Minimum number of 47Ω resistors in series across 9V without burning up

[soul5]

Homework Statement

A 47 resistor can dissipate up to 0.25-W of power without burning up. What is the smallest number of such resistors that can be connected in series across a 9.0-V battery without anyone of them burning up>

Homework Equations

V=IR
P=IV

The Attempt at a Solution

I have no clue what to do.

 

[dynamicsolo]

Homework Statement

A 47 resistor can dissipate up to 0.25-W of power without burning up. What is the smallest number of such resistors that can be connected in series across a 9.0-V battery without anyone of them burning up>

You could start by considering what would happen with a single 47-ohm resistor connected to the 9-V battery. How much current would be flowing through it? How much power would it be dissipating? (The equations you list will tell you this.)

Now consider connecting two of these resistors in series to the battery. How much resistance would be in the circuit? How much current would be flowing through either resistor? How much power would each resistor be dissipating?

You could jump to N of these resistors in series and again answer the questions in the preceding paragraph using an expression involving N. Now, using your expression for the power being dissipated by anyone of the N resistors in series, set that expression equal to 0.25 W and solve for N. (If necessary, round up to the nearest integer.) This will be the smallest number of resistors required. Any larger number of them will reduce the current further and thus the power being dissipated in each resistor.