1. A deep philosophical question to consider
A deep philosophical question to consider. Do you always eat your food from the outside in?

2. Example: For the circuit below, calculate the current drawn from the battery and the current in the 6  resistor.
10 
8 
6 
3 
8 
1 
9 V
The next three slides show the strategy for solving this. I will work the example at the blackboard in lecture.

3. In a few minutes, we will learn a general technique for solving circuit problems. For now, we break the circuit into manageable bits. “Bite-sized chunks.”
10 
8 
6 
3 
8 
1 
9 V
Replace the parallel combination (green) by its equivalent.
Do you see any bite-sized chunks that are simple series or parallel?

4. Any more “bite-sized chunks
Any more “bite-sized chunks?” Remember that everything inside the green box is equivalent to a single resistor.
10 
8 
6 
3 
8 
1 
9 V
Replace the series combination (blue box) by its equivalent.

5. We are left with an equivalent circuit of 3 resistors in series, which is easy handle.
10 
8 
6 
3 
8 
1 
9 V
Next bite-sized chunk. Inside the blue box is “a” resistor. Replace the parallel combination (orange) by its equivalent.

6. Let’s shrink the diagram a bit, and work this a step at a time.
R3 and R4 are in parallel.
R3=8 
R2=6 
R5=
3 
R4=8 
R6=1 
=9 V
R1=10 
R2 and R34 are in series.
R2=6 
R34=4 
R5=
3 
R6=1 
=9 V

7. Let’s shrink the diagram a bit, and work this a step at a time.
R1 and R234 are in parallel.
R234=10 
R5=
3 
R6=1 
=9 V
R1234, R5 and R6 are in series.
R1234=5 
R5=
3 
R6=1 
=9 V

8. Calculate the current drawn from the battery.
3 
I=1 A
R6=1 
=9 V

9. Find the current in the 6  resistor.
There are many ways to do the calculation. This is just one.
R2=6 
R5=
3 
R4=8 
I=1 A
R6=1 
=9 V
R1=10 
V1 = V234 = V1234 (parallel).
R234=10 
R5=
3 
I=1 A
R6=1 
=9 V

10. Find the current in the 6  resistor.
V1234 = I R1234 = (1)(5) = 5 V
R5=
3 
V1 = V234 = 5 V
I=1 A
R6=1 
=9 V
R1=10 
V234 = I234 R234
I234
I234 = V234 / R234 = 5/10
R234=10 
R5=
3 
I234 = 0.5 A
I=1 A
R6=1 
=9 V

11. Find the current in the 6  resistor.
I2=I234
I234 = I2 = I34 = 0.5 A
R2=6 
R5=
3 
R4=8 
I2 = 0.5 A
I=1 A
R6=1 
=9 V

12. Find the current in the 6  resistor.
A student who has taken a circuits class will probably say
R234=10 
R5=
3 
R1 = R234
so I1 = I234 = I/2 = 0.5 A
I=1 A
R6=1 
=9 V
If you want to do this on the exam, make sure you write down your justification on the exam paper, and don’t make a mistake! If you don’t show work and make a mistake, we can’t give partial credit.
Answers without work shown generally receive no credit.