Download presentation

Presentation is loading. Please wait.

Published byElijah Calhoun Modified over 4 years ago

1
**Applied Physics and Chemistry Circuits Lecture 1**

Current Electricity Applied Physics and Chemistry Circuits Lecture 1

2
**Electric Current Current is flow of charges**

There must be a complete loop (closed) for flow to occur Flow from high potential to low potential As potential decreases, work is done by charges Conventional current is flow of positive charges Do positive charges actually flow?

3
**Electric Current Current may be direct or alternating**

Direct current: flows in one direction Alternating current: flows back and forth

4
**Electric Circuit Required for current electricity**

Closed loop or conducting path from high to low potential Must have four parts: Source of charges (area of high potential) Path for charges Device that reduces potential energy Sink (area of low potential)

5
**Electric Circuit Energy is conserved in the circuit**

Charge is conserved in the circuit Energy carried by current depends on charge and potential difference ΔE = qV Remember, V is the potential difference 1 V = 1 J/C

6
**Electric Current Current is rate of flow of electric charge**

Measured in Coulombs / sec 1 C/s = 1 A (Ampere) Symbol for current is I

7
Electric Power Rate of energy transfer Measured in Watts 1W = 1 J/s

8
**Electric Power Since P = E/t And E = qV And q= It Then P = VI**

Power is current times the potential difference.

9
Equation for Amber In a circle! p i v

10
Electric Power If the current through a motor is 3 A and the potential difference is 120 V, what is the power of the motor? Known: I=3A V=120 V Equation: P=IV P=(3A)(120V)=360 W

11
**Electric Power Problem 2**

A 6 V battery delivers 0.5 A of current to an electric motor. What is the power rating of this motor? What we know: V = 6V I = 0.5 A Equation: P = IV Substitute: P = (0.5 A)(6 V) Solve! P = 3 W

12
**Electric Power Problem 2 Contd.**

How much energy does the motor use in 5.0 minutes? What we know: P = 3 W t = 5.0 min = 300 s Equation: P = W/t which means P = E/t and E = Pt Substitute: E = (3 W)(300 s) Solve! E = 300 J

Similar presentations

OK

Chapter 20: Circuits Current and EMF Ohm’s Law and Resistance

Chapter 20: Circuits Current and EMF Ohm’s Law and Resistance

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google