Find the Min. Series Resistors for 9V & 0.25W

[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.

 

[Moetasim]

Based on the given information, we can use the equation P=IV to calculate the current (I) that will flow through the resistor. We know that the voltage (V) is 9V and the power (P) is 0.25W, so we can rearrange the equation to solve for current:

I = P/V = 0.25W/9V = 0.0278A

Now, we can use the equation V=IR to calculate the resistance (R) needed for this current to flow through the resistor. We know that the voltage (V) is 9V and the current (I) is 0.0278A, so we can rearrange the equation to solve for resistance:

R = V/I = 9V/0.0278A = 323.74Ω

Since we are looking for the smallest number of resistors, we can use the smallest commonly available resistor value, which is 330Ω. This means that we would need at least two 330Ω resistors in series to handle the 9V and 0.25W without burning up.

To confirm this, we can calculate the total power dissipated by the two resistors in series:

P = IV = (0.0278A)(9V) = 0.2502W

This is very close to the maximum power of 0.25W that the resistors can handle, so we know that two resistors in series will be enough.

Therefore, the minimum number of 330Ω resistors needed in series to handle 9V and 0.25W is two.