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Power

Recall that power is a measure of how quickly energy is transformed, measured in [J/s = Watts = W]. The power dissipate by a resistor is
 
...(6)
where / is the current through the resistor and  is the potential difference across it.
 

 

Because of Ohm’s law, we may also write
 
Pres = I2R = (ΔV)2/R

 

(It is better to remember how to derive this equation than to memorize it.) The power provided by a DC cell to a circuit is
 
...(7)
where / is the current through the cell and  is the potential difference across it.
 

 

Example

Light bulbs that you use around the house are designed to have 120 V across their terminals. For instance, a 120-W bulb uses 120 Watts of power when placed in a socket with a 120-V potential difference. (Actually it is a little more complicated. See Section F.) With that in mind,

  1. what is the resistance of a “120-Watt” bulb?
  2. What is the resistance of a “30-Watt” bulb?
  3. If these two bulbs are connected in series, and plugged into a wall outlet, which bulb would be brighter (Figure 15-20)?

..\art 15 jpg\figure 15-ts.jpg
Figure 15-20

 
Solution
  1. Figure 15-21 shows the circuit diagram for a single 120-W bulb plugged into a potential source. The equation for power is

P = IΔV
120W = I(120V)
I = 1A

 

By Ohm’s law, we have
 

ΔV = IR
120V = (1A)RR = 120 Ω.

 

  1. A similar calculation gives a resistance 480 Ω for the “30-W” bulb.
  2. If we place the bulbs in series (Figure 15-22), then the current through the bulbs is the same. Since

P = I2R

 

the power is proportional to the resistance. So the 480-Ω bulb (that is, the “30-W” bulb) is brighter because the bulbs are in series. (Usually the “120-W” bulb is brighter because the bulbs are connected in parallel.)
 
..\art 15 jpg\figure 15-tt.jpg
Figure 15-21
..\art 15 jpg\figure 15-tu.jpg
Figure 15-22
 





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