Electric Current: Currents, Resistance, and Power

Let’s Review…

You’ve already learned a lot in this section. It’s time to apply this information to some problem solving. First, take a look at the vocabulary review so you know you are ready to take on those terms in the problems. Then, move on from there for an example and some self-check problems.

self check Vocabulary Self Check

Now, you try. Match the following terms with the proper definitions before moving on to your problem solving.

 

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tutorial Tutorial: Using Coulomb’s Law

In this example, you will see how to use Coulomb’s Law to solve a problem. Select the play button to begin the tutorial, and then use the navigation buttons to pause/stop, continue, or reset the tutorial. View the presentation as often as you would like, and take notes as you follow along. Be sure to set your volume at a reasonable level before you begin.

Download the script as a PDF.


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self check More Problems Self Check

Now, you try. Complete the self-check activity by looking at the questions below, working them out, and then clicking on the question to review the solution.

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A hair dryer draws 7.5 A when plugged into a 120-V line. (a) What is its resistance? (b) How much charge passes through it in 15 min? (Assume direct current.)
(a) Use Ohm’s Law to find resistance.

V equals I R which leads to R equals V over I equals 120 Volts over 7 point 5 Amps equals 16 ohms.

(b) Use the definition of current:

I equals delta Q over delta t which leads to Delta Q equals I times delta t equals 7 point 5 amps times 15 minutes times sixty seconds per minute equals six point 8 times ten to the third Coulombs.

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An electric device draws 6.50 A at 240 V. (a) If the voltage drops by 15%, what will be the current, assuming nothing else changes? (b) If the resistance of the device were reduced by 15%, what current would be drawn at 240 V?
a) If the voltage drops by 15%, and the resistance stays the same, then by V equals I R , the current will also drop by 15%.

I sub final equals zero point eight five times I initial equals zero point eight five times 6 point five zero amps equals 5 point 5 two 5 amps equals 5 point 5 Amps.

(b) If the resistance drops by 15% (the same as being multiplied by 0.85), and the voltage stays the same, then by V equals I R, the current must be divided by 0.85.

I sub final equals I sub initial over zero point eight five equals 6 point 5 zero amps over zero point eight five equals seven point 6 four 7 amps equals 7 point 6 Amps.

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A 12-V battery causes a current of 0.60 A through a resistor. (a) What is its resistance, and (b) how many joules of energy does the battery lose in a minute?
Use Ohm’s Law to find the resistance.

R equals V over I equals 12 volts over zero point six zero amps equals 20 ohms.

(b) An amount of charge delta Q loses a potential energy of Delta Q times V as it passes through the resistor. The amount of charge is found by:

Delta PE equals delta Q times V equals I times delta t times V equals zero point six zero amps times 60 seconds times 12 volts equals 430 Joules.

 

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