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Handout #14: OHM’S LAW and JOULE’S LAW WORKSHEET

V = IR
Volts = Current x Resistance  –  or in units: Volts =Amps x Ohms (Ω)

V is the symbol for Volts, a measure of the amount of energy each charge has.  So if you have two negative charges, the voltage is a measure of how much the two charges don’t want to be near each other.  It is the energy available in the electrical field (provided by the battery) for each charge.  It is like the tension in the space between the charges.  Units of volts are… um, volts.

I is the symbol for Current, a measure of how many charges pass a point in a circuit in a given unit of time.  The unit for current is amps, or amperage.  One amp is defined as the current that exists when one coulomb of charge is passing a point in a circuit in one second.  1 amp = 1 C / s

R is Resistance, a measure of how hard it is for charges to move through a medium, be it a wire, a battery, or any other device in the circuit.  Units of resistance are ohms (W).  When charges move through a resistor, they lose some of their energy, or voltage.  Thus across a resistor there is a drop in voltage, in an amount that satisfies Ohm’s Law.  Where does that energy go?  It is transferred from the charges to the resistor, where it is changed into other forms (light, heat, motion, etc.).  Resistors are things like light bulbs, motors, or any other device that “uses” electricity.

New to your world is the concept of POWER.  Power is a combination of how much charges want to move in the presence of an electrical field (voltage), and how many of them actually do move past a point in a second (amperage).  Power is the rate at which energy is transported

Think of our circuit as a traffic situation:  each car is a negative charge, and is driven with a certain amount of energy (it’s speed, or voltage) along the highway (wire), where it encounters resistance (construction zones, speed limit signs, etc.).  The voltage is like the degree to which each car’s driver wants to get to work, which manifests in how hard they push on the gas pedal, or their speed.

Now suppose an enterprising person wants to set up a ‘resistor’ that uses some of the energy contained in the motion of the cars (their speed) to power a device – maybe this person wants to power a mill because he is a Miller.  So he puts a big treadmill in the road over which you are forced to drive.  As you drive over the treadmill you lose some of your speed (voltage), because the treadmill takes some of your energy into its own turning.  The treadmill belt turns a gear which turns a mill to make flour.

How can the Miller increase the turning of his mill if he is allowed to change the voltage (speed of the cars) and the current (number of cars passing over the treadmill per second)?  We will assume in this case that the resistance of the treadmill is constant.

Well, he can get more energy in the treadmill in two ways: he can increase the speed of the cars as they drive over the treadmill, or he can have more cars drive over the treadmill every second, or he can have both in any combination simultaneously.  The combination of these two quantities determines how much POWER is available to move the treadmill and the mill itself (the device powered by the circuit).

More cars per second at a particular speed means more pushes are given to the treadmill.  Cars with a higher speed give bigger pushes to the treadmill.  In both cases, the mill gets more total energy.

This results in the relationship known as Joule’s Law:

P = I x V
Power = Current x Voltage  –  or in units: Watts = Amps x Volts

Luckily our proportionality constant has a value of one when we use the units of volts, amperes, and watts.  A volt times an ampere is defined as one WATT, which is the standard unit of power (symbol P) in electronics.  Thus, when you plug your 60-watt light bulb into a 120 volt outlet, it uses ½ an amp (60 watts = I x 120 volts, so I = ½ amp).  Of course you could also get 60 watts of power if you had a circuit that provided 60 volts at one amp, or 30 volts at 2 amps, but in these cases your resistor (the light bulb) would need to be calibrated to operate correctly at these voltages and amperages (otherwise it might not work or it might short-circuit or explode!).

Using these two equations you should be able to solve all the problems in this worksheet.

 1. How many amperes does a 100 watt bulb require when operating on 120 volts?     2. A 250 watt radio will draw how many amps when operating on 240 volts?     3. How many ohms of resistance are present in a light bulb that draws 2 amperes at 120 volts?     4. A discman that is powered by two 1.5 volt batteries draws 0.0015 amperes.  What is the total resistance of the discman?     5. An iron that heats with the power of 1200 watts at 120 volts is turned on.  How much current does it draw?  Will a 15 ampere fuse blow?     6. How many ohms resistance are in a 120 volt hair dryer that draws 8 amps?  How many watts of power are used?     7. You have two equations, V = IR and P = IV.  Can you perform some substitutions and come up with new equations?  If you can, what might they mean?     8. A stereo needs 200 watts of power, and you know your house provides 120 volts of AC electricity.  How much current does the stereo draw?     9. How much resistance does the stereo in problem #8 give to the charges in the wire?     10. If the fuse on your circuit breaker can only take 10 amps, will plugging in your stereo from #8 blow the fuse if you also plug in your disco ball that draws 60 watts of power?

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