Simplifying Circuits - Finding equivalent resistance

1. UNIT 2
Electric Circuits
AGORA International School
Technology
3 ESO

2. SERIES AND PARALLEL CIRCUITS
 Step 1. Analyze the circuit to find a section in which all resistors are either series or parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
 Step 4. Reconstruct the circuit step-by-step while analyzing individual resistors finding voltage
(V) and current (I).
SIMPLIFYING CIRCUITS
Most circuits are not in a basic series or parallel configuration, but rather consist of a complex
combination of series and parallel resistances. The key to simplifying circuits is to combine complex
arrangements of resistors into one main resistor. The general rules for solving these types of problems
are as follows:
Tip: Redraw the schematic after every step so you don't miss
an opportunity to simplify

3. R4
R2
R3
R1
ProblemB

4. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.

5. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R2 R1

6. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R2 R1
We can the equivalent resistance:

7. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:

8. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:

9. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series.
R3 R4

10. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series.
R3 R4
RE 3-4 = R3+ R4 = 4 Ω + 5 Ω = 9 Ω
We can the equivalent resistance:

11. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series.
R3 R4
RE 3-4 = R3+ R4 = 4 Ω + 5 Ω = 9 Ω
We can the equivalent resistance:
Redraw the simplified circuit:
RE 1-2
RE 3-4

12. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series.
R3 R4
RE 3-4 = R3+ R4 = 4 Ω + 5 Ω = 9 Ω
We can the equivalent resistance:
Redraw the simplified circuit:
RE 1-2
RE 3-4

13. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series. 3) Resistors E1-2 and E2-3 are in parallel.
R3 R4
RE 3-4 = R3+ R4 = 4 Ω + 5 Ω = 9 Ω
We can the equivalent resistance:
Redraw the simplified circuit:
RE 1-2
RE 3-4
𝑅 𝐸1−2
𝑅 𝐸3−4

14. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series. 3) Resistors E1-2 and E2-3 are in parallel.
R3 R4
RE 3-4 = R3+ R4 = 4 Ω + 5 Ω = 9 Ω
We can the equivalent resistance:
Redraw the simplified circuit:
RE 1-2
RE 3-4
We can the equivalent resistance:
1
𝑅 𝑇
=
1
𝑅 𝐸1−2
+
1
𝑅 𝐸3−4
=
1
4Ω
+
1
9Ω
1
𝑅 𝑇
= 0,36
1
Ω
→ RT =
1
0,36
1
Ω
= 2,8 𝜴
𝑅 𝐸1−2
𝑅 𝐸3−4

15. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series. 3) Resistors E1-2 and E2-3 are in parallel.
R3 R4
RE 3-4 = R3+ R4 = 4 Ω + 5 Ω = 9 Ω
We can the equivalent resistance:
Redraw the simplified circuit:
RE 1-2
RE 3-4
We can the equivalent resistance:
1
𝑅 𝑇
=
1
𝑅 𝐸1−2
+
1
𝑅 𝐸3−4
=
1
4Ω
+
1
9Ω
1
𝑅 𝑇
= 0,36
1
Ω
→ RT =
1
0,36
1
Ω
= 2,8 𝜴
RT
Redraw the simplified circuit:
𝑅 𝐸1−2
𝑅 𝐸3−4

16. ProblemB
R4
R2
R3
R1
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
1) Resistors 1 and 2 are in series.
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
RE 1-2 = R1 + R2 = 2 Ω + 2 Ω = 4 Ω
R4
RE 1-2
R3
R2 R1
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in series. 3) Resistors E1-2 and E2-3 are in parallel.
R3 R4
RE 3-4 = R3+ R4 = 4 Ω + 5 Ω = 9 Ω
We can the equivalent resistance:
Redraw the simplified circuit:
RE 1-2
RE 3-4
We can the equivalent resistance:
1
𝑅 𝑇
=
1
𝑅 𝐸1−2
+
1
𝑅 𝐸3−4
=
1
4Ω
+
1
9Ω
1
𝑅 𝑇
= 0,36
1
Ω
→ RT =
1
0,36
1
Ω
= 2,8 𝜴
RT
Redraw the simplified circuit:
𝑅 𝐸1−2
𝑅 𝐸3−4

17. ProblemC
R3
R4
R1
R2

18. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2

19. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
R1
R2

20. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴

21. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2

22. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in parallel.
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2
R3
R4

23. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in parallel.
We can the equivalent resistance:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2
R3
R4
1
𝑅 𝐸3−4
=
1
𝑅3
+
1
𝑅4
=
1
4Ω
+
1
5Ω
1
𝑅 𝐸3−4
=
0,45
Ω
→ 𝑹 𝑬 𝟑−𝟒
=
1
0,45
Ω
= 2,2 𝜴

24. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2
R3
R4
1
𝑅 𝐸3−4
=
1
𝑅3
+
1
𝑅4
=
1
4Ω
+
1
5Ω
1
𝑅 𝐸3−4
=
0,45
Ω
→ 𝑹 𝑬 𝟑−𝟒
=
1
0,45
Ω
= 2,2 𝜴
RE 1-2RE 3-4

25. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2
R3
R4
1
𝑅 𝐸3−4
=
1
𝑅3
+
1
𝑅4
=
1
4Ω
+
1
5Ω
1
𝑅 𝐸3−4
=
0,45
Ω
→ 𝑹 𝑬 𝟑−𝟒
=
1
0,45
Ω
= 2,2 𝜴
RE 1-2RE 3-4

26. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in parallel. 3) Resistors E1-2 and E2-3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2
R3
R4
1
𝑅 𝐸3−4
=
1
𝑅3
+
1
𝑅4
=
1
4Ω
+
1
5Ω
1
𝑅 𝐸3−4
=
0,45
Ω
→ 𝑹 𝑬 𝟑−𝟒
=
1
0,45
Ω
= 2,2 𝜴
RE 1-2RE 3-4
RE 1-2RE 3-4

27. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in parallel. 3) Resistors E1-2 and E2-3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
We can the equivalent resistance:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2
R3
R4
1
𝑅 𝐸3−4
=
1
𝑅3
+
1
𝑅4
=
1
4Ω
+
1
5Ω
1
𝑅 𝐸3−4
=
0,45
Ω
→ 𝑹 𝑬 𝟑−𝟒
=
1
0,45
Ω
= 2,2 𝜴
RE 1-2RE 3-4
𝑅 𝑇 = 𝑅 𝐸1−2
+𝑅 𝐸3−4
= 1𝛺 + 2,2𝛺
𝑹 𝑻 = 3,2 𝛀
RE 1-2RE 3-4

28. ProblemC R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R3
R4
R1
R2
1) Resistors 1 and 2 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 3 and 4 are in parallel. 3) Resistors E1-2 and E2-3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
We can the equivalent resistance:
Redraw the simplified circuit:
R1
R2
1
𝑅 𝐸1−2
=
1
𝑅1
+
1
𝑅2
=
1
2Ω
+
1
2Ω
1
𝑅 𝐸1−2
= 1
1
Ω
→ 𝑹 𝑬 𝟏−𝟐
=
1
1
1
Ω
= 1 𝜴
R3
R4
RE 1-2
R3
R4
1
𝑅 𝐸3−4
=
1
𝑅3
+
1
𝑅4
=
1
4Ω
+
1
5Ω
1
𝑅 𝐸3−4
=
0,45
Ω
→ 𝑹 𝑬 𝟑−𝟒
=
1
0,45
Ω
= 2,2 𝜴
RE 1-2RE 3-4
𝑅 𝑇 = 𝑅 𝐸1−2
+𝑅 𝐸3−4
= 1𝛺 + 2,2𝛺
𝑹 𝑻 = 3,2 𝛀
RT
RE 1-2RE 3-4

29. ProblemD
R4
R2R3 R1

30. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.R4
R2R3 R1

31. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
R4
R2R3 R1
R2R3

32. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺

33. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺
R4
𝑅 𝐸2−3 R1

34. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3

35. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel
We can the equivalent resistance:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
6Ω
+
1
5Ω
1
𝑅 𝐸2−3−4
=
0,37
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,37
Ω
= 2,7 𝜴
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3

36. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel
We can the equivalent resistance:
Redraw the simplified circuit:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
6Ω
+
1
5Ω
1
𝑅 𝐸2−3−4
=
0,37
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,37
Ω
= 2,7 𝜴
𝑅1𝑅 𝐸2−3−4
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3

37. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel 3) Resistors 𝑅 𝐸2−3−4
and 𝑅1 are in
series
We can the equivalent resistance:
Redraw the simplified circuit:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
6Ω
+
1
5Ω
1
𝑅 𝐸2−3−4
=
0,37
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,37
Ω
= 2,7 𝜴
𝑅1𝑅 𝐸2−3−4
R1𝑅 𝐸2−3−4
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3

38. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel 3) Resistors 𝑅 𝐸2−3−4
and 𝑅1 are in
series
We can the equivalent resistance:
Redraw the simplified circuit:
We can the equivalent resistance:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
6Ω
+
1
5Ω
1
𝑅 𝐸2−3−4
=
0,37
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,37
Ω
= 2,7 𝜴
𝑅1𝑅 𝐸2−3−4
𝑅 𝑇 = 𝑅 𝐸2−3−4
+𝑅1= 2,7𝛺 + 2𝛺
𝑹 𝑻 = 4,7 𝛀
R1𝑅 𝐸2−3−4
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3

39. ProblemD
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in series.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel 3) Resistors 𝑅 𝐸2−3−4
and 𝑅1 are in
series
We can the equivalent resistance:
Redraw the simplified circuit:
We can the equivalent resistance:
Redraw the simplified circuit:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
6Ω
+
1
5Ω
1
𝑅 𝐸2−3−4
=
0,37
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,37
Ω
= 2,7 𝜴
𝑅1𝑅 𝐸2−3−4
𝑅 𝑇 = 𝑅 𝐸2−3−4
+𝑅1= 2,7𝛺 + 2𝛺
𝑹 𝑻 = 4,7 𝛀
RT
R1𝑅 𝐸2−3−4
R4
R2R3 R1
R2R3
𝑅 𝐸2−3
= 𝑅2 +𝑅3= 2𝛺 + 4𝛺
𝑅 𝐸2−3
= 6 𝛺
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3

