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.