1. Series and Parallel Resistors

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3. Different Ways of Combining Resistors+ - 5V + -
470W + - 5V 5V
1000W
220W
470W
1000W
10kW
1000W
Series
Parallel
Series and Parallel
Electrons flow in sequence through the resistors
Electrons divide to pass through each resistor
For analysis, multiple resistors must often be combined as a single equivalent resistance.

4. Series Resistors Find three resistors: 220Ω, 470Ω, & 1000Ω
Place them in series on the breadboard with two jumper wires
220W
470W
1000W
Do not plug up power!

5. Measure Resistance Combinations
220Ω
220Ω
220Ω
220Ω
470Ω
470Ω
470Ω
470Ω
1000Ω
1000Ω
1000Ω
1000Ω
𝑅 𝑒𝑞 =220Ω+470Ω+1000Ω=1690Ω
𝑅 𝑒𝑞 = 𝑅 1 + 𝑅 2 +⋯+ 𝑅 𝑛 = 𝑖=1 𝑛 𝑅 𝑖
Generalizing the Equivalent Resistance:

6. Measuring Resistance for Resistors in Parallel
Do not plug up power!
220Ω
470Ω
1000Ω
𝑅 𝑒𝑞 = Ω Ω Ω =130.3Ω
𝑅 𝑒𝑞 = 𝑅 𝑅 2 +⋯+ 1 𝑅 𝑛 = 1 𝑖=1 𝑛 1 𝑅 𝑖
Generalizing the Equivalent Resistance:
Now measure resistance across each individual resistor. What happens?

7. Class Problem: Consider the circuit below.
What is the total current supplied to the resistors (current that leaves the power source)?
How much power is consumed by the resistors?
Solution: + - 3V 2W 6W a.
𝑅 𝑒𝑞 = Ω + 1 6Ω = = = 6 4 =1.5Ω
𝑉=𝐼∙𝑅 ⟹𝐼= 𝑉 𝑅 = 3𝑉 1.5Ω =2𝐴 b.
𝑃=𝐼∙𝑉=3𝑉∙2𝐴=6𝑊

8. Combined Series & Parallel Resistors
Procedure:
STEP 1: Combine adjacent series resistors in series.
STEP 2: Compute the equivalent resistance of each “group” of parallel resistors.
STEP 3: Redraw the circuit with intermediate equivalent resistances.
STEP 4: Repeat steps 1, 2, and 3 until Req is found. R1 R2 + - V R3 R4

9. Find Req for the given circuit
50W
150W + - 5V
100W
100W
STEP 1:
STEP 2 and 3:
𝑅 3,4 = Ω Ω =50Ω
𝑅 1,2 =50Ω+150Ω=200Ω
R1,2 R3 R4 V
R1,2
R3,4 V
STEP 4:
𝑅 𝑒𝑞 = 𝑅 1,2 + 𝑅 3,4 =200Ω+50Ω=250Ω V
Req