40. ProblemE
R2
R3
R4
R1

41. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
R2
R3
R4
R1

42. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
R2
R3
R4
R1
R2
R3

43. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

44. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
R4
𝑅 𝐸2−3 R1
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

45. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

46. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel
We can the equivalent resistance:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
1,3 Ω
+
1
5 Ω
1
𝑅 𝐸2−3−4
=
0,97
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,97
Ω
= 1,0 𝜴
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

47. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel
We can the equivalent resistance:
Redraw the simplified circuit:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
1,3 Ω
+
1
5 Ω
1
𝑅 𝐸2−3−4
=
0,97
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,97
Ω
= 1,0 𝜴
𝑅1𝑅 𝐸2−3−4
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

48. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel 3) Resistors 𝑅 𝐸2−3−4
and 𝑅1 are in
series
We can the equivalent resistance:
Redraw the simplified circuit:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
1,3 Ω
+
1
5 Ω
1
𝑅 𝐸2−3−4
=
0,97
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,97
Ω
= 1,0 𝜴
𝑅1𝑅 𝐸2−3−4
R1𝑅 𝐸2−3−4
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

49. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel 3) Resistors 𝑅 𝐸2−3−4
and 𝑅1 are in
series
We can the equivalent resistance:
Redraw the simplified circuit:
We can the equivalent resistance:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
1,3 Ω
+
1
5 Ω
1
𝑅 𝐸2−3−4
=
0,97
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,97
Ω
= 1,0 𝜴
𝑅1𝑅 𝐸2−3−4
𝑅 𝑇 = 𝑅 𝐸2−3−4
+𝑅1= 1,0𝛺 + 2𝛺
𝑹 𝑻 = 3 𝛀
R1𝑅 𝐸2−3−4
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

50. ProblemE
R1 = 2Ω
R2 = 2Ω
R3 = 4Ω
R4 = 5Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 2 and 3 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑅 𝐸2−3
and 4 are in parallel 3) Resistors 𝑅 𝐸2−3−4
and 𝑅1 are in
series
We can the equivalent resistance:
Redraw the simplified circuit:
We can the equivalent resistance:
Redraw the simplified circuit:
1
𝑅 𝐸2−3−4
=
1
𝑅 𝐸2−3
+
1
𝑅4
=
1
1,3 Ω
+
1
5 Ω
1
𝑅 𝐸2−3−4
=
0,97
Ω
→ 𝑹 𝑬 𝟐−𝟑−𝟒
=
1
0,97
Ω
= 1,0 𝜴
𝑅1𝑅 𝐸2−3−4
𝑅 𝑇 = 𝑅 𝐸2−3−4
+𝑅1= 1,0𝛺 + 2𝛺
𝑹 𝑻 = 3 𝛀
RT
R1𝑅 𝐸2−3−4
R4
𝑅 𝐸2−3 R1
R4
𝑅 𝐸2−3
R2
R3
R4
R1
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴

51. ProblemF
R2
R3
R4
R1

52. ProblemF
R1 = 2 Ω
R2 = 2 Ω
R3 = 4 Ω
R4 = 5 Ω
 Step 1. Analyze the circuit to find a section in which all resistors are either series or
parallel.
 Step 2. Reduce series and parallel configurations into equivalent resistances (RE).
 Step 3. Continue combining resistors until one, total resistance (RT) remains.
1) Resistors 𝑹 𝟐 and 𝑹 𝟑 are in parallel.
We can the equivalent resistance:
Redraw the simplified circuit:
2) Resistors 𝑹 𝑬 𝟐−𝟑
and 𝑹 𝟏 are in series 3) Resistors 𝑅 𝐸2−3−4
and 𝑅1 are in
parallel
We can the equivalent resistance:
Redraw the simplified circuit:
We can the equivalent resistance:
Redraw the simplified circuit:
𝑅 𝐸1−2−3
= 𝑅 𝐸2−3
+ 𝑅1= 1,3 𝛺 + 2𝛺
𝑹 𝑻 = 2,3 𝛀
RT
R1𝑅 𝐸2−3
R2
R3
1
𝑅 𝐸2−3
=
1
𝑅2
+
1
𝑅3
=
1
2Ω
+
1
4Ω
1
𝑅 𝐸2−3
=
0,75
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,75
Ω
= 1,3 𝜴
R2
R3
R1
R4
𝑅 𝐸2−3 R1
R4
R4
𝑅 𝐸1−2−3
1
𝑅 𝑇
=
1
𝑅 𝐸1−2−3
+
1
𝑅4
=
1
2,3Ω
+
1
5Ω
1
𝑅 𝐸2−3
=
0,63
Ω
→ 𝑹 𝑬 𝟐−𝟑
= 1
0,63
Ω
= 1,6 𝜴
𝑅 𝐸1−2−3
R4