Kirchhoff law Kirchhoff current law Kirchhoff voltage law Assessment

2. KIRCHHOFF’S
LAWS
Engr. Fawad Humayun
Fawadhumayun.ucp@gmail.com
University of Central Punjab

3. Kirchhoff’s laws:
current law:
voltage law:
Equations

4. The current law is also known as
the junction rule.
A junction is a place where three
or more wires come together.
This figure shows an enlargement
of the junction at the top of the
circuit.
Kirchhoff’s current law

5. Current Io flows INTO the junction.
Currents I1 and I2 flow OUT of the
junction.
What do you think the current law
says about I, I1, and I2?
Kirchhoff’s current law

6. Kirchhoff’s current law
Kirchhoff’s current law:
The current flowing INTO a junction
always equals the current flowing OUT
of the junction.

7. Example:
Kirchhoff’s current law

8. Conservation of charge
Why is this law always true?
It is true because electric charge
can never be created or destroyed.
Charge is ALWAYS conserved.

9. This series circuit has NO
junctions.
The current must be the same
everywhere in the circuit.
Current can only change at a
junction.
Applying the current law

10. A 60 volt battery is connected
to three identical 10 Ω
resistors.
What are the currents
through the resistors?
Applying the current law
60 V
10 Ω
10 Ω
10 Ω

11. Req = 30 Ω
I = 60 V/30 Ω
= 2 amps through each resistor
Applying the current law
60 V
10 Ω
10 Ω
10 Ω
A 60 volt battery is connected
to three identical 10 Ω
resistors.
What are the currents
through the resistors?

12. Applying the current law
?
This series circuit has two junctions. Find the missing current.

13. Applying the current law
This series circuit has two junctions. Find the missing current.
2 amps
I2 = 2 A

14. Applying the current law
How much current flows into the upper junction?

15. Applying the current law
I = 4 A
How much current flows into the upper junction?
4 amps

16. The voltage law is also known as the loop rule.
A loop is any complete path around a circuit.
This circuit has only ONE loop.
Pick a starting place. There is only ONE
possible way to go around the circuit and
return to your starting place.
Kirchhoff’s voltage law

17. This circuit has more than one loop.
Charges can flow up through the
battery and back through R1.
That’s one loop.
Can you describe a second loop
that charges might take?
Kirchhoff’s voltage law

18. Charges can flow up through the battery
and back through R2. That’s another loop.
Kirchhoff’s voltage law
This circuit has more than one loop.
Charges can flow up through the
battery and back through R1.
That’s one loop.
Can you describe a second loop
that charges might take?

19. Kirchhoff’s voltage law
Kirchhoff’s voltage law says that
sum of the voltage gains and drops
around any closed loop must equal
zero.

20. If this battery provides a 30 V gain,
what is the voltage drop across each
resistor?
Assume the resistors are identical.
Kirchhoff’s voltage law
30 V

21. Kirchhoff’s voltage law
-10 V
-10 V
-10 V
+30 V
If this battery provides a 30 V gain,
what is the voltage drop across each
resistor?
Assume the resistors are identical.
10 volts each!

22. Applying Kirchhoff’s voltage law
A 60 V battery is connected in series
with three different resistors.
Resistor R1 has a 10 volt drop.
Resistor R2 has a 30 volt drop.
What is the voltage across R3? 60 V
-10 V
-30 V
?

23. Applying Kirchhoff’s voltage law
60 V
-10 V
-30 V
-20 V
20 volts
A 60 V battery is connected in series
with three different resistors.
Resistor R1 has a 10 volt drop.
Resistor R2 has a 30 volt drop.
What is the voltage across R3?

24. What if a circuit has more than one loop?
Applying Kirchhoff’s voltage law
Treat each loop separately.
The voltage gains and drops
around EVERY closed loop
must equal zero.

25. A 30 V battery is connected in
parallel with two resistors.
What is the voltage across R1?
Applying Kirchhoff’s voltage law
30 V

26. Applying Kirchhoff’s voltage law
A 30 V battery is connected in
parallel with two resistors.
What is the voltage across R1?
30 V
R1 must have a 30 V drop.

27. Applying Kirchhoff’s voltage law
A 30 V battery is connected in
parallel with two resistors.
What is the voltage across R1?
30 V
R1 must have a 30 V drop.
What is the voltage across R2?

28. R1 must have a 30 V drop.
Applying Kirchhoff’s voltage law
A 30 V battery is connected in
parallel with two resistors.
What is the voltage across R1?
30 V
What is the voltage across R2?
R2 also has a 30 V drop.

29. Why is this law always true?
This law is really conservation of energy for circuits.
All the electric potential energy gained
by the charges must equal the energy
lost in one complete trip around a loop.
Why is the voltage law true?

30. Assessment
1. A current I = 4.0 amps flows
into a junction where three
wires meet.
I1 = 1.0 amp. What is I2?

31. Assessment
Use the junction rule: I2 = 3.0 amps
1. A current I = 4.0 amps flows
into a junction where three
wires meet.
I1 = 1.0 amp. What is I2?

32. Assessment
2. A 15 volt battery is connected in
parallel to two identical resistors.
a) What is the voltage across R1?
b) If R1 and R2 have different
resistances, will they have
different voltages?

