8A

YOUR ANSWER: Volt.

You are not correct.

Earlier in your studies of basic electricity, you were told that the "volt" is a unit of

electromotive force. Now suppose you pushed against the bulkhead over there, surely you

are applying a pressure; but unless you moved that bulkhead, you could not do work; thus, no

work, no time rate for doing work. The same is true of that wall plug there in the bulkhead.

We know that there is approximately 110 volts available there, 110 volts of electromotive force

ready and able to do electrical work; but unless we use this voltage to make current flow, no

electrical work can be done; thus, no power can be consumed.

Return to frame 5A and select the correct answer.

8B

YOUR ANSWER: 3 amperes.

Very good. Just a little transposition of our basic power formula, P = E x I.

I= P and E = P.

E

I

P=E x l.

P = I2R

P = E2.

R

A certain soldering iron offers 200 ohms of resistance to 3 amperes of current flow.

The power used, then, is P = I2R = 3 x 3 x 200 = 9 x 200 = 1800 watts.

How much power is consumed by a circuit that has .5 ampere of current flowing through

a resistance of 500 ohms?

125 watts

frame 11A

250 watts

frame 13A

1-9

IT0348

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