1. Today, we learn some tools derived from the 3 basic laws:
Equivalent resistance
Voltage division
Current division
They help you develop intuition about circuits:
To see relationship between variables
To see several steps ahead
To plan a solution to a circuit
Two terminal circuit with all resistors + 𝑣 − 𝑖
𝑅 𝑒𝑞 + 𝑣 − 𝑖
Behaves like a single resistor
𝑅 𝑒𝑞 ≜ 𝑣 𝑖

2. By KCL, same current flows through elements in series
Ways of connection
Series connection:
Two or more elements are in series if every connected pair exclusively share a single node, i.e., one node connects only two elements. i1 i2 i3
i1= i2= i3
By KCL, same current flows through elements in series
R1 and R2 are not in series because the node connecting them also connects a third element. R1 R2
Parallel connection:
Two or more elements are in parallel if they are connected between the same two nodes.
Node 1
Node 2
By KVL, same voltage across parallel elements:
v1=v2=v3

3. 𝑖 Determine series and parallel connection: R2 and R3 are in parallel,
How about 𝑅 1 𝑎𝑛𝑑 𝑅 2 ?
R1 and R2 are not in parallel.
𝑅 1 is between “a” and “c”, 𝑅 2 between “b” and “d”
Any series connection? R1 R2 R3 R4 R5
+ 𝑣 − 𝑖 𝑎 𝑏
𝑅 1 𝑎𝑛𝑑 𝑅 5 are not in series
𝑅 1 𝑎𝑛𝑑 𝑅 4 are not in series 𝑐 𝑑
No series connection between any two elements.
These wires cannot be thrown away.
They must be connected to somewhere to draw power

4. Which pair of resistors are in series? Which pair in parallel?
Node 1
Node 4
10Ω 𝑎𝑛𝑑 5Ω?
Not parallel, Not series 9 5
10
18
30V 𝐼𝑠 + - 4
4Ω 𝑎𝑛𝑑 9Ω?
Not parallel, Not series
4Ω 𝑎𝑛𝑑 5Ω?
In Series
− 𝑣 9 +
9Ω 𝑎𝑛𝑑 18Ω?
In parallel
Node 2
How many nodes in the circuit?
+ 𝑣 18 −
Only 4 nodes 9 5
10
18
30V 𝐼𝑠 + - 4
Node 3
𝑣 9 = 𝑣 18

5. § 2.5 Series resistors and voltage division
Given R1, R2.
What is 𝑣/𝑖 ?
By KCL, same current in R1,R2
By Ohm’s law, 𝑣1=𝑅1𝑖, 𝑣2=𝑅2𝑖
By KVL, 𝑣 = 𝑣1+𝑣2.
Putting together:
𝑣 =𝑣1+𝑣2=𝑅1𝑖+𝑅2𝑖=(𝑅1+𝑅2)𝑖
 𝑣 = (𝑅1+𝑅2)𝑖 (1)
𝑅𝑒𝑞
𝑣 = 𝑅𝑒𝑞𝑖 (2)
𝑅 𝑒𝑞 =?
Compare (1) and (2): 𝑣
𝑅𝑒𝑞=𝑅1+𝑅2

6. If Req=R1+R2+….+RN, same v i relationship.
In general: R1 R2 RN Rk
Req
If Req=R1+R2+….+RN, same v i relationship.
Req is called the equivalent resistance.
Req=R1+R2+….+RN

7. Voltage division R1 R2 Rk RN
Voltage division: Total voltage v is divided among the resistors in direct proportion to the resistances. Larger resistance takes more voltage.
Special case with two resistors: R1 R2 R1 R2

8. § 2.6 parallel resistors and current division
What is 𝑅 𝑒𝑞 =𝑣/𝑖 ?
By KVL, same voltage across 𝑅1, 𝑅2
By ohm’s law: 𝑖1=𝑣/𝑅1, 𝑖2=𝑣/𝑅2
By KCL: 𝑖=𝑖1+𝑖2
Req
Put together:
𝑅 𝑒𝑞 =?

9. L5
In general R1 R2
….. RN
Req

10. Equivalent resistance for parallel resistors:
Notation:
Special cases:
N=2:
R1=R2=…=RN=R:
Simple combinations:
Simple rule:
3//6=
12//6=
15//10=
20//5= 2
𝛼 𝑅 1 //𝛼 𝑅 2 = 𝛼 𝑅 1 ×𝛼 𝑅 2 𝛼 𝑅 1 +𝛼 𝑅 2 = 𝛼 𝑅 2 𝑅 2 𝑅 1 + 𝑅 2 =𝛼 (𝑅 1 // 𝑅 2 ) 4 6 4
27//54=
9(3//6)
=9×2=18

11. Adding more resistor to existing parallel ones reduces Req:
Common sense:
Adding more resistor to existing parallel ones reduces Req:
Equivalent resistance for parallel connection is less than any individual resistance
Equivalent conductance:

12. Current division: Given total 𝑖, what is 𝑖1,𝑖2 ? Req
The other resistance on top
The current is shared by resistors in inverse proportion to resistance. Larger resistor takes less current.

