Calculating Current Flow in a Short-Circuited Battery

[kirsten_2009]

Homework Statement

[/B]
A 1.5 volt battery is short-circuited by a 2–meter long wire having a resistance of 0.02 ohm per meter. How large is the current flowing through the wire (before the wire or the battery burn out)?

Homework Equations

Voltage = Ampere x Resistance
Ampere = Voltage/Resistance

The Attempt at a Solution

Ampere = 1.5 V / 0.02 ohm x 2 m (0.04 ohm/m)
Ampere = 37.5 Amperes

 

[phinds]

"Ampere = 1.5 V / 0.02 ohm x 2 m (0.04 ohm/m)" is not really a good expression but yes, your answer is right and I think you clearly understand what you are doing even if you did not express it well mathematically. Without units, it's just 1.5/.04 = 37.5

You should express it mathematically in a way that makes the units come out right

 

[Mark44]

Homework Statement

[/B]
A 1.5 volt battery is short-circuited by a 2–meter long wire having a resistance of 0.02 ohm per meter. How large is the current flowing through the wire (before the wire or the battery burn out)?

Homework Equations

Voltage = Ampere x Resistance
Ampere = Voltage/Resistance

The Attempt at a Solution

Ampere = 1.5 V / 0.02 ohm x 2 m (0.04 ohm/m)
Ampere = 37.5 Amperes

Your answer is fine, but your work is a little off in the units. In the first line above, you have .02 ohm x 2 m, and get .04 ohm/m. The numbers are right, but the units aren't. They should be .02 ohm/m x 2 m, which results in .04 ohm, not ohm/m.

Some suggestions:
Instead of writing this -- 1.5 V / 0.02 ohm x 2 m (0.04 ohm/m) -- as 1.5 V /(0.02 ohm/m x 2m). In other words, use parentheses around the two factors in the denominator. If you write a/bc, most will interpret this as (a/b) times c, rather than a divided by (bc).

 

[kirsten_2009]

Oh O.k I see...thank you very much, I will adjust my units :)