33. 15 volts (use the loop rule)
a) What is the voltage across R1?
b) If R1 and R2 have different
resistances, will they have
different voltages?
Assessment
2. A 15 volt battery is connected in
parallel to two identical resistors.

34. They will still both have
a 15 V drop.
2. A 15 volt battery is connected in
parallel to two identical resistors.
Assessment
15 volts (use the loop rule)
a) What is the voltage across R1?
b) If R1 and R2 have different
resistances, will they have
different voltages?

35. 3. Two 30 Ω resistors are connected in parallel with a 10 volt battery.
a) What is the total resistance of the circuit?
a) What is the voltage drop across each resistor?
Assessment
c) What is the current flow through each resistor?

36. 3. Two 30 Ω resistors are connected in parallel with a 10 volt battery.
a) What is the total resistance of the circuit? 15 ohms
a) What is the voltage drop across each resistor?
Assessment
c) What is the current flow through each resistor?

37. 3. Two 30 Ω resistors are connected in parallel with a 10 volt battery.
a) What is the total resistance of the circuit? 15 ohms
a) What is the voltage drop across each resistor? 10 volts
Assessment
Each resistor is in its own loop with the 10 V battery,
so each resistor has a voltage drop of 10 V.
c) What is the current flow through each resistor?

38. Assessment
Each resistor is in its own loop with the 10 V battery,
so each resistor has a voltage drop of 10 V.
c) What is the current flow through each resistor? 0.33 amps
3. Two 30 Ω resistors are connected in parallel with a 10 volt battery.
a) What is the total resistance of the circuit? 15 ohms
a) What is the voltage drop across each resistor? 10 volts

39. Assessment
4. Two 5.0 Ω resistors are connected in series with a 30 volt battery.
a) What is the total resistance of the circuit?
a) What is the current flow through each resistor?
c) What is the voltage drop across each resistor?

40. Assessment
4. Two 5.0 Ω resistors are connected in series with a 30 volt battery.
a) What is the total resistance of the circuit? 10 ohms
a) What is the current flow through each resistor?
c) What is the voltage drop across each resistor?

41. 4. Two 5.0 Ω resistors are connected in series with a 30 volt battery.
a) What is the total resistance of the circuit? 10 ohms
a) What is the current flow through each resistor? 3.0 amps
c) What is the voltage drop across each resistor?
The circuit has only one branch,
so current flow is the same
everywhere in the circuit.
Assessment

42. 4. Two 5.0 Ω resistors are connected in series with a 30 volt battery.
a) What is the total resistance of the circuit? 10 ohms
a) What is the current flow through each resistor? 3.0 amps
c) What is the voltage drop across each resistor? 15 volts
Use the loop rule:
Assessment
The circuit has only one branch,
so current flow is the same
everywhere in the circuit.

43. METHOD OF BRANCH CURRENTS
Loop equations:
V1 – I1R1 – (I1+I2) R3 = 0
VR1
= I1R1 VR2
= I2R2 VR3
= (I1+I2)R3
VR3
= I3R3
V2 – I2R2 – (I1+I2) R3 = 0
Fig: Application of
Kirchhoff’s laws to a
circuit with two sources
in different branches.

44. METHOD OF BRANCH CURRENTS
1 3
2 3
R R
R R
Loop 1:
84 V V = 0
Loop 2:
2I V V = 0
 
 
Fig. 9-5

45. METHOD OF BRANCH CURRENTS
1
2
3
1 2 3
R 1 1 1 1
R 2 2 2 2
R 1 2 3 1 2
Using the known values of R , R and R to specify the IR voltage drops,
V = I R = I 12 = 12 I
V = I R = I 3 = 3 I
V = (I I ) R = 6(I I )
Substituting these values in the voltage eq


 
1 1 2
uation for loop 1
84 12I 6(I I ) = 0
  

46. METHOD OF BRANCH CURRENTS
Also, in loop 2,
2I − 3I2 − 6 (I1 + I2) = 0
Multiplying (I1 + I2) by 6 and combining terms
and transposing, the two equations are
18I1 − 6I2 = −84
−6I1 − 9I2 = −21
Divide the top equation by −6 and the bottom
by −3 which results in simplest and positive
terms
3I1 + I2 = 14
2I1 + 3I2 = 7

47. METHOD OF BRANCH CURRENTS
Solving for currents
Using the method of elimination, multiply the top equation
by 3 to make the I2 terms the same in both equations
9I1 + 3I2 = 42
1I1 + 3I2 = 7
Subtracting
7I1 = 35
I1 = 5A
To determine I2, substitute 5 for I1
2(5) + 3I2 = 7
3I2 = 7 − 10
3I2 = −3
I2 = −1A

48. METHOD OF BRANCH CURRENTS
Calculating the Voltages
VR1 = I1R1 = 5 x 12 = 60V
VR2 = I2R2 = 1 x 3 = 3V
VR3 = I3R3 = 4 x 6 = 24V

49. METHOD OF BRANCH CURRENTS
Checking the Solution
At point C: 5A = 4A + 1A
At point D: 4A + 1A = 5A
Around the loop with V1
clockwise from B,
84V − 60V − 24V = 0
Around the loop with V2
counterclockwise from F,
21V + 3V − 24V = 0

50. 50
FORMULA WHEEL
If you have trouble remembering these formula,
here is a useful tool.