Register to reply

RL circuit

Share this thread:
reising1
#1
Apr3-10, 04:22 PM
P: 55
1. The problem statement, all variables and given/known data

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?

3. 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!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
LeonhardEuler
#2
Apr3-10, 05:20 PM
PF Gold
LeonhardEuler's Avatar
P: 864
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:
Quote Quote by reising1
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?
reising1
#3
Apr3-10, 05:34 PM
P: 55
Oh, wait. That is wrong.

With respect to I, you get:

R + LI = 0

So I = -R/L?

reising1
#4
Apr3-10, 05:34 PM
P: 55
RL circuit

Deriving with respect to I. Sorry.
LeonhardEuler
#5
Apr3-10, 05:36 PM
PF Gold
LeonhardEuler's Avatar
P: 864
This is still not right because you differentiated one term and integrated the other.
LeonhardEuler
#6
Apr3-10, 05:38 PM
PF Gold
LeonhardEuler's Avatar
P: 864
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?
reising1
#7
Apr3-10, 05:38 PM
P: 55
What do you mean?

d/di(RI) = R
d/di(L(di/dt)) = LI
reising1
#8
Apr3-10, 05:38 PM
P: 55
differential equation
LeonhardEuler
#9
Apr3-10, 05:39 PM
PF Gold
LeonhardEuler's Avatar
P: 864
Quote Quote by reising1 View Post
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.
LeonhardEuler
#10
Apr3-10, 05:39 PM
PF Gold
LeonhardEuler's Avatar
P: 864
Quote Quote by reising1 View Post
differential equation
Yes!


Register to reply

Related Discussions
RF transmit circuit/RF receive circuit? (Hard to explain; I made a diagram.) Electrical Engineering 8
RC Circuit/Current/Resistance Problem. Circuit Gurus come here! Introductory Physics Homework 15
Experts please: DC CIRCUIT (PARALLEL-SERIES CIRCUIT) QUESTION Introductory Physics Homework 3
CIRCUIT ANALYSIS: FInd the Thevenin equivalent of the circuit. 2 res - 1 IVS - 1 VCVS Introductory Physics Homework 2
Equiv. stiffness matrix (Mech. circuit) for hydraulic flow circuit? Mechanical Engineering 0