H5.3a Seriecircuits TV3B

5.3a Seriescircuits
Connecting resistors
1 / 18
next
Slide 1: Slide
NatuurkundeMiddelbare schoolvwoLeerjaar 3

This lesson contains 18 slides, with interactive quizzes and text slides.

time-iconLesson duration is: 45 min

Items in this lesson

5.3a Seriescircuits
Connecting resistors

Slide 1 - Slide

Last week
  • (I,U)-diagram is a graph in which the current is plotted against the voltage.
  • A resistor obeys Ohm's law, resistance is always constant.
  • A light bulb does not, because the resistance of the filament increases as the wire heats up.

Slide 2 - Slide

The package of an LED light bulb reads: 12 V/ 100 mA.
Calculate the resistance of the light bulb when it is lit at the correct voltage.

Slide 3 - Open question

Elaboration:
Given:
U = 12 V
I = 100 mA = 0,100 A

Asked:
R = ?

Elaboration: R =      =              = 120 Ω
IU
0,10012

Slide 4 - Slide

This lesson
Lesson goals:
  • you know what a series circuits is.
  • you can calculate the voltage across each resistor in a series circuit.
  • you can calculate the equivalent resistance of the resistors in a series circuit.

Slide 5 - Slide

Serie circuit
Parallel circuit
Mixed circuit

Slide 6 - Slide

Serie circuit

Slide 7 - Slide

Current
  • rule: current flows from + to - (elektrons from - to +)
  • The current is the same everywhere in a series circuit (because the current can only go through 1 'path')
  • I = I1 = I2 = I3 = ...

serie circuits

Slide 8 - Slide

Voltage
  • The voltage divides among the various components in a series circuit.
  • Utot = U1 + U2 + ...
serie circuits

Slide 9 - Slide

Equivalent resistance
  • more resistors in an circuit  --> bigger resistance
  • Req = R1 + R2 + R3 + ...
  • Req = equivalent resistance = the total resistance in Ω
serie circuits

Slide 10 - Slide

=
10
Utot = U1 + U2 + ...
I = I1 = I2 = I3 = ...
R3 = U/I = 6/0.12 = 50Ω
R3 = R1 + R2
1,2 V
4,8 V
1,2 V
0,12 A
0,12 A
40
10

Slide 11 - Slide

=

Slide 12 - Slide

Circuitry Analysis
  1. Simplify all parallel 'blocks'.
  2. Calculate Req.
  3. Calculate Isource.
  4. Calculate the voltage across each 'block'.
  5. Calculate the individual current through e.ach resistor

Slide 13 - Slide

Exercise 
Resistance (Ω)
Current (A)
Voltage (v)
R1
5000
R2
3300
R3
500
R4
100
Assignement: Complate the table shown on the bottom of this slide. Analyze the circuit according to the instructions given.
Req = R1 + R2 + R3 + .....
Isource = I1 = I2 = I3 =.....
Usource = U1 + U2 + U3......
Req = R1 + R2 + R3 + R4 = 5000+3300+500+100 = 8900Ω
Isource = Usource/Req = 24/8900 = 0,00269

Stap 2: 
Req
Stap 3:
Isource
Stap 4: 
U1...
U = I x R
U1 = 0,00269 x 5000 = 13,5 V
U2 = 0,00269 x 3300 = 8,9 V
U3 = 0,00269 x 500 = 1,35 V
U4 = 0,00269 x 100 = 0,27 V

Slide 14 - Slide

Summary
  • Current is the same everywhere in series circuit.
  • The voltage distributes itself among the various components
  • Equivalent resistance (Req) can be calculated by adding the resistances together.

Slide 15 - Slide

Individuel assignment
Stap 2:
Calculate Req
Stap 3:
Calculate Isource
Stap 4:
Calculate the voltage across each 'block'.
Resistance (Ω)
Current (A)
Voltage (V)
R1
2000
R2
1470
R3
2200
R4 
1850
Rtot = R1 + R2 + R3 + .....
Ibron = I1 = I2 = I3 =.....
Ubron = U1 + U2 + U3......
Homework: 1, 3, 4 
Questions? Send in the chat or put your mic on. 

Slide 16 - Slide

Individuele Assignment

Slide 17 - Open question

Homework
5.3:  3ab, 4, 5, 6

Slide 18 - Slide