Introductory Physics Volume Two

3.9 More Examples 73
First combine R 1 and R 2 . Since they are in parallel:
1
= 1 + 1 = 1
R 12 R 1 R 2 10Ω + 1
15Ω −→ R 12 = 6Ω
The circuit has been reduced to:
R = 30Ω
4
A
R = 6Ω
12
R = 4Ω
3
Now combine R 12 and R 3 , which are in series:
This leaves the effective circuit:
R 123 = R 12 + R 3 = 6Ω + 4Ω = 10Ω.
R = 30Ω
4
B
A
R = 10Ω
123
B
Finally, combine the last two resistances in parallel:
1
= 1 + 1 = 1
R AB R 123 R 4 10Ω + 1
30Ω −→ R 12 = 7.5Ω
Example
A 12V battery is connected across the points AB in the circuit from the
previous example. What current flows through the R 3 = 4Ω resistor?
First find the current that flows from A to B:
I AB = V AB
= 12V
R AB 7.5Ω = 1.6A
The incoming current splits between the 30Ω resistor and the effective
resistance R 123 :
A
I AB
I 4
I 123
R = 30Ω
4
R = 10Ω
123
B
We can easily compute the current through the 30Ω resistor, since the
potential difference across it is V AB :
I 4 = V AB
= 12V
R 4 30Ω = 0.4A
Using Kirchoff’s junction rule:
I 123 = I AB − I 4 = 1.6A − 0.4A = 1.2A.