NCEA Level 3 Physics Electricity AS91526 Kirchhoff's Laws

1. Kirchhoff’s Laws
What goes in must come
out

2. Kirchhoff’s Current Law
• The current that flows into a junction must equal the
current flowing out of a junction
3 A
(out)
1 2 3 I  I  I 5 A
(in)
2 A
(out)
5  3 2
That all makes perfect sense; you can’t get more current out
that you put in

3. Kirchhoff’s Voltage Law
• The sum of the voltages around any circuit loop is zero
(voltage produced must equal the voltage used in a loop)
1 2 3 0 V V V 
14V
(produced)
7V
(used)
7V
(used)
Call this voltage
produced positive
+14V
1477  0
Call these
voltages used
negative -7V
This makes sense too:
you can’t be using more
energy that you are
producing

4. Getting more complex
• Some loops involve more than one cell or battery and
several components.
• It is important to work your way around a loop
systematically in one direction (doesn’t matter which way you go
as long as you keep going in the same direction around the whole loop)

5. The Rules -Cells
1. Passing through a cell in the same direction as
conventional current is a positive voltage (energy
produced)
12V
I
2. Passing through a cell in the opposite direction to
conventional current 12V
is a negative voltage (energy used)
I
+12V
-12V

6. The Rules -Components
1. Passing through a component in the same direction as
conventional current is a negative voltage (energy used)
-12V
2. Passing through a component in the opposite direction
to conventional current is a positive voltage (energy
gained)
+12V

7. Putting it all together
• Working anticlockwise
around the loop ABCDA
AB +15V
BC –(2X11)V
(V=IR)
CD +(4X3)V -5V
so:
+15 - 22 +12 – 5 = 0
2A
15V
4 5V
11
B A
C D
5A

8. Try the other direction
5A
2A
15V
4 5V
11
B A
C D
• Now clockwise around
the loop BADCB
BA -15V
DC +5V –(4X3)V
(V=IR)
CB +(11X2)V
so:
-15 + 5 - 12 + 22 = 0

9. Try this;
• Working anticlockwise
around the loop ABCDA
AB +15V
BC –(1.5X40)V
(V=IR)
CD +(2.5X20)V -5V
so:
+15 - 60 +50 – 5 = 0
4A
15V
20 5V
40
B A
2.5A
C D

10. What if something is missing?
A B
7A
3A
5V
4 12V
?
D C
• Now clockwise around
the loop ABCDA
AB -5V
CD +12V –(4X4)V
(V=IR)
DA +(3XR)V
so:
-5 + 12 – 16 + 3R = 0
3R=9
R=3 

11. or this one…
• Now clockwise around
the loop ABCDA
AB -15V
CD + V –(0.5X80)V
(V=IR)
DA +(1.5X30)V
so:
-15 + V – 40 + 45 = 0
V=10V
15V
80 ?V
0.5A
30
A B
D C
2A

12. Now some Exercises
• Try ESA, Activity 13C, Pg 214
• ABA, Pg 134-136

13. Homework