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

In summary, the conversation is about analyzing a circuit with capacitors and resistors. The participants discuss the calculations for current and voltage in different scenarios, including when the switch is open and closed. They also discuss the charges on the capacitors in both cases. The summary concludes that in the new steady state with the switch closed, the voltages across the capacitors would be the same as the voltages across their corresponding resistors, and the charges on the capacitors would be different than in the open switch scenario.
  • #1
subzero0137
91
4
1.
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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.
 
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  • #2
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.
 
  • #3
subzero0137 said:
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?
 
  • #4
cnh1995 said:
I don't see where scott said that.
Sorry I misread it, I deleted the post.
 
  • #5
ehild said:
Sorry I misread it, I deleted the post.
I'll delete mine.
 
  • #6
scottdave said:
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.
 
  • #7
cnh1995 said:
What is the current through the capacitors in the new steady state?

Wouldn't it be zero again?
 
  • #8
subzero0137 said:
Wouldn't it be zero again?
Yes.
 
  • #9
cnh1995 said:
Yes.

So the answer is the same as before? Or am I missing something again?
 
  • #10
subzero0137 said:
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?
 
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  • #11
ehild said:
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?
 
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  • #12
subzero0137 said:
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.
 

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

1. What is an RC circuit?

An RC circuit is an electrical circuit that contains a resistor (R) and a capacitor (C) connected in series. It is used to control the flow of electric current and store electrical energy.

2. How do I solve an RC circuit problem using the equations V=IR and Q=CV?

To solve an RC circuit problem, you can use the equations V=IR and Q=CV. First, calculate the total resistance (R) of the circuit by adding the individual resistances of each component. Then, use the equation V=IR to find the voltage (V) across the circuit. Next, use the equation Q=CV to find the charge (Q) stored in the capacitor. Finally, you can use these values to solve for any other unknown variables.

3. What is the relationship between voltage (V) and current (I) in an RC circuit?

In an RC circuit, the voltage and current are directly proportional to each other. This means that as the voltage increases, the current also increases. Similarly, if the voltage decreases, the current will also decrease.

4. How does the capacitance (C) affect an RC circuit?

The capacitance (C) in an RC circuit determines how much electrical energy can be stored in the capacitor. A higher capacitance means that the capacitor can store more charge, which can affect the overall voltage and current in the circuit.

5. Can the equations V=IR and Q=CV be applied to any RC circuit problem?

Yes, the equations V=IR and Q=CV can be applied to any RC circuit problem as long as the circuit is made up of only a resistor and a capacitor connected in series. These equations are fundamental laws of electricity and can be used to solve any circuit that follows Ohm's Law and the equation for capacitance.

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