Okay, think I got it
You're right it should be +5I3 because it's with the current, just like I2 works out in the first loop equation
L1=2-4I1+4I2=0
Using this to solve for I1
I1=I2+(1/2)
Now I did this wrong, so going back through the derivation of this loop equation...
Using the one know current, the current would be the same at a correct?
If so you couls just do
Va-Vb=-24V+(I)(10)
=>-24+(0.0706)(10)=-23.294
Which implies
Vb-Va=+23.294V
Or is this still the wrong path?
Do I need to use loop rules and find the unknown values of the current when it...
Okay, re-attempted using loop one and loop two, does, this seem more accurate?
L1=12V-I1(2)+I2(1)-10V+I2(3)-I1(1)=0
=>
L1=2V-4I1+4I2=0
L2=-3(I2)+10V-I2(1)-I3(1)-8V-I3(2)=0
=>
L2=2V-4I2-5I3=0
=>
L2=2-4I2-5(I1-I2)=0
=>
L2=2-4I2-5I1+5I2=0
=>
L2=2-5I1+I2=0
So from there...
This is useful, in a case like mine you wouldn't want multiple I notations of current cause then your introducing variables you don't know right?
should I go opposite the current from b→a?
like
ΔV=-(0.0706A)(10Ω)+12V=12.706V
Since that is the only value of I given?
Homework Statement
Find the potential difference between points a and b.
Picture of Circuit involved.
Homework Equations
V=IR
ΔV=Vb-Va
The Attempt at a Solution
Pretty much completely lost, due to the only one current shown I'm not sure which way the current flows through...
Homework Statement
Apply the loop rule to two of the three loops to calculate I1,I2,and I3.
Picture of DC circuit involved
Homework Equations
V=IR
The Attempt at a Solution
L1=2v+2I2-4I1=0 (simplified)
and
L3=4-4I1-5I3=0
But if you work through the algebra you'd get
I2=1...
Am I just being an idiot? is the -4r2 supposed to be under the root?
I mean that makes sense to me in terms of a quadratic formula -4AC being under the root.
I've tried this it kinda looks like
R=1/2[(2.885)-0.76]±√((2.885-0.76)^2)-0.5776
What messes with me is that the root term should just cancel most itself and become the first number again.
My professor just got back to me saying that his answer for the root term was 0.9895. which if this...
Homework Statement
A resistor with resistance R is connected to a battery that has emf 15.0[V] and internal resistance r = 0.38[Ω].
For what two values of R will the power dissipated in the resistor be 78.0 W?
Homework Equations
P=I2R=(ε2/(R+r)2)R
The Attempt at a Solution...