The previous section of Lesson iii elaborated upon the dependence of electric current upon the electric potential deviation and the resistance. The current in an electrical device is directly proportional to the electrical potential departure impressed across the device and inversely proportional to the resistance of the device. If this is the case, and then the rate at which that device transforms electrical energy to other forms is as well dependent upon the electric 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 ability in terms of current, electric potential difference and resistance.
New Equations for Ability
In Lesson ii, the concept of electrical power was introduced. Electrical power was defined as the rate at which electrical energy is supplied to a excursion or consumed by a load. The equation for calculating the power delivered to the circuit or consumed past a load was derived to exist
P = ΔV • I
(Equation one)
The two quantities that ability depends upon are both related to the resistance of the load by Ohm's constabulary. The electric potential difference ( ΔV ) and the electric current ( I ) tin can exist expressed in terms of their dependence upon resistance as shown in the following equations.
If the expressions for electric potential difference and electric current are substituted into the power equation, 2 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 = ΔVtwo / R |
We at present have three equations for electric power, with two derived from the first using the Ohm'south law equation. These equations are oftentimes used in issues involving the computation of power from known values of electric potential difference (ΔV), current (I), and resistance (R). Equation ii relates the charge per unit at which an electrical device consumes energy to the current at the device and the resistance of the device. Annotation the double importance of the current in the equation as denoted past the foursquare of electric 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, and then information technology will exist necessary to either use one of the other two equations to calculate power or to utilise the Ohm'south law equation to calculate the quantity needed in gild to use Equation ii.
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 equally denoted by the foursquare of ΔV. Equation 3 can be used to calculate the ability provided that the resistance and the voltage driblet are known. If either ane is not known, so it will be of import to either utilise one of the other 2 equations to calculate ability or to use the Ohm'south law equation to summate the quantity needed in order to use Equation three.
Concepts Come First
While these three equations provide one with user-friendly formulas for calculating unknown quantities in physics problems, ane must be careful to non misuse them past 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, so how many times greater would the electric current exist in that lamp circuit? Using equation 2, i might reason (incorrectly), that the doubling of the power ways that the Iii quantity must be doubled. Thus, current would take to increment by a cistron 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 divergence between a 60-Watt bulb and a 120-Watt bulb is non the current that is in the seedling, but rather the resistance of the seedling. It is the resistances that are different for these ii bulbs; the difference in current is just 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 electrical potential deviation is around 120 Volts. The ΔV would be the aforementioned for each seedling. The 120-Watt bulb has the lower resistance; and using Ohm'due south law, ane would look it also has the college electric current. In fact, the 120-Watt bulb would have a current of 1 Amp and a resistance of 120 Ω; the lx-Watt seedling would have a electric 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 5) I = i Amp ΔV = I • R R = ΔV / I R = (120 V) / (1 Amp) R = 120 Ω | Calculations for lx-Watt Bulb P = ΔV • I I = P / ΔV I = (60 W) / (120 Five) I = 0.v Amp ΔV = I • R R = ΔV / I R = (120 V) / (0.5 Amp) R = 240 Ω |
At present using equation 2 properly, i can see why twice the power means that in that location would be twice the current since the resistance also changes with a bulb change. The adding of current below yields the same issue as shown higher up.
| Calculations for 120-Watt Bulb P = Itwo • R I2 = P / R Itwo = (120 West) / (120 Ω) I2 = ane Westward / Ω I = SQRT ( i W / Ω ) I = 1 Amp | Calculations for 60-Watt Bulb P = I2 • R Itwo = P / R Iii = (60 Westward) / (240 Ω) I2 = 0.25 W / Ω I = SQRT ( 0.25 West / Ω ) I = 0.5 Amp |
Check Your Understanding
i. Which would be thicker (wider) - the filament of a 60-Watt light bulb or the filament of a 100-Westward light bulb? Explicate.
two. Summate the resistance and the current of a vii.5-Watt night light bulb plugged into a United states household outlet (120 V).
three. Summate the resistance and the current of a 1500-Watt electric pilus dryer plugged into a US household outlet (120 Five).
four. The box on a table saw indicates that the amperage at startup is 15 Amps. Decide the resistance and the power of the motor during this fourth dimension.
five. The sticker on a meaty 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 histrion?
6. A 541-Watt toaster is connected to a 120-V household outlet. What is the resistance (in ohms) of the toaster?
seven. 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? And what is the power (in Watts) of the Idiot box gear up?