Transcript Goal: To understand how to solve circuits with multiple

Goal: To understand how to solve
circuits with multiple power
sources
Objectives:
1) To learn Kirchhoff’s Laws
2) To use those laws in order to
play Physics Chutes &
Ladders
Suppose we have the following
circuit:
I1
10 V
5 Ohms
2 Ohms
3 Ohms
I2
8 Ohms
6 Ohms
5V
I3
Kirchhoff’s Laws
• You have already learned them!
• Law 1: The current going into a point must
be the current going out of the point.
• So Iin = Iout
• Law 2: If you do a full loop, the net
change in voltage is 0.
• We can use this to our advantage!
At the cross points, Iin = Iout
I1
10 V
5 Ohms
2 Ohms
3 Ohms
I2
8 Ohms
6 Ohms
5V
I3
If we do any loops then the net
voltage is 0. That is the up voltage
= the down voltage
I1
10 V
5 Ohms
2 Ohms
3 Ohms
I2
8 Ohms
6 Ohms
5V
I3
Physics Chutes and Ladders
• When you look at a circuit, first plot the potential
raises and drops.
• In the direction of the current for each section,
the voltages for from low (-) to high (+)
potentials.
• Resistors go from + to –
• What direction should the current go? Well don’t
worry about it. If you choose wrong you will just
get a negative current, and the is okay. Just be
consistant.
To solve the circuit:
•
•
•
•
In this example you have 4 equations.
The first is the easiest.
Find a point where the 3 currents meet.
Iin = Iout is the equation.
• The next 3 are similar.
Current Loops
• Each way you can go in a full circle is a loop.
• In this example there are 3 loops (full perimeter, bottom circle, top
circle).
• For each pick a direction (either one).
• On the left side of the equation you will write in the value of the
voltage when you go UP.
• For a power source this is some # of volts.
• For a resister the voltage is IR. Be sure to include the correct value
of I.
• Also, note that you can’t add up all the resistances in the loop
because they have different currents.
• If any have the same current you can add those, but only if they are
the same.
• On the right side of the equation you put in all the downs.
• Do this 3 times and you have a total of 4 equations.
Solve for the currents:
I1
10 V
5 Ohms
2 Ohms
3 Ohms
I2
8 Ohms
6 Ohms
5V
I3
Equations
• I1 = I2 + I3
• Top: 10 V = 7 Ohms I1 + 3 Ohms I2
• Bottom: 5V + 3 Ohms I2 = 14 Ohms I3
• Perimeter: 5V + 10V = 7 Ohms I1 + 14 Ohms I3
• Now we solve…
• Easiest to solve for 1 value in 3 different
equations. Then you can get rid of it.
• Lets do it for I1…
Equations
• I1 = I2 + I3
• Top: I1 = 10/7 A - 3/7 I2
• Perimeter: I1 = 15/7 A - 2 I3
• Now we substitute I1 from first equation into the 2nd. Then we solve
for say I2.
• I2 + I3 = 10/7 A – 3/7 I2
• I2 = 1 A – 0.7 I3
•
•
•
•
•
Okay, now we substitute I1 and I2 into the 3rd equation for I1…
I1 = I2 + I3 = 1 A + 0.3 I3
1A + 0.3 I3 = 15/7 A – 2 I3
So, 2.3 I3 = 8/7 A
Or I3 = 0.50 A
Finish Up
• Now we just solve for the rest.
• I1 = I2 + I3
• Top: I1 = 10/7 A - 3/7 I2
• Perimeter: I1 = 15/7 A - 2 I3
• I3 = 0.5 A
•
•
•
•
So, I1 = 15/7 A – 2 * 0.5 A
I1 = 1.15 A
1.15 A = I2 + 0.50 A
I2 = 0.65 A
Try a less difficult one:
I1
10 V
5 Ohms
2 Ohms
4 Ohms
I2
0 Ohms
0 Ohms
4V
I3
Set it up
• I1 = I2 + I3
• Top: 10 V = 7 Ohms I1 + 4 Ohms I2
• Bottom: 4V + 4 Ohms I2 = 0
• Perimeter: 4V + 10V = 7 Ohms I1
• This one can we worked out with a lot less
steps…
Set it up
• I1 = I2 + I3
• Top: 10 V = 7 Ohms I1 + 4 Ohms I2
• Bottom: 4V + 4 Ohms I2 = 0
• Perimeter: 4V + 10V = 7 Ohms I1
•
•
•
•
This one is a lot easier…
I2 = -1 A
I1 = 2 A
I3 = I1 – I2 = 3A
Conclusion:
• Current in = Current out
• The net voltage around a loop is 0 so
therefore the ups = the downs.
• This allows us to play the most boring
game EVER invented: Physics Chutes and
Ladders
• By using this to create our equations for
the loops we are able to solve the circuit.