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 (ΔV) and the current (I) can be expressed in terms of their dependence
upon resistance as shown in the following equations.
ΔV = (I • R) |
I = ΔV / 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.
Equation 2: P = ΔV • I P = (I • R) • I P = I2 • R |
Equation 3: P = ΔV • I P = ΔV • (ΔV / R) P = ΔV2 / R |
We 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 (ΔV), 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 ΔV. 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 conceptualprinciples 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 ΔV 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 Ω; the 60-Watt
bulb would have a current of 0.5 Amp and a resistance of 240 Ω.
Calculations for 120-Watt Bulb P = ΔV • I I = P / ΔV I = (120 W) / (120 V) I = 1 Amp ΔV = I • R R = ΔV / I R = (120 V) / (1 Amp) R = 120 Ω |
Calculations for 60-Watt Bulb P = ΔV • I I = P / ΔV I = (60 W) / (120 V) I = 0.5 Amp ΔV = I • R R = ΔV / I R = (120 V) / (0.5 Amp) R = 240 Ω |
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.
Calculations for 120-Watt Bulb P = I2 • R I2 =
P / R I2 =
(120 W) / (120 Ω) I2 =
1 W / Ω I = SQRT ( 1 W / Ω ) I = 1 Amp |
Calculations for 60-Watt Bulb P = I2 •
R I2 =
P / R I2 =
(60 W) / (240 Ω) I2 =
0.25 W / Ω I = SQRT ( 0.25 W / Ω ) I = 0.5 Amp |
1. Which would be thicker (wider) - the filament of a 60-Watt light bulb or the
filament of a 100-W light bulb? Explain.
The resistance
of a light bulb filament is effected by both length and cross-sectional area.
Thicker wires have less resistance. A 120-Watt bulb has a greater current and a
smaller resistance. Thus, a 120-Watt bulb must have a thicker filament than a
60-Watt bulb (assuming the lengths of the filaments are identical).
2. Calculate the resistance and the current of a 7.5-Watt
night light bulb plugged into a US household outlet (120 V).
Answers: I = 0.0625 Amp and R = 1920 Ω
Given:
V = 120 V (true for US
household outlets)
P = 7.5 W
Find: I and R
Use the P = V • I
equation to calculate the current:
I = P / V = (7.5 W) /
(120 V)
I = 0.0625 Amp
Now since current is known, use the Ohm's law equation (V = I • R) to
calculate the resistance:
R = V / I= (120
V) / (0.0625 Amp)
R = 1920 Ω
3. Calculate the resistance and the current of a 1500-Watt
electric hair dryer plugged into a US household outlet (120 V).
Answers: I = 12.5 Amp and R = 9.6 Ω
Given:
ΔV = 120 V
(true for US household outlets)
P = 1500 W
Find: I and R
Use the P = ΔV • I equation to calculate the current:
I = P / ΔV
= (1500 W) / (120 V)
I = 12.5 Amp
Now since current is known, use the Ohm's law equation (ΔV
= I • R) to calculate the resistance:
R
= ΔV / I = (120 V) / (12.5 Amp)
R = 9.6 Ω
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.
Answers: R = 8 Ω and P = 1800 W
Given:
ΔV = 120 V
(true for US household outlets)
I = 15 A
Find: R and P
Use the Ohm's law equation (ΔV = I • R) to calculate the
resistance:
R = ΔV
/ I= (120 V) / (15 Amp)
R = 8 Ω
Now calculate power using any one of the three equations:
P
= I • ΔV = (15 A) • (120 V)
P = 1800 W
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?
Answer: P = 2.59 W
Given:
ΔV = 9 V
I = 288 mA =
0.288 Amp
Find: P
P
= I • ΔV = (0.288 A) • (9 V)
P = 2.59 W
6. A 541-Watt toaster is connected to a 120-V household
outlet. What is the resistance (in ohms) of the toaster?
Answer: P = 2.59 W
Given:
ΔV = 9 V
I
= 288 mA = 0.288 Amp
Find:
P
P
= I • ΔV = (0.288 A) • (9 V)
P
= 2.59 W
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? And what is the power (in Watts) of the
TV set?
Answers: R = 60.3 Ω and P = 239 W
Given:
ΔV = 120 V
(true for US household outlets)
I = 1.99 A
Find: R and P
First use the Ohm's law equation (ΔV = I • R) to calculate
the resistance:
R = ΔV
/ I= (120 V) / (1.99 Amp)
R = 60.3 Ω
Now any one of the three power equations can be used to
determine the power. The P = I2 • R equation will be used here:
P = I2 •
R = (1.99 A)2 • (60.3 )
P = 239 W