A model for a standard two D-cell flashlight is shown. Find the power dissipated in the lamp.
Image from: Irwin, J. David., and R. M. Nelms. Basic Engineering Circuit Analysis, Tenth Edition. N.p.: John Wiley & Sons, 2010.
Solution:
We can see from the diagram that all elements in this circuit are in series. Using our equation, V=IR, where V is voltage, I is current and R is resistance, we can figure out the current going through the 1\Omega lamp.
Thus, we can write:
3V=(I)(1\Omega)
I=3A
Now, we can figure out the power dissipated by using the following equation.
P=I^2R where P is power, I is current, and R is resistance.
Substituting the values we know gives us the following:
P=(3)^2(1)
P=9W
This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 2, question 2.7.
