1. SPH4UW
Kirchhoff’s Laws
Lecture timing is fine. No need to rush 1

2. Kirchhoff’s Rules Kirchhoff’s Voltage Rule (KVR):
Sum of voltage drops around a loop is zero.
Kirchhoff’s Current Rule (KCR):
Current going in equals current coming out.
Students running works pretty well.

3. Kirchhoff’s Rules
Between a and b a b
=-IR I a b
=IR I a b
=+E a b
=-E

4. Kirchhoff’s Laws e1- I1R1- I2R2-e2=0 (1) Label all currents
Choose any direction
Choose loop and direction
Must start on wire, not element. R4
Write down voltage drops
-Batteries increase or decrease
according to which end you encounter first.
-Resistors drop if going with current.
-Resistors increase if gong against current.
Have students label I5, since it isn’t shown in their drawing
e1- I1R1- I2R2-e2=0
For inner loop

5. Practice ε1- IR1 - ε2 - IR2 = 0 Find I:
Label currents
Choose loop
Write KVR
Practice B
R1=5 W I
Find I:
ε1- IR1 - ε2 - IR2 = 0
I I = 0
I = +2 Amps
e1= 50V A
R2=15 W
e2= 10V
What if only went from A to B?, Find VB-VA
e2= 10V
R1=5 W I
e1= 50V
R2=15 W A B
VB - VA = e1 - IR1
= 5 = 40 Volts or
VB - VA = +IR2 + e2
= 2 = +40 Volts
Therefore B is 40V higher than A

6. Understanding Resistors R1 and R2 are:
R1=10 W
Resistors R1 and R2 are:
1) in parallel ) in series 3) neither
E2 = 5 V I2
R2=10 W
Definition of parallel:
Two elements are in parallel if (and only if) you can make a loop that contains only those two elements. IB - +
E1 = 10 V
Note that nothing is in series or in parallel!
Definition of series:
Two elements are in series if (and only if) every loop that
Contains R1 also contains R2

7. Understanding: Voltage Law
Practice
slide 7
Calculate the current through resistor 1. I1
R1=10 W
I1 = 0.5 A
I1 = 1.0 A
I1 = 1.5 A
E2 = 5 V I2
R2=10 W IB
<--Start
E1 = 10 V
Understanding: Voltage Law
How would I1 change if the switch was closed?
1) Increase 2) No change ) Decrease

8. Starting at Star and move clockwise around loop
Understanding
slide 7
Calculate the current through resistor 2. I1
R1=10 W
I2 = 0.5 A
I2 = 1.0 A
I2 = 1.5 A
E2 = 5 V I2
R2=10 W IB
E1 = 10 V
Starting at Star and move clockwise around loop

9. Kirchhoff’s Junction Rule
slide 7
Kirchhoff’s Junction Rule
Current Entering = Current Leaving
I1 = I2 + I3 I1 I2 I3
Understanding
I1=1.0A
R=10 W
IB = I1 + I2
= 1.0A A
= 1.5 A
IB = 0.5 A
IB = 1.0 A
IB = 1.5 A
E = 5 V
R=10 W
I2=0.5 IB - +
E1 = 10 V

10. Kirchhoff’s Laws (1) Label all currents Choose loop and direction
Choose any direction
Choose loop and direction
Your choice! R4 R1 E1 R2 R3 E2 E3 I1 I3 I2 I4 R5 A B
Write down voltage drops
Follow any loops
Have them go back to this slide and fill in (5)
Write down node equation
Iin = Iout

11. Ohm’s Law

12. Resistor Series Parallel Wiring Voltage Current Resistance R1 R1 R2 R2
Each resistor on the same wire.
Each resistor on a different wire.
Wiring
Different for each resistor.
Vtotal = V1 + V2
Same for each resistor.
Vtotal = V1 = V2
Voltage
Same for each resistor
Itotal = I1 = I2
Different for each resistor
Itotal = I1 + I2
Current
Increases
Req = R1 + R2
Decreases
1/Req = 1/R1 + 1/R2
Resistance

13. Understanding 1 2 3
R R R/2
Which configuration has the smallest resistance? 1 2 3
Which configuration has the largest resistance? 1 2 3