13. Case 1: A resistor in parallel with a short circuit
Two extreme cases:
Case 1: A resistor in parallel with a short circuit R1
R2=0
Case 2: A resistor in parallel with an open circuit R1
R2=∞

14. Equivalent resistance Examples4 2 1 5 3 6 8
Req 6 2 4 2 6 8
Req 4
2.4 8
Req
Req= =14.4Ω

15. Equivalent resistance Examples
10 3 1 5 4 12
Req 6
10 3 1 5 4 12
Req 6 1
10 2 6 3
Req
10 2
Req 1
Req=10+2//(1+2)
=10+2//3
=10+1.2
=11.2Ω

16. Need to find the equivalent resistance with respect to 40V
30Ω
18Ω 8Ω 9Ω
14Ω
50Ω
40V
20Ω + −
𝐼 𝑠
Example: Compute 𝐼 𝑠
𝑅 𝑒𝑞1 = 30//(20+50)
= 30//70
= 21Ω
𝑅 𝑒𝑞1
𝑅 𝑒𝑞2 = 14+9//18 =
= 20Ω
𝑅 𝑒𝑞2
Need to find the equivalent resistance with respect to 40V
21Ω 8Ω
20Ω
40V 9Ω + −
𝐼 𝑠
𝑅 𝑒𝑞
𝑅 𝑒𝑞 =8+20//(21+9)
= 8+12 =20Ω
𝐼 𝑠 = =2𝐴

17. Example: Compute 𝑅 𝑒𝑞 𝑅 𝑒𝑞 = 10// (12+4//12) = 10 // 15 =6 Ω 12Ω 10Ω
Req
12Ω 4Ω
10Ω
12Ω
10Ω 4Ω
Req
12Ω
10Ω
12Ω 4Ω
𝑅 𝑒𝑞 = 10// (12+4//12)
= 10 // 15 =6 Ω
Req

18. Practice 6: Find 𝑅 𝑒𝑞 Req Req Practice 7: Find 𝐼 𝑠 𝐼𝑠 4 10 5 + -3
20 4 4 5
15
Req
15
Req
18 9 6 8 9 5
10
18
30V 𝐼𝑠 + - 4
Practice 7: Find 𝐼 𝑠

19. Practice 8: Find 𝑣𝑠 Practice 8: Find 𝑣𝑠 15 10 15 10 14 30 2A 143
30
Practice 8: Find 𝑣𝑠
14
10
15 2A
15
10
14
30 2A 3
Practice 9: Find 𝑣 1 , 𝑣 2
Practice 10: Find 𝑖 1 , 𝑖 2
+ 𝑣 1 −
+ 𝑣 2 −
𝑖 1 4 4 6 4 2 3𝐴
15V
+ -
𝑖 2

20. Three useful tools derived from basic laws:
Equivalent resistance
Voltage division
Current division
Used together to solve circuit problems R1 R2
Voltage division: R1 R2 R1 R2
Current Division: 𝑖
𝑅 1
𝑅 2

21. Approach 1: Assign auxiliary variable 𝐼.4 6 3
Example: Find i0 and v1.
Approach 1: Assign auxiliary variable 𝐼.
Use equivalent resistance with respect to 12V to find I, then v1=4I, and i0 can be computed by current division.
Req= 4 + 6//3 =4+2=6
By Ohm’s Law, I=12/Req=12/6=2A
Thus v1= 4I=8V.
Req
By current division:
Be Careful: In this equivalent circuit, only I is the same as in the original circuit. Neither v1 nor i0, can be found in it. You need to go back to the original circuit to find v1 and i0.

