1. Kirchhoff’s Rules Limits of Progressive Simplification
Simple Examples
More complicated examples
Capacitors in series/parallel

2. Progressive Simplification
Simplify down
Rtot = 10.3 Ω
Ibat = 0.87 A
Build back up
Voltage across top
=9 𝑉− .87 𝐴 .5 Ω − 𝐴 5Ω
Current through 10 Ω
Current through 8.7 Ω
Current through 6 Ω
Voltage across 4/8 Ω

3. Kirchhoff’s Rules
Progressive simplification doesn’t work on some circuits.
Can combine 1 Ω and 20 Ω, and 1 Ω and 40 Ω, but that’s it!

4. Kirchhoff’s Rules Declare each branch has it’s own separate current
At any junction, current entering = current leaving
For any complete loop, sum of voltage rises/drops equals 0
Currents I1, I2, I3
Canal analogy

5. Example – Problem 23 Only 1 current I Just like we’ve done all along.
9 𝑉−𝐼 8 Ω−𝐼 12 Ω−𝐼 2 Ω=0
𝐼= 9 𝑉 22 Ω =0.41 𝐴
Just like we’ve done all along.

6. Example – Problem 24 Two batteries backwards (charging circuit)
Only 1 current I
18 𝑉−𝐼 6.6 Ω−12 𝑉−𝐼 2 Ω−𝐼 1 Ω=0
𝐼= 6 𝑉 9.6 Ω =0.625 𝐴
Note we went across 12 V battery backwards.

7. Example – Problem 24 (1) Branch point!
Draw currents I1, I2, I3 as shown
Guess at direction
Kirchhoff Rule #1 at point a
𝐼 1 = 𝐼 2 + 𝐼 3
Kirchhoff Rule #1 same at point b I1 I2 I3 a b

8. Example – Problem 24 (2) Kirchhoff Rule #2 around upper loop
9 𝑉− 𝐼 Ω− 𝐼 Ω=0
Note separate I1 I2
Kirchhoff Rule #2 around lower loop
6 𝑉+ 𝐼 Ω=0
Note (+) going across resistor backwards
Kirchhoff Rule #2 around outer loop
9 𝑉+6 𝑉− 𝐼 Ω=0
General rule: battery (+) and resistor (-), but flip sign when you go across backwards I1 I2 I3 a b

9. Example – Problem 24 (3) 6 𝑉+ 𝐼 2 15 Ω=0 Outer loop
9 𝑉+6 𝑉− 𝐼 Ω=0
𝐼 1 = 15 𝑉 22 Ω =0.68 𝐴
Lower loop
6 𝑉+ 𝐼 Ω=0
𝐼 2 = −6 𝑉 15 Ω =−0.4 𝐴 *
Current Rule
𝐼 1 = 𝐼 2 + 𝐼 𝐼 3 = 𝐼 1 − 𝐼 2 =1.08 𝐴
Upper loop
9 𝑉− 𝐼 Ω− 𝐼 Ω=0
9 𝑉− −0.4 𝐴 15 Ω− 0.68 𝐴 22 Ω=0
*assumed wrong direction for I2 but that’s OK I1 I2 I3 a b

10. Example 19-3 (1) 45 𝑉 − 𝐼 3 1Ω− 𝐼 3 40 Ω− 𝐼 1 30 Ω=0
Draw I1, I2, I3
Set node rule (Kirchhoff 1)
Set 2 loops (Kirchhoff 2)
Solve simultaneous equations
Node rule from diagram
𝐼 3 = 𝐼 1 + 𝐼 2
Upper loop
45 𝑉 − 𝐼 3 1Ω− 𝐼 Ω− 𝐼 Ω=0
Lower loop
80 𝑉 − 𝐼 2 1Ω− 𝐼 Ω+45 𝑉 − 𝐼 3 1Ω− 𝐼 Ω=0
3 variables, 3 simultaneous equations

11. Example 19-3 (2) 3 variables, 3 simultaneous equations 𝐼 3 = 𝐼 1 + 𝐼 2
45 𝑉 − 𝐼 3 1Ω− 𝐼 Ω− 𝐼 Ω=0
80 𝑉 − 𝐼 2 1Ω− 𝐼 Ω+45 𝑉 − 𝐼 3 1Ω− 𝐼 Ω=0
Simplify
45 𝑉 − 𝐼 3 41Ω− 𝐼 Ω=0
125 𝑉 − 𝐼 Ω− 𝐼 Ω=0
My suggestion
𝐼 1 = − 𝐼 3
𝐼 2 = − 𝐼 3

12. Example 19-3 (3) Calculating 𝐼 3 = 𝐼 1 + 𝐼 2 (1) 𝐼 1 =1.5−1.37 𝐼 3 (2)
𝐼 3 = 𝐼 1 + 𝐼 (1)
𝐼 1 =1.5−1.37 𝐼 (2)
𝐼 2 =5.95−1.95 𝐼 (3)
Substitute eqns. (2) and (3) in (1)
𝐼 3 = 1.5−1.37 𝐼 −1.95 𝐼 3
Simplify
4.32 𝐼 3 =7.45
𝐼 3 =1.72 𝐴
Plug in (2) and (3) above
𝐼 1 =1.5−1.37 𝐼 3 = −0.86 𝐴 (assumed wrong direction, that’s OK)
𝐼 2 =5.95−1.95 𝐼 3 = 𝐴

13. Example 19-3 (textbook solution)
Draw I1, I2, I3
Set node rule (Kirchhoff 1)
Set 2 loops (Kirchhoff 2)
Solve simultaneous equations
Node rule from diagram
𝐼 3 = 𝐼 1 + 𝐼 2
Upper loop
45 𝑉 − 𝐼 3 1Ω− 𝐼 Ω− 𝐼 Ω=0
Outer loop
80 𝑉 − 𝐼 2 1Ω− 𝐼 Ω+ 𝐼 1 30Ω=0
This solution gives same answer!
Note sign flip, going across resistor backwards (against assumed current)

14. Elevation analogy

15. Capacitors in series and parallel
Charges add
Voltage same
𝑄 𝑡𝑜𝑡 = 𝑄 1 + 𝑄 2 + 𝑄 3 = 𝐶 1 𝑉+ 𝐶 2 𝑉 +𝐶 3 𝑉
= 𝐶 1 + 𝐶 2 + 𝐶 3 𝑉= 𝐶 𝑒𝑞 𝑉
Equivalent parallel capacitance 𝐶 𝑡𝑜𝑡 =𝐶 1 + 𝐶 2 + 𝐶 3
Series
Voltages add
Charge same
𝑉 𝑡𝑜𝑡 =𝑉 1 + 𝑉 2 + 𝑉 3 = 𝑄 𝐶 1 + 𝑄 𝐶 2 + 𝑄 𝐶 2
=𝑄 1 𝐶 𝐶 𝐶 3 = 𝑄 𝐶 𝑒𝑞
Equivalent series capacitance 𝐶 𝑒𝑞 = 1 𝐶 𝐶 𝐶 3
series/parallel rules reversed from resistors!

16. Summary – resistors/capacitors in series/parallel
𝑅 𝑠𝑒𝑟𝑖𝑒𝑠 = 𝑅 1 + 𝑅 2 + 𝑅 3
1 𝑅 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 1 𝑅 𝑅 𝑅 3
Capacitors
𝐶 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 𝐶 1 + 𝐶 2 + 𝐶 3
1 𝐶 𝑠𝑒𝑟𝑖𝑒𝑠 = 1 𝐶 𝐶 𝐶 3

17. Examples – Capacitors in series and parallel