14. Parallel + Series Tests
Resistors R1 and R2 are in series if and only if every loop that contains R1 also contains R2
Resistors R1 and R2 are in parallel if and only if you can make a loop that has ONLY R1 and a loop that ONLY contains R2

15. Understanding Determine the voltage and current in each resistor
First we notice the voltage drop through these two resistor groupings total 10V.
Let’s combine to find REQ
R3=17W
e0=10V
R4=5W
R2=25W
R1=15W
Let’s now add R4 to this REQ

16. Understanding Let’s determine the current, IT e0=10V e0=10V R4: R1:
R3=17W
e0=10V
R4=5W
R2=25W
R1=15W
e0=10V
RT=11.04W
R1, R2, R3, experience the same voltage
This current will flow through each of the resistor groupings
R4:
R1:
R2:
R3:
Since voltage drop has to total 10V, VEQ=10V-4.5V=5.5V

17. Determine the Voltage, Current, and Resistance
The V-I-R Chart
Determine the Voltage, Current, and Resistance 5Ω 7Ω
12V
10Ω
Step 1:
Fill out the table with known resistors and the Total Voltage for circuit V I R R1 5 R2 7 R3 10
Total 12 V I R R1 R2 R3
Total

18. Determine the Voltage, Current, and Resistance
The V-I-R Chart
Determine the Voltage, Current, and Resistance 5Ω 7Ω
12V
10Ω
Step 2:
Using resistor laws, determine total resistance of circuit. V I R R1 5 R2 7 R3 10
Total 12
9.117 V I R R1 5 R2 7 R3 10
Total 12

19. Determine the Voltage, Current, and Resistance
The V-I-R Chart
Determine the Voltage, Current, and Resistance 5Ω 7Ω
12V
10Ω
Step 4:
Since initial current amount will also pass through resistor R1, we can determine its voltage drop.
Step 3:
Using Ohm’s law, determine the Current of circuit. V I R R1
6.6
1.32 5 R2 7 R3 10
Total 12
9.117 V I R R1 5 R2 7 R3 10
Total 12
9.117 V I R R1 5 R2 7 R3 10
Total 12
1.32
9.117

20. Determine the Voltage, Current, and Resistance
The V-I-R Chart
Determine the Voltage, Current, and Resistance 5Ω 7Ω
12V
10Ω
Step 5:
Since R2 and R3 have the same Voltage drop, we then must have 12V-6.6V=5.4V. V I R R1
6.6
1.32 5 R2
5.4 7 R3 10
Total 12
9.117 V I R R1
6.6
1.32 5 R2 7 R3 10
Total 12
9.117

21. Determine the Voltage, Current, and Resistance
The V-I-R Chart
Determine the Voltage, Current, and Resistance 5Ω 7Ω
12V
10Ω V I R R1
6.6
1.32 5 R2
5.4 7 R3 10
Total 12
9.117 V I R R1
6.6
1.32 5 R2
5.4
0.77 7 R3
0.54 10
Total 12
9.117
Step 6:
Use ohm’s law to find currents

22. Question
If a 60 watt light bulb operates at a voltage of 120V, what is the resistance of the bulb? 2Ω
30Ω
240Ω
720Ω
7200Ω

23. Question
Three resistors are connected to a 10-V battery as shown below. What is the current through the 2.0 Ω resistor?
4.0Ω
2.0Ω
ε=10V
0.25A
0.50A
1.0A
2.0A
4.0A
Since all resistors are in series, the amount of current that passes through any one of them is the same. So we need to simply the circuit to determine that current.

24. Question Determine the equivalent resistance between points a and b?
0.167Ω
0.25 Ω
0.333 Ω
1.5 Ω
2 Ω
12Ω 3Ω 4Ω 3Ω

25. Question
What is the voltage drop across the 12 ohm resistor in the portion of the circuit shown?
24V
36V
48V
72V
144V 8Ω
Since the top branch has 4Ω of resistance and the bottom branch has 12Ω of resistance, therefore 3 times as much current will flow through the 4Ω resistor than the 12Ω resistor. This breaks down the 12A into 9A up and 3A down. So form V=IR we have V=(3A)(12Ω)=36V 4Ω 2Ω 8Ω
12A
12Ω
We could also have produced a RT and found 3Ω value, therefore a voltage drop of (3Ω)(12A)=36V on both the upper and lower branch

