Page 90 - Teach Yourself Electricity and Electronics

70 Basic dc circuits

values can be assigned for the purpose of creating sample Ohm’s Law problems. While
calculating the current in the following problems, it is necessary to mentally “cover up”
the meter.

4-7 Circuit for working Ohm’s Law problems.

Problem 4-1
Suppose that the dc generator (Fig. 4-7) produces 10 V, and that the potentiometer is
set to a value of 10 . Then what is the current?
This is easily solved by the formula I E/R. Just plug in the values for E and R; they
are both 10, because the units were given in volts and ohms. Then I 10/10 1 A.

Problem 4-2
The dc generator (Fig. 4-7) produces 100 V and the potentiometer is set to 10 KΩ.
What is the current?
First, convert the resistance to ohms: 10 K Ω 10,000 . Then plug the values in:
I 100/10,000 0.01 A. This might better be expressed as 10 mA.

Engineers and technicians prefer to keep the numbers within reason when speci-
fying quantities. Although it’s perfectly all right to say that a current is 0.01 A, it’s best if
the numbers can be kept at 1 or more, but less than 1,000. It is a little silly to talk about
a current of 0.003 A, or a resistance of 107,000 Ω, when you can say 3 mA or 107 KΩ.

Problem 4-3
The dc generator (Fig. 4-7) is set to provide 88.5 V, and the potentiometer is set to 477
M . What is the current?
This problem involves numbers that aren’t exactly round, and one of them is
huge. But you can use a calculator. The resistance is first changed to ohms, giving
477,000,000 . Then you plug into the Ohm’s Law formula: I E/R 88.5/
477,000,000 0.000000186 A 0.186 uA. This value is less than 1, but there isn’t much