22. Approach 2: Use equivalent resistance of 6//3, denoted as R’eq.4 6 3 a b
Approach 2: Use equivalent resistance of 6//3, denoted as R’eq. 4
R’eq
R’eq=3//6=2. By voltage division, a
Be careful, i0 cannot be found in the equivalent circuit. i0≠ I. You have to use the original circuit to find i0. b
Where is 𝑣 2 in the original circuit?
Since the voltage across 3 is v2, by Ohm’s Law, i0=v2/3=4/3A

23. Approach 1: Which is the voltage across 3A ?
Example: Find v1, v2, 4
15 8
10 3A 3A
𝑅 1
𝑅 2
4+8=
=15//10 3A
𝑅 1
𝑅 2
Approach 1: Which is the voltage across 3A ?
𝑣 1 𝑜𝑟 𝑣 2 ?
It is 𝑣 If you know the Req w.r.t 3A, you can obtain 𝑣 2 by ohm’s law.
𝑅 𝑒𝑞 = 𝑅 1 // 𝑅 2
𝑣 2 = 𝑅 𝑒𝑞 ×3
=4×3=12V 3A
𝑅 𝑒𝑞 = (4+8)//15//10
How to find 𝑣 1 ?
𝑅 𝑒𝑞
Need the original circuit.
= 12//6
By voltage division,
= 4Ω
𝑣 1 = × 𝑣 2 = 4 12 ×12=4𝑉

24. Be careful, you cannot find 𝑣 1 in the simplified circuit.
𝐼 1
𝐼 2
Be careful, you cannot find 𝑣 1 in the simplified circuit.
Have to use the original circuit to find 𝑣 1 4
15 8
10 3A
Approach 2: Use current division. Assign 𝐼 1 , 𝐼 2 .
𝑅 1 =4+8=12Ω 3A
𝑅 1
𝑅 2
𝐼 1
𝐼 2
𝑅 2 =15//10=6Ω
+ 𝑣 2 −
By current division,
𝐼 1 = 𝑅 2 𝑅 1 + 𝑅 2 ×3= 6 18 ×3=1𝐴
𝐼 2 = 𝑅 1 𝑅 1 + 𝑅 2 ×3= ×3=2𝐴
By ohm’s law, 𝑣 1 =4 𝐼 1 =4𝑉; 𝑣 2 = 𝑅 2 𝐼 2 =6×2=12𝑉.

25. Outline of the solution:
Example: Find 𝐼 𝑠 , 𝐼 1 , 𝑣 𝑎𝑏
𝐼 1 8
𝐼 2
25
𝐼 𝑠
+ 𝑣 1 −
− 𝑣 2 +
15 3 a
+ 𝑣 𝑎𝑏 − b
10
20 + -
25
27V
15
12
Outline of the solution:
Find 𝐼 𝑠 first, then use current division to find 𝐼 1 .
To find 𝑣 𝑎𝑏 , need to find 𝑣 1 , 𝑣 2 , then use KVL
𝑣 2 + 𝑣 𝑎𝑏 + 𝑣 1 =0⇒ 𝑣 𝑎𝑏 = −𝑣 2 − 𝑣 1
To find 𝑣 1 , need the current through 15Ω.
Assign 𝐼 2 ,
𝐼 2 can be computed by current division on 𝐼 1

26. Type equation here.Type equation here.
Step 1: Compute I s
Type equation here.Type equation here.
𝐼 1 8
𝑅 𝑒𝑞1
25
𝐼 𝑠
15
𝑅 𝑒𝑞2 3
10
20 + -
𝑅 𝑒𝑞1 =10+8=18Ω
25
27V
15
12
𝑅 𝑒𝑞2 =12+(15+25)\\( )
=12+40\\ 60 =12+24=36Ω
𝑅 𝑒𝑞1 3
27V + -
𝐼 𝑠
𝐼 1
𝑅 𝑒𝑞2
𝑅 𝑒𝑞 w.r.t. 27V:
𝑅 𝑒𝑞 =3+ 𝑅 𝑒𝑞1 \\ 𝑅 𝑒𝑞2 = \\18
=3+6 6\\3 =3+6×2=15Ω
𝐼 𝑠 = 27 𝑅 𝑒𝑞 = =1.8𝐴
𝑅 𝑒𝑞1 𝑅 𝑒𝑞1 + 𝑅 𝑒𝑞2 × 𝐼 𝑠
Step 2: By current division: 𝐼 1 =
= ×1.8=0.6𝐴

27. Step 3: Go back to the original circuit to find 𝑣 𝑎𝑏 :
𝐼 1 8
25
𝐼 2
+ 𝑣 1 −
𝐼 𝑠
− 𝑣 2 +
15 3 a
+ 𝑣 𝑎𝑏 − b
10
20
𝐼 𝑠 =1.8𝐴;
𝐼 1 =0.6𝐴 + -
25
27V
15
12
𝐼 1 is divided by 15Ω+25Ω=40Ω and 25Ω+20Ω+15Ω=60Ω
𝐼 2 = ×0.6=0.36𝐴
𝑣 1 =15 𝐼 2 =15×0.36=5.4𝑉
𝑣 2 =3 𝐼 𝑠 =3×1.8=5.4𝑉
v ab + v 1 + v 2 =0
𝑣 𝑎𝑏 =− 𝑣 1 − 𝑣 2 =−5.4−5.4=−10.8𝑉

