Current Through A Resistor That Is In Parallel With A Capacitor

In summary, Kirchoff's voltage law can be used to find the voltage of a capacitor as a function of time, and current can be found as a function of time by dividing the voltage by a constant. However, the time constant (Tau) does not depend on the resistance of the light bulb in parallel with the capacitor, so using Thevenin's Equivalence would be the easiest way to find this value.
  • #1
solour
9
0

Homework Statement


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My question is regarding part C of the question.

Homework Equations


V = IR
V(t) = V(1-e^(-t/tau))

The Attempt at a Solution


My idea is to use Kirchoff's Voltage Law and find the voltage of the capacitor as a function of time, then since the voltage across capacitor is the same as voltage across resistor I can simply divide that by a constant R and obtain current as a function of time.

The problem I am running into is: I am unsure what to put as R in the time constant.
To my understanding time constant is the amount of time it takes to charge the capacitor to about 60%, and from my instinct it does not depend on the resistance of light bulb that is in parallel with the capacitor. Therefore Tau(time constant) = 50*Capacitance.

However, I am unsure of what I said above, and would like to know if there's a more definitive way to find the R value for time constant. I did see one approach which uses Thevenin's Equivalence but it was very confusing.
 
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  • #2
Thevenin is the way.

Show what you've tried and where it gets confusing.
 
  • #3
solour said:

Homework Statement


http://imgur.com/a/KnYI2
http://imgur.com/a/KnYI2
My question is regarding part C of the question.

Homework Equations


V = IR
V(t) = V(1-e^(-t/tau)

The Attempt at a Solution


My idea is to use Kirchoff's Voltage Law and find the voltage of the capacitor as a function of time, then since the voltage across capacitor is the same as voltage across resistor I can simply divide that by a constant R and obtain current as a function of time.

The problem I am running into is: I am unsure what to put as R in the time constant.
To my understanding time constant is the amount of time it takes to charge the capacitor to about 60%, and from my instinct it does not depend on the resistance of light bulb that is in parallel with the capacitor. Therefore Tau(time constant) = 50*Capacitance.
gneill said:
Thevenin is the way.

Show what you've tried and where it gets confusing.
I am taking first year physics and Thevenin is not in the curriculum so I'm not sure how to calculate it. I will learn how it works then!
 
  • #4
solour said:
I am taking first year physics and Thevenin is not in the curriculum so I'm not sure how to calculate it. I will learn how it works then!
Okay. After you've done some research come back with any questions.
 
  • #5
solour said:
I am taking first year physics and Thevenin is not in the curriculum so I'm not sure how to calculate it. I will learn how it works then!
Thevenin would be the easiest and most practical way of doing this problem. But if it is not in your curriculum, I am not sure if you'll get full credit for this question if it is a part of your assignment or exam. After solving this problem using Thevenin, you might want to try the usual mesh analysis KVL method.

But the Thevenin method will surely be very useful for you to analyse circuits with increased complexity.
 

FAQ: Current Through A Resistor That Is In Parallel With A Capacitor

What is the equation for calculating the current through a resistor in parallel with a capacitor?

The equation for calculating the current through a resistor in parallel with a capacitor is I = V/R, where I is the current, V is the voltage, and R is the resistance.

How does the current change when a capacitor is added in parallel with a resistor?

When a capacitor is added in parallel with a resistor, the total current will increase. This is because the capacitor allows more current to flow through the circuit, effectively decreasing the overall resistance.

What is the relationship between current and voltage in a parallel circuit with a resistor and a capacitor?

In a parallel circuit with a resistor and a capacitor, the current is divided between the two components, but the voltage stays the same across both. This means that the current through the resistor and the current through the capacitor are not equal, but the voltage drop across each component is equal.

How does the current through a resistor in parallel with a capacitor change over time?

The current through a resistor in parallel with a capacitor will initially be high, but it will decrease over time as the capacitor charges. Eventually, the current will reach a steady state where it is equal to the current through the capacitor.

What happens to the current through a resistor in parallel with a capacitor when the frequency of the applied voltage changes?

The current through a resistor in parallel with a capacitor will change depending on the frequency of the applied voltage. At lower frequencies, the current will be higher because the capacitor has more time to charge and discharge. At higher frequencies, the current will be lower because the capacitor has less time to charge and discharge.

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