Peoples Physics Concepts

http://www.ck12.org Chapter 13. Electric Circuits: Batteries and Resistors

13.2 Ohm’s Law

• Describe the parts of a basic electric circuit and apply Ohm’s law.

Students will learn a basic understanding of electric circuits and how to apply Ohm’s law.

Key Equations

Ohm’s Law

V=IR; Voltage drop equals current multiplied by resistance.

Power

P=IV ; Power released is equal to the voltage multiplied by the current.

Guidance

• Ohm’s Law is the main equation for electric circuits but it is often misused. In order to calculate the voltage
  drop across a light bulb use the formula:Vlightbulb=IlightbulbRlightbulb. For thetotalcurrent flowing out of the
  power source, you need thetotalresistance of the circuit and thetotalvoltage:Vtotal=ItotalRtotal.


• The equations used to calculate thepowerdissipated in a circuit isP=IV. As with Ohm’s Law, one must
  be careful not to mix apples with oranges. If you want the power of the entire circuit, then you multiply the
  totalvoltage of the power source by thetotalcurrent coming out of the power source. If you want the power
  dissipated (i.e. released) by a light bulb, then you multiply thevoltage dropacross the light bulb by thecurrent
  going through that light bulb.


• Power is the rate that energy is released. The units for power are Watts(W), which equal Joules per second
  [W] = [J]/[s]. Therefore, a 60 W light bulb releases 60 Joules of energy every second.

Example 1

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/320

Example 2

A small flash light uses a single AA battery which provides a voltage of 1.5 V. If the bulb has a resistance of 2Ω,

how much power is dissipated by the light bulb.