26. Question
A 100Ω, 120Ω, and 150Ω resistor are connected to a 9-V battery as in the circuit shown below. Which of the three resistors dissipates the most power?
We can also answer this question faster by noticing that the voltage drops across the 100Ω and then across the parallel resistors. Since the 100Ω has a higher value than the parallel combination, it will have a larger voltage drop than the combination so via P=IV, it dissipates the most power.
The 100Ω resistor
The 120Ω resistor
The 150Ω resistor
Both the 120Ω and the 150Ω
All dissipate the same power
100Ω
120Ω 9V
150Ω
Therefore the 100Ω resistor dissipates the most amount of power.

27. Question Given the following circuit 10Ω a 10Ω 20Ω 40Ω ε=120V 100Ω b
At what rate does the battery deliver energy to the circuit?
Determine the current through the 20 Ω resistor.
i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher?
Determine the energy dissipated by the 100 Ω resistor in 10 s

28. Question Given the following circuit
At what rate does the battery deliver energy to the circuit?
10Ω
20Ω
100Ω
40Ω
ε=120V
Recall: Rate is Power RT
We need to write this as a simple circuit with a RT so that we can determine the current, I by using V=IR.

29. Question Given the following circuit
b) Determine the current through the 20 Ω resistor.
We can determine the voltage drop .
Recall: from Question a) we have a 2A current
10Ω 2A
10Ω
120V-(2A)(10Ω)-(2A)(10Ω)-(2A)(10Ω)
=60V
20Ω
40Ω
The ratio of the resistance of left side to right side is 40:120 or 1:3
ε=120V
100Ω
Then we use V=IR to find each individual current. 2A
10Ω
Therefore ¼ of 2 A goes through the right and ¾ of 2A goes through the left.
Thus (1/4)(2A)=0.5A passes through the 20Ω resistor. OR

30. Question Given the following circuit
c) i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher?
10Ω
20Ω
100Ω
40Ω
ε=120V a b i)
ii) Point a is at a higher potential. Since current flows from a high potential to a low potential.

31. Question Given the following circuit
d) Determine the energy dissipated by the 100 Ω resistor in 10 s
10Ω
20Ω
100Ω
40Ω
ε=120V a b
Recall: Energy equals Power multiplied by Time

32. Understanding Let’s follow a conventional current path through R1
R3=17W
e0=10V
R4=5W
R2=25W
R1=15W
You can pick any path through the circuit and the total voltage increases and decrease will balance
You can reverse the direction of the current and thus the signs, (batteries increase the voltage, resistors drop the voltage) and obtain the same results.

33. Let us calculate the Current and the Power (used/generated) by the elements of the following circuit.
What happens to the Power delivery and consumption if another identical bulb is place in parallel or in series with the first?

34. Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in parallel. 2A 2A 4A
Because the bulbs (resistors) are in parallel, we use the parallel law to determine total resistance of the circuit.

35. Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in series. 1A
Because the bulbs (resistors) are in series, we use the series law to determine total resistance of the circuit.

36. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
Simplify the above circuit so that it consists of one equivalent resistor and the battery.
What is the total current through this circuit?
Find the voltage across each resistor.
Find the current through each resistor.
The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same.

37. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
REQ
Simplify the above circuit so that it consists of one equivalent resistor and the battery.

38. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
REQ
=342.2Ω
b) What is the total current through this circuit?

39. Question
200Ω
ε= 9V
300Ω
400Ω
500Ω
Find the voltage across each resistor.
Find the current through each resistor.
Let’s use a VIR chart V I R
200Ω
300Ω
400Ω
500Ω
Total 9V
0.0263A
342.2Ω
3.16V
0.0158A
3.16V
0.0105A
5.84V
0.0146A
5.84V
0.0117A

40. Question
200Ω
e) The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same.
300Ω
ε= 9V
400Ω
500Ω
By removing a resistor from a parallel set, we actually increase the resistance of the total circuit.
Therefore by Ohm’s law if the voltage remains the same and the resistance increases, the total current must decrease
Now through the 200Ω set, the total resistance remains the same, yet the current decreases, therefore the voltage across each resistor decreases as well as the current .