What is the voltage drop across the resistor in a discharging RC circuit?

[brenfox]

Homework Statement

Determine the voltage drop across the resistor when the capacitor has been discharging for 10μs.
Capacitance=400 pico farad and is charged to 10v through a 50kΩ resistor.

Homework Equations

Vr= V*e -t/CR

The Attempt at a Solution

Vr = V-Vc at the instant of 10μs
Vc=Ve-t/CR
I get 3.935 volts? (without showing all my workings out)

 

[gneill]

Homework Statement

Determine the voltage drop across the resistor when the capacitor has been discharging for 10μs.
Capacitance=400 pico farad and is charged to 10v through a 50kΩ resistor.

Homework Equations

Vr= V*e -t/CR

The Attempt at a Solution

Vr = V-Vc at the instant of 10μs
Vc=Ve-t/CR
I get 3.935 volts? (without showing all my workings out)

It's not clear from what you've presented whether the capacitor is charging up from 0V to 10V, or discharging from 10V down to 0V. Does the capacitor start out at 0V or 10V?

 

[brenfox]

Fair point.The capacitor is fully charged to 10volts. So i believe the capacitor starts out at 10 volts?

 

[gneill]

Fair point.The capacitor is fully charged to 10volts. So i believe the capacitor starts out at 10 volts?

So the circuit looks something like this?

The capacitor is initially charged to 10V and at time t = 0 the switch closes allowing current to start to flow, discharging the capacitor?

 

[brenfox]

The capacitor is fully charged to 10volts.Calculate the voltage drop when the capacitor has been discharging for 10 micro seconds? I thought the switch must be open to allow the capacitor to discharge?

 

[gneill]

The capacitor is fully charged to 10volts.Calculate the voltage drop when the capacitor has been discharging for 10 micro seconds?

Okay then according to that statement the diagram fits the scenario. You will want to revisit your Attempt at a Solution now. Can you show more of your working?

 

[brenfox]

Ok. here is my workings out.

Vr= V-Vc at the instant of 10μs.
Vc=V*e-t/CR
-t/CR = -0.5
Vc=10*-0.5
Vc=6.065volts.
Vr = V-Vc
Vr= 10-6.065
Vr = 3.935 volts?
Hope this helps!

 

[gneill]

Ok. here is my workings out.

Vr= V-Vc at the instant of 10μs.

Can you justify the above equation? Why would Vr = V - Vc?

Vc=V*e-t/CR
-t/CR = -0.5
Vc=10*-0.5
Vc=6.065volts.

That's a good value for Vc.

 

[brenfox]

This is how i see this: Voltage is 10v from the battery. The resistor is in series with the capacitor. The capacitor drops 6.065 volts so that must mean the resistor drops 3.935 volts which equates to a total of 10 volts dropped from the battery. But i am confused now! . This would be the case in a steady state circuit but not after 10 micro seconds.

 

[gneill]

Are you given a circuit diagram for the problem?

If there is a battery in the circuit and all three components form a series circuit, then it likely that the capacitor is meant to start out uncharged (0 V), and charges up over time.

 

[brenfox]

No diagram. The capacitor is set to the maximum 400 pico farad and is charged to 10 volt through a 50k resistor. Determine the voltage drop across the resistor when the capacitor has been discharging for 10 micro seconds. This is the info i have recieved. Hope this helps!

 

[gneill]

No diagram. The capacitor is set to the maximum 400 pico farad and is charged to 10 volt through a 50k resistor. Determine the voltage drop across the resistor when the capacitor has been discharging for 10 micro seconds. This is the info i have recieved. Hope this helps!

Okay then. So you can justify your equation for Vr by doing KVL around the loop, and your calculations for Vr are correct.

 

[brenfox]

i thought Vr= V- Vc. At this instant of 10 micro seconds the voltage is 6.065 volts so i see the voltage must be 10 - Vc?? This would be the voltage across the resistor?

 

[gneill]

i thought Vr= V- Vc. At this instant of 10 micro seconds the voltage is 6.065 volts so i see the voltage must be 10 - Vc?? This would be the voltage across the resistor?

Right. That's KVL for the loop for that instant in time.

 

[brenfox]

So the voltage drop is 6.065volts because the voltage across the resistor is 3.935 volts??

 

[gneill]

So the voltage drop is 6.065volts because the voltage across the resistor is 3.935 volts??

Just write KVL for the loop using V for the battery voltage, Vr for the potential drop across the resistor and Vc for the potential drop across the capacitor. KVL applies always, at all times. You'll then have the equation that relates all the potential drops around the circuit.