28. Example: Find 𝑣 0 and the power absorbed by the dependent current source.
+ -
𝑯𝒊𝒏𝒕: 𝑈𝑠𝑒 𝐾𝐶𝐿 𝑡𝑜 𝑚𝑎𝑘𝑒 𝑎𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑣 0 2
4.5V 3
+ 𝑣 0 −
1.8A 2
0.2 𝑣 0 4 5
Solution:
Assign current 𝐼 1 2 3 4
0.2 𝑣 0
1.8A
+ -
4.5V
+ 𝑣 0 − 5
𝐼 1 𝒂
𝐼 1 is the same current through 2Ω
By Ohm’s Law,
𝐼 1 = 𝑣 0 2 =0.5 𝑣 0
KCL at node 𝒂:
𝑣 0 = 𝐼 1 =0.5 𝑣 0
⇒ 1.8=0.3 𝑣 0
⇒ 𝑣 0 =6𝑉

29. 𝐼 1 W𝑒 ℎ𝑎𝑣𝑒 : 𝑣 0 =6𝑉 Power absorbed by the dependent current source?𝒂
+ -
− 𝑣 3 + 2
4.5V
𝐼 1 3
Need voltage across 0.2 𝑣 0
+ 𝑣 0 −
− 𝑣 𝑠 +
1.8A 2
0.2 𝑣 0
Assign 𝑣 𝑠
− 𝑣 4 +
To apply KVL correctly,
Assign 𝑣 3 , 𝑣 4 5 4
KVL around right side loop:
𝑣 3 =?
𝑣 3 =3×0.2 𝑣 0
𝑣 𝑠 + 𝑣 𝑣 0 + 𝑣 4 =0
=3×0.2×6=3.6𝑉
𝑣 𝑠 =− 𝑣 3 −4.5 − 𝑣 0 − 𝑣 4
𝐼 1 = 𝑣 0 2 = 6 2 =3𝐴
𝑣 4 =?
𝑣 4 =4 𝐼 1 ,
=−3.6−4.5−6−12
=−26.1𝑉
𝑣 4 =4×3=12𝑉
𝑝 0.2 𝑣 0 = v s ×0.2 𝑣 0 =−26.1×0.2×6 =−31.32𝑊

30. Variations of the circuit + -2
4.5V 3
+ 𝑣 0 − 2 3 4
0.2 𝑣 0
1.8A
+ -
4.5V
+ 𝑣 0 − 5
1.8A 2
0.2 𝑣 0 4 5 2
99V
0.8 𝑣 0
1.8A
+ 𝑣 0 − 5 + -
+ -
15V 7
𝐼 1
𝐼 2
Same idea!
Express Branch current 𝐼 1 , 𝐼 2 In terms of 𝑣 0 ,
Then apply KCL
𝐼𝑓 power by a current source is needed, compute its voltage by using KVL

31. More current division: 𝐼 3 = 10 10+15 × 𝐼 1 =1.08𝐴
Example: Find v1 and v2.
Alternatively:
12
Ohms Law:
𝑣 𝑥 =6 𝐼 1 =16.2𝑉 2 8
Voltage division:
+ 𝑣 1 −
+ 𝑣 2 −
𝑣 1 = × 𝑣 𝑥 =9.72V 8 9
10 5
4.5A 6
More current division:
𝐼 3 = × 𝐼 1 =1.08𝐴 9 8 2
4.5A 6
10
12 5
+ 𝑣 1 −
+ 𝑣 2 −
𝐼 1
𝐼 2
𝐼 3
𝐼 4
𝑣 1 =9 𝐼 3 =9.72𝑉
+ 𝑣 𝑥 −
𝐼 4 = × 𝐼 2 =0.36𝐴
𝑣 2 =8 𝐼 4 =2.88𝑉
𝑅 𝑒𝑞2
𝑅 𝑒𝑞1
4.5A
𝐼 1
𝐼 2
𝑅 𝑒𝑞1
𝑅 𝑒𝑞2
Current division:
𝐼 1 = ×4.5=2.7𝐴
8 =
= 12
𝐼 2 = ×4.5=1.8𝐴

