What is the current through Resistor on RC circuit

In summary, the conversation discusses a diagram of an RC-circuit with given values for C, R1, R2, and V0, and poses two questions regarding the current through R1 and R2 after the switch S is closed. The solution for part A is correctly calculated using the equation I=E/R. For part B, the initial solution is incorrect due to a missing 0 after the decimal, but is later corrected. The conversation ends with a humorous exchange between the participants.
  • #1
MeMoses
129
0

Homework Statement



The diagram below depicts an RC-circuit where C = 1.70 μF, R1 = 18.0 Ω, R2 = 10.0 Ω, and V0 = 2.0 V.
a.) What is the current through R1 immediately after the switch S is closed?
b.) What is the current through R2 after the switch S has been closed for a very long time? Assume that the battery does not go dead.

Homework Equations


I=E/R


The Attempt at a Solution


I got part A. I divided 2V by 18Ω to get 0.111A which is correct. I know when the switch is initially closed current flows through the capacitor and when it is charge current no longer flows through. I figured part B you would just find the current through 2/(18+10)=0.0714 which turned out incorrect so I'm not sure what to try next. Also I don't think the capacitance is relevant, so I hope I'm not wrong there. Any help would be great
 

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  • #2
Your method and current value looks correct. Do they want any particular units, notation, or significant figures in the answer format?
 
  • #3
It looks like I forgot to enter the 0 after the decimal. Well, that's embarassing
 
  • #4
MeMoses said:
It looks like I forgot to enter the 0 after the decimal. Well, that's embarassing

Been there, done that :smile:
 
  • #5
.

I would like to first clarify that the current through a resistor in an RC circuit is not a constant value. It changes over time as the capacitor charges or discharges.

For part A, you correctly calculated the initial current through R1 when the switch is closed. This is because at the moment the switch is closed, the capacitor acts as a short circuit and all the current flows through R1.

For part B, after a very long time, the capacitor will be fully charged and no more current will flow through it. This means that the current through R1 and R2 will be the same. Using Ohm's Law (I = V/R), we can calculate the current through R2 as 2V/(18Ω+10Ω) = 0.0769A. This is slightly different from your attempt, which may have been a calculation error.

It is also worth mentioning that the capacitance is indeed relevant in this circuit. It affects the rate at which the capacitor charges and discharges, which in turn affects the current through the resistors.

I hope this helps clarify the concepts and calculations involved in an RC circuit.
 

1. What is a resistor in an RC circuit?

A resistor is an electronic component that restricts the flow of electric current in a circuit. In an RC circuit, it is used to control the current flowing through the circuit.

2. How does the current flow through a resistor in an RC circuit?

The current flows through a resistor in an RC circuit according to Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). This means that the higher the resistance, the lower the current flow will be.

3. What is the relationship between the current and the resistor in an RC circuit?

The current and the resistor have a direct relationship in an RC circuit. This means that as the resistance increases, the current decreases, and vice versa. This relationship is described by Ohm's Law, as mentioned in the previous question.

4. How can the current through a resistor in an RC circuit be calculated?

The current through a resistor in an RC circuit can be calculated using Ohm's Law. You will need to know the voltage across the resistor and its resistance. Then, simply divide the voltage by the resistance to get the current.

5. Can the current through a resistor in an RC circuit change over time?

Yes, the current through a resistor in an RC circuit can change over time. This is because the resistor and capacitor in the circuit have different effects on the current. At the start, the current will be high, but it will decrease over time as the capacitor charges up and the resistance increases.

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