Page 50 - Teach Yourself Electricity and Electronics

30 Electrical units

2-5 Total resistances and conductances for a wire having 10 ohms of
resistivity per kilometer.

Look again at the diagram of Fig. 2-4. There is a certain voltage across the resistor,
not specifically given in the diagram. There’s also a current flowing through the resis-
tance, not quantified in the diagram, either. Suppose we call the voltage E and the cur-
rent I, in volts and amperes, respectively. Then the power in watts dissipated by the
resistance, call it P, is the product E I. That is: P EI.
This power might all be heat. Or it might exist in several forms, such as heat, light
and infrared. This would be the state of affairs if the resistor were an incandescent light
bulb, for example. If it were a motor, some of the power would exist in the form of me-
chanical work.
If the voltage across the resistance is caused by two flashlight cells in series, giving
3 V, and if the current through the resistance (a light bulb, perhaps) is 0.1 A, then
E 3 and I 0.1, and we can calculate the power P, in watts, as:

(watts) EI 3 0.1 0.3 W

Suppose the voltage is 117 V, and the current is 855 mA. To calculate the power,
we must convert the current into amperes; 855 mA 855/1000 0.855 A. Then

P (watts) 117 0.855 100 W