A
|-----2 ohm r-----------2 ohm r--|
| | |
___ 5 ohm r 3 ohm r
- | |
| ___ _
2 ohm r - ___
| | |
| I3 |
| | |
|-----<I1<---------------<I2<----|
B
Hopefully my drawing came out right because I can't find a way to attach the picture of the circuit.
Key:
2 ohm r= 2 ohm resistor
I1, I2, and I3 are currents at three branches. (I1 and I2 are to the left, I3 is downward)
Batteries marked with their usual symbol have voltages (left to right) 12 V, 7 V, 13 V.
1) Find the currents in the three branches.
2) What is the change in potential V_A-V_B?
|-----2 ohm r-----------2 ohm r--|
|*****************|**************|
___************5 ohm r********3 ohm r
-****************|**************|
|****************___*************_
2 ohm r***********-*************___
|*****************|**************|
|****************I3**************|
|*****************|**************|
|-----<I1<---------------<I2<----|
Ignore the stars as blank spaces... same problem.
|-----2-----------------2--------|
|*****************|**************|
___***************5**************3
*-****************|**************|
|****************___*************_
2*****************-*************___
|*****************|**************|
|****************I3**************|
|*****************|**************|
|-----<I1<---------------<I2<----|
"2" instead of "2 ohm r"
In the last drawing you can sort of get the idea of how my circuit is drawn (its still flawed obviously) but if you redraw on a piece of paper it is easy to see. Any input is appreciated. Sorry for all the posts. I can't tell how it will come out until it is actually posted.
To find the currents in the three branches, we can apply Kirchhoff's laws. Let's call the current flowing from A to the junction above the 2 ohm resistor as I1, the current flowing from the junction above the 2 ohm resistor to the junction below the 5 ohm resistor as I2, and the current flowing downward as I3.
1) Applying Kirchhoff's current law at the junction above the 2 ohm resistor:
I1 = I2 + I3
2) Applying Kirchhoff's voltage law in the loop above the resistors and battery symbols:
-12V + (2Ω)I1 + (2Ω)(I1 - I2) = 0
Simplifying the equation above:
4I1 - 2I2 = 12 ... (Equation 1)
3) Applying Kirchhoff's voltage law in the loop right to left through the resistors, batteries, and the junction:
-12V - 7V + (2Ω)I1 + (5Ω)(I2 - I3) + 13V = 0
Simplifying the equation above:
2I1 - 5I2 + 5I3 = 6 ... (Equation 2)
4) Applying Kirchhoff's voltage law in the loop downward through the resistors and the junction:
-13V + (3Ω)I3 + (2Ω)(I2 - I3) = 0
Simplifying the equation above:
-2I2 + 5I3 = 13 ... (Equation 3)
Now we have a system of three equations (Equations 1, 2, and 3) with three unknowns (I1, I2, and I3). Solving this system of equations will give us the currents in the three branches.
To find the change in potential V_A - V_B, we need to calculate the potential difference between points A and B.
5) From Equation 1, we can solve for I1:
I1 = (2I2 + 12) / 4
6) Substitute the value of I1 from Equation 5 into Equations 2 and 3 to eliminate I1:
(2(2I2 + 12) / 4) - 5I2 + 5I3 = 6
-2I2 + 5I3 = -(I2 + 7) ... (Equation 4)
7) Substitute the value of I1 from Equation 5 into Equation 3 to eliminate I1:
-2I2 + 5I3 = 13 - (2I2 + 12) / 4
-8I2 + 20I3 = 40 - 2I2 - 12
6I2 + 20I3 = 28 ... (Equation 5)
Now, we have Equation 4 and Equation 5, which is a system of two equations with two unknowns (I2 and I3). Solving this system will give us the values of I2 and I3.
8) Solve the system of Equation 4 and Equation 5.
Once we have obtained the values of I1, I2, and I3, we can find the change in potential V_A - V_B by calculating the potential difference between points A and B, where V_A represents the voltage at point A and V_B represents the voltage at point B.