32. Example: Find the currents i1, …, i5+ -
18V 4 3
18 9 i3 i1 i2 i5 i4
Approach1: Use equivalence resistance and current division.
Req
𝑖 1
𝑅 𝑒𝑞 =4+3//9//18=6Ω
𝑖 1 = 18 6 =3𝐴 + -
18V 4 3 6 i3 i1 i2
𝑖 2 = × 𝑖 1 = 6 9 ×3=2𝐴
𝑖 3 = × 𝑖 1 = 3 9 ×3=1𝐴
𝑖 4 = × 𝑖 3 = ×1= 2 3 𝐴
𝑖 5 = × 𝑖 3 = 9 27 ×1= 1 3 𝐴

33. Example: Find the currents i1, …, i5+ -
18V 4 3
18 9 i3 i1 i2 i5 i4
Approach2: Use voltage division and Ohm’s law . Assign 𝑣 2 .
By voltage division 𝑣 2 = ×18=6𝑉 + -
18V 4 i1
By Ohm’s law, 𝑖 1 = 𝑣 2 2 = 6 2 =3𝐴
𝑖 2 = 𝑣 2 3 = 6 3 =2𝐴
3//9//18=2
𝑖 5 = 𝑣 = 6 18 = 1 3 𝐴
𝑖 4 = 𝑣 2 9 = 6 9 = 2 3 𝐴
𝑖 3 = 𝑖 4 + 𝑖 5 = =1𝐴

34. In all three circuits, 𝑖 1 , 𝑖 2 , 𝑖 3 , 𝑖 4 are the same
Example: Find 𝑖 1 , 𝑖 2 , 𝑖 3 , 𝑖 4 , 𝐼 𝑥
20
10
30 3A
𝑖 1
𝑖 2
𝑖 4
𝑖 3
𝑖 2
𝑖 1
10
20
extend 3A
Shrink to one point 𝐼𝑥
20
30
𝑖 3
𝑖 4
20
10
30 3A
𝑖 1
𝑖 2
𝑖 4
𝑖 3
𝐼 0
In all three circuits, 𝑖 1 , 𝑖 2 , 𝑖 3 , 𝑖 4 are the same
How about 𝐼 𝑥 , 𝐼 0 ?
By KCL, 𝐼 0 = 𝑖 1 + 𝑖 2 =?
𝐼 0 = 𝑖 1 + 𝑖 2 = 𝑖 3 + 𝑖 4 =3𝐴
𝐼 𝑥 = 𝑖 1 − 𝑖 3 = 𝑖 4 − 𝑖 2
By current division:
𝑖 1 = ×3=2𝐴; 𝑖 2 =1𝐴;
𝑖 3 =1.8𝐴; 𝑖 4 =1.2𝐴;
𝐼 𝑥 = 𝑖 1 − 𝑖 3 =0.2𝐴
𝐼 𝑥 = 𝑖 4 − 𝑖 2 =0.2𝐴

35. Test 1 will be given on Sept 30 (Monday), 11-11:50am
Test 1 will be given on Sept 30 (Monday), :50am. In Ball Hall 210 There will be two versions in different colors.
Please arrive 5-10 minutes earlier
A practice exam will be given on 9/25/2019(Wednesday), 11-11:50am in Ball Hall 210
Solution to practice exams will be posted at website
Solution to all practice problems in lecture note will be posted.

36. R6
30V 4 6 3
15 + -
+ 𝑣 4 −
𝐼 1
𝐼 2
Practice 12: Find 𝐼 1 , 𝐼 2 , 𝑣 4
Practice 13: Find 𝐼 0 , 𝑣 4
𝐼 0 5 8 6
12 4
3.5A

37. R6
Practice 14: Find 𝐼 0 , 𝑣 2 6 6
𝐼 0
20𝑉 + - 6
10
15

38. Practice 15: Find the voltage 𝑣𝑥
+ -
30V
12 6
18
30V
10
Practice 16: Find 𝑣0, 𝐼 𝑥 , 6
15
12 4
𝐼 𝑥 + -

39. R6
Practice 16a: Find v1, I0, 9 5
20
18 5A
15 I0

40. R7
Practice 17: Find 𝑣 𝑥 , 𝐼,
14 6
15
24 + -
𝑣 𝑥 −
36V
16
12 𝐼

41. 8 4 3 3A 8 24V + - 4 3 R7 Practice 18: Find i0 i0

42. Practice 20: Find R for the circuit6 R
20V + -
30V 4 6
𝑉 𝑠 + - 4
Practice 21: Find Vs for the circuit
10 9
I1=2A

43. Practice 22: Find R so that 𝐼 𝑠 is 10A
10
Is=10A 4 + -
8/3
64𝑉 R