Download presentation

Presentation is loading. Please wait.

Published byElfrieda Phelps Modified over 2 years ago

1
Physics Mrs. Coyle

2
Kirchhoff’s Rules Series Circuits Equivalent Resistance Voltage Drop Across Resistors Brightness of Bulbs in a Series Circuit

3
There is one current path. All resistors have the same current.

5
Positive charges are “pumped” by the battery from low to high potential. V>0 When traversing a resistor with the current, there is a decrease in potential. V<0

6
1 st Rule: (Junction Theorem): At a junction (node), current in= current out 2 nd Rule: (Loop Theorem): In a closed loop the sum of the voltages is zero.

7
In a series circuit the total voltage drop across the resistors equals the sum of the individual voltages. V = V 1 + V 2 + V 3

8
If the battery’s voltage is 12V and the voltage across R 1 is 5 V, and across R 2 is 4V, find the voltage across R 3. Answer: 3V

9
V = V 1 + V 2 + V 3 Using Ohm’s Law: IR eq = IR 1 +IR 2 +IR 3 Equivalent resistance R eq = R 1 + R 2 + R 3

10
If the battery’s voltage is 12V and R 1 = 1 Ω R 2 = 2 Ω R 3 = 3 Ω Find the equivalent resistance. Find the current. Find the voltage across each resistor. Answer: 6 Ω, 2A, 2V, 4V, 6V

11
The greater the power actually used by a light bulb, the greater the brightness. Note: the power rating of a light bulb is indicated for a given voltage, at room temperature and the bulb may be in a circuit that does not have that voltage.

12
P= I V P=I 2 R P=V 2 / R

13
Find the total resistance. Find the current. Find the power dissipated in each lamp. Which light bulb will be the brightest and why? Find the total power. How does the total power compare to the powers of the individual bulbs. Ans: 450Ω, 0.027A, 0.18W, 0.036W, 0.109W, 250 Ω, 0.324W 250Ω 150Ω 50Ω 12 V

14
Parallel Circuits Equivalent Resistance Brightness of Light Bulb Combination Circuits

15
There is more than one current path. The voltage across the resistors is the same. http://www1.curriculum.edu.au/sciencepd/energy/images/energy_ill112.gif

16
I = I 1 + I 2 + I 3 V =V 1 =V 2 =V 3 Using Ohm’s Law: V/R eq = V/R 1 +V/R 2 + V/R 3 Equivalent Resistance: 1/R eq = 1/R 1 +1/R 2 + 1/R 3

18
Find the R eq, I’s. How does R eq compare with each R? Ans: 0.55 Ω, I= 22A, (12A, 6A, 4A) =1Ω =3Ω =2Ω 12V

19
Why should you not plug in too many appliances in the same outlet in a home?

21
Ans: 11 Ω, 1.8A, V 1 =9V, V 2 =11V, I 2 =1.1A, I 3 =0.7A =20V =10Ω =15Ω =5Ω

22
http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/DC/00123.png

23
R eq 1 = 71.4 Ω R eq 2 = 127.3 Ω R eq = 198.7 Ω I=0.12A V 1 = 8.6V V 2 = 15.3V

Similar presentations

OK

Circuits Series and Parallel. Series Circuits Example: A 6.00 Ω resistor and a 3.00 Ω resistor are connected in series with a 12.0 V battery. Determine.

Circuits Series and Parallel. Series Circuits Example: A 6.00 Ω resistor and a 3.00 Ω resistor are connected in series with a 12.0 V battery. Determine.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on c language history Ppt on p&g products rebates list Ppt on polymerase chain reaction Full ppt on electron beam machining set Ppt on electronic payment systems Ppt on swami vivekananda teachings Ppt on various types of web browsers and their comparative features Ppt on environmental pollution in hindi Ppt on operational research Ppt on transportation of substances in plants and animals