Page 93 - Teach Yourself Electricity and Electronics

Resistances in series 73

Problem 4-10
Suppose that the voltmeter reads 12 V and the ammeter shows 50 mA. What is the
power dissipated by the potentiometer?
Use the formula P EI. First, convert the current to amperes, getting I 0.050 A.
(Note that the zero counts as a significant digit.) Then P EI 12 0.050 0.60 W.

You might say that this is 600 mW, although that is to three significant figures. It’s
not easy to specify the number 600 to two significant digits without using a means of
writing numbers called scientific notation. That subject is beyond the scope of this
discussion, so for now, you might want to say “600 milliwatts, accurate to two significant
figures.” (You can probably get away with “600 milliwatts” and nobody will call you on
the number of significant digits.)

Problem 4-11
If the resistance in the circuit of Fig. 4-7 is 999 and the voltage source delivers 3 V,
what is the dissipated power?
2
Use the formula P E /R 3 3/999 9/999 0. 009 W 9 mW. You are justi-
fied in going to only one significant figure here.
Problem 4-12
Suppose the resistance is 47 K and the current is 680 mA. What is the power dissi-
pated by the potentiometer?
2
Use the formula P I R, after converting to ohms and amperes. Then P 0.680
0.680 47,000 22,000 W 22 kW.

This is a ridiculous state of affairs. An ordinary potentiometer, such as the one you
would get at an electronics store, dissipating 22 kW, several times more than a typical
household. The voltage must be phenomenal. It’s not too hard to figure out that such a
voltage would burn out the potentiometer so fast that it would be ruined before the lit-
tle “Pow!” could even begin to register.
Problem 4-13
Just from curiosity, what is the voltage that would cause so much current to be driven
through such a large resistance?
Use Ohm’s Law to find the current: E IR 0.680 47, 000 32,000 V 32 kV.
That’s the sort of voltage you’d expect to find only in certain industrial/commercial ap-
plications. The resistance capable of drawing 680 mA from such a voltage would surely
not be a potentiometer, but perhaps something like an amplifier tube in a radio broad-
cast transmitter.

Resistances in series

When you place resistances in series, their ohmic values simply add together to get the
total resistance. This is easy to see intuitively, and it’s quite simple to remember.