Calculating Current in a Single-Loop Circuit with Resistor and Capacitor

In summary: That's correct. In summary, the conversation discusses a battery connected to a circuit with two resistors and a capacitor. After the battery is removed, the circuit contains only two resistors and a capacitor. The question is about the current going through one of the resistors. The conversation also includes a solution attempt using equations for an RL circuit, but there are several errors and suggestions to rethink the strategy.
  • #1
reising1
54
0

Homework Statement



There is a Battery connected to a single loop circuit containing two resistors, R1 and R2, and one capacitor L.

After a long time, the battery is removed, so there is a single loop circuit with just two resistors and a capacitor.

What is the current going through R1?

The Attempt at a Solution



This is what I thought:

So, the EMF is removed. Thus, using a loop rule, we have the formula:

0 = IR + L(di/dt) where R is R1+R2

Integrating, we have

0 = (1/2)(I^2)(R) + (L)(I)

Dividing everything by I, we have

0 = (1/2)(I)(R) + L

Thus, I = (2L) / R

However, this is incorrect.

Any ideas?

Thanks!
 
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  • #2
There are a few problems here. You call this an RL circuit, and you use the equations for an RL circuit, but you say it contains a capacitor instead of an inductor. I assume you are just using the wrong word.

The second problem is here:
reising1 said:
0 = IR + L(di/dt) where R is R1+R2

Integrating, we have

0 = (1/2)(I^2)(R) + (L)(I)
Are you sure you did that correctly? What variable are you integrating with respect to?
 
  • #3
Oh, wait. That is wrong.

With respect to I, you get:

R + LI = 0

So I = -R/L?
 
  • #4
Deriving with respect to I. Sorry.
 
  • #5
This is still not right because you differentiated one term and integrated the other.
 
  • #6
You might want to rethink the whole strategy of trying to take an integral or a derivative.
You have both I and dI/dt in this equation. What kind of equation does that make it?
 
  • #7
What do you mean?

d/di(RI) = R
d/di(L(di/dt)) = LI
 
  • #8
differential equation
 
  • #9
reising1 said:
What do you mean?

d/di(RI) = R
d/di(L(di/dt)) = LI

Taking a derivative of a derivative doesn't make the derivative do away.
 
  • #10
reising1 said:
differential equation

Yes!
 

What is a single-loop circuit?

A single-loop circuit is a circuit that has only one path for the electric current to flow. It typically consists of a power source, a resistor, and a capacitor connected in a loop.

How do I calculate the current in a single-loop circuit?

To calculate the current in a single-loop circuit, you can use Ohm's Law (I=V/R) to calculate the current through the resistor, and use the equation I=C*dV/dt to calculate the current through the capacitor. Then, add these two currents together to get the total current in the circuit.

What is the role of a resistor in a single-loop circuit?

A resistor is used to limit the flow of electric current in a circuit. It helps to control the amount of current that flows through the circuit and prevents damage to components.

How does a capacitor affect the current in a single-loop circuit?

A capacitor stores electric charge and can affect the current in a single-loop circuit by either allowing or blocking the flow of current, depending on the frequency and voltage of the current.

Can I use the same formula to calculate the current in any single-loop circuit?

Yes, the same formula can be used to calculate the current in any single-loop circuit, as long as the circuit consists of a power source, a resistor, and a capacitor connected in a loop.

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