# Introduction to series circuits

```INTRODUCTION TO SERIES CIRCUITS
Welcome Back! We hope you had a restful spring break.
Seed Question:
What is a circuit?
What are the 5 basic parts of a circuit?
SERIES CIRCUITS ONLY HAVE ONE POSSIBLE
PATHWAY FOR ELECTRONS.
IN ADDITION TO THE 5 BASIC PARTS THAT WE HAVE
ALREADY DISCUSSED, THERE ARE TWO VERY
IMPORTANT MEASUREMENT DEVICES USED FOR
CIRCUIT LABS.
SERIES CIRCUITS OBEY SOME SPECIFIC RULES
REGARDING RESISTANCE, VOLTAGE, AND CURRENT.
TODAY IN THE VIRTUAL LAB YOU WILL TRY TO FIGURE
OUT WHAT THESE RULES ARE.
AS YOU ADD RESISTORS IN SERIES, THE
EQUIVALENT (TOTAL) RESISTANCE INCREASES.
ππ = π1 + π2 + π3 … … … … …
YOU CAN ALSO LOOK AT TOTAL EQUIVALENCY
(AKA EQUIVALENT RESISTANCE) AS SIMPLIFYING THE
CIRCUIT TO THE SIMPLEST FORM.
CURRENT IS THE SAME AT ALL LOCATIONS IN
THE CIRCUIT.
πΌπ = πΌ1 = πΌ2 = πΌ3 = πΌ4
AS YOU ADD RESISTORS IN SERIES
THE TOTAL AMOUNT OF CURRENT
DECREASES.
1.
2.
3.
π
πΌ=
π
THE VOLTAGE GAINED ACROSS THE BATTERY IS
EQUAL TO THE SUM OF THE VOLTAGE DROPS
ACROSS THE RESISTORS.
ππ = π1 + π2 + π3 … … …
TRY TO FIGURE OUT VOLTAGE DROP ON YOUR
OWN.
• Two 5Ω resistors are connected in series to a 12V battery.
Calculate the voltage drop across each resistor.
• A 5Ω and 10Ω resistor are connected in series to a 15V power
supply. Calculate the voltage drop across each resistor.
SOLVING SERIES CIRCUITS REQUIRES
ORGANIZATION.
Resistor
R
I
V
1
5Ω
0.5 A
2.5 V
2
1Ω
0.5 A
0.5 V
Total
6Ω
0.5 A
3V
TRY TO SOLVE A SERIES CIRCUIT ON YOUR
OWN.
3Ω
10 V
2Ω
Resistor
R
I
V
1
2Ω
2A
4V
2
3Ω
2A
6V
Total
5Ω
2A
10 V
JUST ONE MORE!
Resistor
R
I
V
1
150 Ω
0.013 A
2V
2
300 Ω
0.013 A
4V
Total
450 Ω
0.013 A
6V
```