Two Loops, Two Batteries: Find the Currents

In summary, to find the current in the circuit shown, you can use the junction rule and the loop rule to set up equations. After simplifying and solving the equations, the values for I1, I2, and I3 are 0.070A, -0.010A, and 0.080A, respectively. It is important to watch out for sign errors when substituting and solving the equations.
  • #1
JP4Fun
4
0

Homework Statement


Question: Find the current in the circuit shown.
http://i1307.photobucket.com/albums/s599/JustOn4Fun/PhysicsFigure_zps39d4c66d.png

Homework Equations


Not sure if equations are necessary since it's a question testing the use of the following rules:
Junction Rule
The Loop Rule

The Attempt at a Solution



At Junction A: I1-I2-I3=0
At Junction B:-I1+I2+I3=0
Loop 1:15V-I3R-I1R=0
Loop 2:-9V-I2R+I3R=0

I1=I2+I3

15V-100I3-100I1=0
15/100=I3+I1=.15

-9V-100I2+I3=0
I3-I2=9/100=.09

(I3+I1)-(I3-I2)=.06
I1+I2=.06

I think I'm having a brain fart or something because it feels like I'm just messing with the problem but not seeing something that's right under my nose. Thank you for the help!
 
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  • #2
Cancel one of the currents from the equations. For example, substitute I2+I3 for I1 into the first loop equation.

ehild
 
  • #3
I tried what you said and got the following:
15V-I3R-(I2+I3)R=0
and when the I3R cancels out, I'm left with 15V-I2R=0 giving me I2=.15

The answers are: I1=0.070A, I2=-0.010A, and I3=0.080A

What am I doing wrong?
 
  • #4
JP4Fun said:
I tried what you said and got the following:
15V-I3R-(I2+I3)R=0
and when the I3R cancels out?, I'm left with 15V-I2R=0 giving me I2=.15

The answers are: I1=0.070A, I2=-0.010A, and I3=0.080A

What am I doing wrong?

I3 does not cancel out.

ehild
 
  • #5
ehild said:
I3 does not cancel out.

ehild

JP4Fun said:
I tried what you said and got the following:
15V-I3R-(I2+I3)R=0
and when the I3R cancels out, I'm left with 15V-I2R=0 giving me I2=.15

The answers are: I1=0.070A, I2=-0.010A, and I3=0.080A

What am I doing wrong?
Ah, yeah bad mistake. I got...
15V-2I3R-I2R=0
15V=2I3R-I2R
Unfortunately, I'm not seeing it haha
 
  • #6
Now you have two equation with two unknowns.
15V=2I3R-I2R
-9V-I2R+I3R=0

Isolate I2 from one of them and substitute into the other one.

ehild
 
  • #7
JP4Fun said:
Ah, yeah bad mistake. I got...
15V-2I3R-I2R=0
[STRIKE]15V=2I3R-I2R[/STRIKE]
Unfortunately, I'm not seeing it haha
A sign error again.
Now you have two equation with two unknowns.
15V-2I3R-I2R=0
-9V-I2R+I3R=0

Isolate I2 from one of them and substitute into the other one.

ehild
 
  • #8
Ah, I got it now. Thank you very much for being patient with me.
 
  • #9
You are welcome:smile:

ehild
 

Related to Two Loops, Two Batteries: Find the Currents

1. What is the purpose of "Two Loops, Two Batteries: Find the Currents"?

The purpose of this experiment is to determine the currents in two separate loops that are connected to two different batteries. This information can be useful in understanding the flow of electricity in a circuit and can be applied in various electrical engineering and physics applications.

2. What materials are needed for this experiment?

To perform this experiment, you will need two loops of wire, two batteries, a multimeter, and connecting wires. You may also need a power source for the batteries, such as a battery holder or power supply.

3. How do I calculate the currents in each loop?

To calculate the currents, you will need to use Ohm's law (I = V/R). Measure the voltage of each battery using the multimeter and then measure the resistance of each loop using the multimeter. Once you have these values, you can plug them into the equation to calculate the currents in each loop.

4. Can this experiment be done with more than two batteries and loops?

Yes, this experiment can be expanded to include more batteries and loops. However, the calculations will become more complex as the number of batteries and loops increases. It is important to carefully plan and label each component to keep track of the different circuits and their currents.

5. Are there any safety precautions I should take when performing this experiment?

Yes, it is important to take proper safety precautions when working with electricity. Be sure to disconnect all power sources before making any adjustments or measurements. Wear protective gear, such as gloves and safety glasses, and be aware of any potential hazards in the surrounding area.

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