# Current homework problem

1. Dec 3, 2009

### captainjack2000

1. The problem statement, all variables and given/known data
A light bulb has a straight metal wire filament of 50 micro metres diameter and 10mm long and when powered with a 5V constant voltage supply it draws 0.5A of current. What happens to the current if the filament is replaced with one of three times the diameter?

2. Relevant equations
I=nAvQ ??

3. The attempt at a solution

2. Dec 3, 2009

### mikelepore

Re: Current

Look at the formula for the resistance of a resistor in terms of its physical dimensions.

3. Dec 3, 2009

### captainjack2000

Re: Current

Thanks

So
R=l * rho /A
where l is the length of the conductor
rho is the electrical resistivity of the material
A is the cross-sectional area of the conductor

So V=IR
I(1) = V/R(1) and I(2)=V/R(2) where I(1) is the inital current
taking a ratio
I(1)/I(2) = [V/R(1)]/[V/R(2)] = R(2)/R(1) = [l(2)*rho*A(1)] /[l(1) * rho* A(2)]
rho is the same for each and l is the same for each so
I(1)/I(2) = A(1)/A(2)
= (pi * r(1) ^2) / (pi* r(2) ^2)
=r(1)^2 / r(2) ^2 where r(1) and r(2) are the corresponding radii of the filaments
=(25*10^-6)^2 / (75*10^-6)^2

so I2=4.5A

Is that right?

4. Dec 5, 2009

### mikelepore

Re: Current

My own method would be:

Step 1. begin by transcribing the phrase "the filament is replaced with one of three times the diameter" into the phrase "the filament is replaced with one with nine times the cross sectional area."

Step 2. Nine times larger area means nine times smaller resistance.

Step 3. Look at I=V/R. If we make the denominator of any fraction nine times smaller, the fraction become nine times larger.

Last edited: Dec 5, 2009