Electrical Resistance
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Power Revisited
The previous section of Lesson 3 elaborated upon the dependence of current upon the electric potential difference and the resistance. The current in an electrical device is directly proportional to the electric potential difference impressed across the device and inversely proportional to the resistance of the device. If this is the case, then the rate at which that device transforms electrical energy to other forms is also dependent upon the current, the electric potential difference and the resistance. In this section of Lesson 3, we will revisit the concept of power and develop new equations that express power in terms of current, electric potential difference and resistance.
In Lesson 2, the concept of electrical power was introduced. Electrical power was defined as the rate at which electrical energy is supplied to a circuit or consumed by a load. The equation for calculating the power delivered to the circuit or consumed by a load was derived to be
(Equation 1)
The two quantities that power depends upon are both
related to the resistance of the load by Ohm's
law. The electric potential difference (
V)
and the current (I) can be expressed in terms of their
dependence upon resistance as shown in the following
equations.
|
V
= (I • R) |
V
/ R |
If the expressions for electric potential difference and current are substituted into the power equation, two new equations can be derived that relate the power to the current and the resistance and to the electric potential difference and the resistance. These derivations are shown below.
|
P = P = (I • R) • I P = I2 • R |
P = P = P = |
V2
/ RWe now have three equations for electrical
power, with two derived from the first using the Ohm's law
equation. These equations are often used in problems
involving the computation of power from known values of
electric potential difference (V),
current (I), and resistance (R). Equation 2 relates the rate
at which an electrical device consumes energy to the current
at the device and the resistance of the device. Note the
double importance of the current in the equation as
denoted by the square of current. Equation 2 can be used to
calculate the power provided that the resistance and the
current are known. If either one is not known, then it will
be necessary to either use one of the other two equations to
calculate power or to use the Ohm's law equation to
calculate the quantity needed in order to use Equation
2.
Equation 3 relates the rate at which an
electrical device consumes energy to the voltage drop across
the device and to the resistance of the device. Note the
double importance of the voltage drop as denoted by
the square of V.
Equation 3 can be used to calculate the power provided that
the resistance and the voltage drop are known. If either one
is not known, then it will be important to either use one of
the other two equations to calculate power or to use the
Ohm's law equation to calculate the quantity needed in order
to use Equation 3.
While these three equations provide one
with convenient formulas for calculating unknown quantities
in physics problems, one must be careful to not misuse them
by ignoring conceptual 
principles
regarding circuits. To illustrate, suppose that you were
asked this question: If a 60-watt bulb in a household lamp
was replaced with a 120-watt bulb, then how many times
greater would the current be in that lamp circuit? Using
equation 2, one might reason (incorrectly), that the
doubling of the power means that the I2 quantity
must be doubled. Thus, current would have to increase by a
factor of 1.41 (the square root of 2). This is an example of
incorrect reasoning because it removes the mathematical
formula from the context of electric circuits. The
fundamental difference between a 60-Watt bulb and a 120-Watt
bulb is not the current that is in the bulb, but rather the
resistance of the bulb. It is the resistances that are
different for these two bulbs; the difference in current is
merely the consequence of this difference in resistance. If
the bulbs are in a lamp socket that is plugged into a
United States wall outlet, then one can be certain that the
electric potential difference is around 120 Volts. The
V
would be the same for each bulb. The 120-Watt bulb has the
lower resistance; and using Ohm's law, one would expect it
also has the higher current. In fact, the 120-Watt bulb
would have a current of 1 Amp and a resistance of 120
;
the 60-Watt bulb would have a current of 0.5 Amp and a
resistance of 240 .
|
P = I = P / I = (120 W) / (120 V) I = 1 Amp
R = R = (120 V) / (1 Amp) R = 120 |
P = I = P / I = (60 W) / (120 V) I = 0.5 Amp
R = R = (120 V) / (0.5 Amp) R = 240 |
Now using equation 2 properly, one can see why twice the power means that there would be twice the current since the resistance also changes with a bulb change. The calculation of current below yields the same result as shown above.
|
P = I2 • R I2 = P / R I2 = (120 W) / (120 I2 = 1 W / I = SQRT ( 1 W / I = 1 Amp |
P = I2 • R I2 = P / R I2 = (60 W) / (240 I2 = 0.25 W / I = SQRT ( 0.25 W / I = 0.5 Amp |
Check
Your Understanding
1. Which would be thicker (wider) - the filament of a 60-Watt light bulb or the filament of a 100-W light bulb? Explain.
2. Calculate the resistance and the current of a 7.5-Watt night light bulb plugged into a US household outlet (120 V).
3. Calculate the resistance and the current of a 1500-Watt electric hair dryer plugged into a US household outlet (120 V).
4. The box on a table saw indicates that the amperage at startup is 15 Amps. Determine the resistance and the power of the motor during this time.
5. The sticker on a compact disc player says that it draws 288 mA of current when powered by a 9 Volt battery. What is the power (in Watts) of the CD player?
6. A 541-Watt toaster is connected to a 120-V household outlet. What is the resistance (in ohms) of the toaster?
7. A color TV has a current of 1.99 Amps when connected to a 120-Volt household circuit. What is the resistance (in ohms) of the TV set?