Understanding RC Circuit Currents: A Homework Question

In summary: I got I2=.19765A, but I'm not sure if that's correct or not. I'll need to do some more research on the subject. Thanks for your help!
  • #1
smithisize
13
0

Homework Statement



I'm really looking for help with A.) because I think that if I could figure out A.) the rest should be ok. So, here are all parts of the question, but I'm really only concerned with A at the moment!

The diagram below depicts an RC-circuit where C = 5.40 F, R0 = 14.5Ω , R1 = 24.5 Ω, R2 = 27.5 Ω, and V = 7.50 Volts. The capacitor is initially uncharged.

6ozouc.gif


This is how I drew my currents etc:

35m2iw5.gif


The circles are for Kirchoff Loops I tried to construct, and they're going clockwise. Along with a loop around the outside of the circuit.

A.) What is the current through R1 immediately after the switch S is closed?
B.) What is the current through R2 immediately after the switch S is closed?
C.) What is the current through the capacitor immediately after the switch S is closed?
D.) What is the current through R1 after the switch S has been closed for a very long time? Assume that the battery does not go dead.
E.)
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



V=IR
C=(Q/V)


The Attempt at a Solution


According to Kirchoff's Laws: I1= I2 + I3
So, since V=IR and R1 and R2 are || : I1 = (Vparallel/R1)+(Vparallel/R2) --> I1= V*(1/R1 + 1/R2)
Therefore: I1/(1/R2 + 1/R1) = Vparallel. Where I1 = V/R0.
Then, I2 = Vparallel/R1

But, that's not correct unfortunately. Where am I going wrong? Thanks!

Side note, here are the eqns I got from kirchhoffs laws (pretty sure I messed something up):

V - I1R0 - I2R1 = 0 Left loop
-I4R4 + I2R1 = 0 Right loop at t=0 (or basically zero) Also, I4=I3.
V-I1R0-I4R2 = 0 Entire loop again at t=0.
 
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  • #2
smithisize said:

Homework Statement



I'm really looking for help with A.) because I think that if I could figure out A.) the rest should be ok. So, here are all parts of the question, but I'm really only concerned with A at the moment!

The diagram below depicts an RC-circuit where C = 5.40 F, R0 = 14.5Ω , R1 = 24.5 Ω, R2 = 27.5 Ω, and V = 7.50 Volts. The capacitor is initially uncharged.

6ozouc.gif


This is how I drew my currents etc:

35m2iw5.gif


The circles are for Kirchoff Loops I tried to construct, and they're going clockwise. Along with a loop around the outside of the circuit.

A.) What is the current through R1 immediately after the switch S is closed?
B.) What is the current through R2 immediately after the switch S is closed?
C.) What is the current through the capacitor immediately after the switch S is closed?
D.) What is the current through R1 after the switch S has been closed for a very long time? Assume that the battery does not go dead.
E.)
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



V=IR
C=(Q/V)


The Attempt at a Solution


According to Kirchoff's Laws: I1= I2 + I3
So, since V=IR and R1 and R2 are || : I1 = (Vparallel/R1)+(Vparallel/R2) --> I1= V*(1/R1 + 1/R2)
Therefore: I1/(1/R2 + 1/R1) = Vparallel. Where I1 = V/R0.
Then, I2 = Vparallel/R1

But, that's not correct unfortunately. Where am I going wrong? Thanks!
Hi smithisize; Welcome to Physics Forums.

I1 ≠ V/R0. The potential across resistor R0 is not V; Potential V is dropped across the sum of R0 and R1||R2, so all of V cannot be across just Ro.

You could use your KVL system of equations to determine the individual currents.

Or calculate the total resistance first and use it to find the current I1. With I1 in hand you can then use your other relationships to proceed.
 
  • #3
gneill said:
Hi smithisize; Welcome to Physics Forums.

I1 ≠ V/R0. The potential across resistor R0 is not V; Potential V is dropped across the sum of R0 and R1||R2, so all of V cannot be across just Ro.

You could use your KVL system of equations to determine the individual currents.

Or calculate the total resistance first and use it to find the current I1. With I1 in hand you can then use your other relationships to proceed.
Thanks, I appreciate it.

So, I think I should probably become more familiar with using the KVL sys of eqns. Not my strong point.

Given the eqns I posted above, I substituted I2+I3 for I1 then solved for I3 (using the 1st eqn) and got I3 = (V - I2R0)/(R0+R2)
Then I plugged that value into my third equation and solved for I2 and got I2 = (R2*V)/(R0+R1*R2+R1*R0)
Evaluated: .19765 A

Does that look correct??

EDIT: Well, I was able to solve for I1, and in turn solve for I2. I2 ended up being .144 A. So, which of my eqns is wrong? Where is my KVL going awry?
 
Last edited:
  • #4
smithisize said:
Thanks, I appreciate it.

So, I think I should probably become more familiar with using the KVL sys of eqns. Not my strong point.

Given the eqns I posted above, I substituted I2+I3 for I1 then solved for I3 (using the 1st eqn) and got I3 = (V - I2R0)/(R0+R2)
Then I plugged that value into my third equation and solved for I2 and got I2 = (R2*V)/(R0+R1*R2+R1*R0)
Evaluated: .19765 A

Does that look correct??
No, something's gone wrong in your manipulations. Note that the denominator of your last expression has a resistance term added to resistance2 terms --- can't add mixed units!
EDIT: Well, I was able to solve for I1, and in turn solve for I2. I2 ended up being .144 A. So, which of my eqns is wrong? Where is my KVL going awry?
0.144A is a good result.
 
  • #5
While you can use KVL & KCL (and it might be good practice to do so) I probably would have treated this as a variant of a potential divider circuit to work out the voltage on R1 and from that the currents

For example, initially the capacitor is discharged so..

VR1 = V * (R1//R2) / {(R1//R2) + R0}

IR1 = VR1 / R1
 
  • #6
Well thanks to both of you for your replies, really helped me out!
 

What is a Resistor Capacitor Circuit?

A resistor capacitor circuit is an electrical circuit that contains both a resistor and a capacitor. These two components work together to regulate the flow of electricity within the circuit.

What is the purpose of a Resistor Capacitor Circuit?

The purpose of a resistor capacitor circuit is to filter or smooth out the flow of electricity within a circuit. This can be useful in applications where a constant and stable current is needed, such as in power supplies or audio equipment.

How does a Resistor Capacitor Circuit work?

A resistor capacitor circuit works by using the properties of both a resistor and a capacitor. The resistor limits the flow of electricity, while the capacitor stores and releases electrical charge. Together, they create a circuit that can regulate the flow of electricity.

What are the key components of a Resistor Capacitor Circuit?

The key components of a resistor capacitor circuit are a resistor, a capacitor, and a power source. The resistor and capacitor can be connected in series or parallel, and the power source provides the electrical energy for the circuit to function.

What are some common applications of Resistor Capacitor Circuits?

Resistor capacitor circuits have many applications, including in power supplies, audio equipment, and electronic filters. They are also commonly used in timing circuits and as part of electronic oscillators.

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