Solve RC Circuit Problem 1: V=IR, Q=CV

[subzero0137]

1.

2. V=IR, Q=CV3.
To calculate the current in the circuit in the case when the switch is open and steady state is reached, I assumed that no current will flow "across" the capacitors and so the current I will simply be emf/(r+R1+R2) = 12 V/(1+3+2) = 2 A. The voltage drop across r would be V=IR=2 A * 1 ohms = 2 V, therefore the voltage across the capacitors would be 10 V. Since C1 and C2 are in series, Q1=Q2=CV where C is the combined capacitance equal to (4/3)F and V=10V. Therefore Q1=Q2=(40/3) C. Is this correct so far?

I'm not sure how to analyze the circuit when the switch closes.

 

[scottdave]

How much current goes through C1 in steady state? What about C2? Now with the switch closed, C1 and R1 are parallel. What can you say about the voltage on components in parallel? Take a look at C2 and R2, as well.
Edit: and yes, you were on the right track for your analysis of the open switch.

 

[cnh1995]

I'm not sure how to analyze the circuit when the switch closes.

What is the current through the capacitors in the new steady state?

 

[ehild]

I don't see where scott said that.

Sorry I misread it, I deleted the post.

 

[cnh1995]

Sorry I misread it, I deleted the post.

I'll delete mine.

 

[subzero0137]

How much current goes through C1 in steady state? What about C2? Now with the switch closed, C1 and R1 are parallel. What can you say about the voltage on components in parallel? Take a look at C2 and R2, as well.
Edit: and yes, you were on the right track for your analysis of the open switch.

Thanks for the reply. I'm not sure how much current goes through the capacitors in steady state. I thought no current can flow through the capacitors in steady state? I can see how C1 and R1 are parallel so the voltage across both will be the same. But again I thought currents can't flow through capacitors in steady state.

 

[subzero0137]

What is the current through the capacitors in the new steady state?

Wouldn't it be zero again?

 

[cnh1995]

Wouldn't it be zero again?

Yes.

 

[subzero0137]

Yes.

So the answer is the same as before? Or am I missing something again?

 

[ehild]

So the answer is the same as before? Or am I missing something again?

The same currents, the same charges?
The currents are the same, zero at steady state, but the charges of the capacitors are different in both cases. When the switch was open, the capacitors were in series, having the same charge. What are the voltages across the capacitors when the switch is closed? What are the charges?

 

[subzero0137]

The same currents, the same charges?
The currents are the same, zero at steady state, but the charges of the capacitors are different in the two cases. When the switch was open, the capacitors were in series, having the same charge. What are the voltages across the capacitors when the switch is closed? What are the charges?

The voltage across C1 would be the same as the voltage across R1, and the voltage across C2 would be the same as the voltage across R2. So Q1 = 2F*6V = 12C and Q2=4F*4V=16C?

 

[ehild]

The voltage across C1 would be the same as the voltage across R1, and the voltage across C2 would be the same as the voltage across R2. So Q1 = 2F*6V = 12C and Q2=4F*4V=16C?

Yes.