to discharge creating a potential difference. The equation I tried to solve is

0.5 = e^(-t/(4.0*R))

The problem I'm having is that I don't know what R, the equivalent resistance is for this circuit.

The answer is that it takes 10.2 μs for the potential to drop to half of its initial value. Working backwards, I get the equivalent resistance is 3.68 ∏. How do I get that?

 

[gneill]

What components comprise the circuit of interest when the switch has opened?

 

[Avalanche]

When the circuit is opened, the left loop should still has current because there is a battery in that loop.

 

[gneill]

When the circuit is opened, the left loop should still has current because there is a battery in that loop.

Is the current in that loop of interest? What part of the overall circuit is the question asking about?

 

[Avalanche]

No ... it's the right loop. When the capacitor discharges, it creates a current that flows around the right loop. I3 becomes zero. Is that correct?

 

[gneill]

No ... it's the right loop. When the capacitor discharges, it creates a current that flows around the right loop. I3 becomes zero. Is that correct?

Yes, that is correct. Look at the components in that loop. How might you determine the time constant for it?

 

[Avalanche]

The time constant = RC.

The only resistor is the 5 ohm resistor. Is that R?

And for C, do I use the equivalent capacitance or just the 4 microfarad?

 

[gneill]

The time constant = RC.

The only resistor is the 5 ohm resistor. Is that R?

Yes. (being the only resistor, it must be!)

And for C, do I use the equivalent capacitance or just the 4 microfarad?

The loop comprises a single circuit, so all the components participate... Find the total equivalent capacitance.

 

[Avalanche]

C = (1/4+1/11)^-1 = 2.9333

Time constant = RC = 5*2.93333 = 14.6667 microseconds

0.5 = e^(-t/14.6667)
t = 10.2 microseconds

Thanks for all the help!

btw ... does it also take 10.2 microseconds for the potential across the 6 microfarad resistor to drop to half of its initial value?

 

[gneill]

C = (1/4+1/11)^-1 = 2.9333

Time constant = RC = 5*2.93333 = 14.6667 microseconds

0.5 = e^(-t/14.6667)
t = 10.2 microseconds

Thanks for all the help!

You're welcome!

btw ... does it also take 10.2 microseconds for the potential across the 6 microfarad resistor to drop to half of its initial value?

Yup.