# Average Current Simple

1. Feb 6, 2008

### BuBbLeS01

Average Current...Simple....

1. The problem statement, all variables and given/known data
A typical American family uses 1430 kWh of electricity a month.
What is the average current in the 110 V power line to the house?

2. Relevant equations

3. The attempt at a solution
Don't I just change kWh to W and divide it by the voltage...
1430 x10^3 = 1430000 / 110 = 13,000A which is def. not right lol.

2. Feb 6, 2008

### Tedjn

Well, $[\text{kW}\cdot\text{h}]$ is not the same thing as a kW, which you can convert to W. Instead, $[\text{kW}\cdot\text{h}]$ is power (kW) multiplied by time (h), which is energy. You have the equation right (i.e. E = iV), but you need to convert to the SI unit of energy, the joule. How would you do that?

3. Feb 6, 2008

### BuBbLeS01

Do 1430x10^3 * 3600s = 5.148x10^9 J

4. Feb 6, 2008

### BuBbLeS01

but what now?

5. Feb 6, 2008

### Tedjn

You had the right idea before with your equation E = iV. You know E, you know V, so you can find i.

6. Feb 6, 2008

### BuBbLeS01

but what is E? Its not measured in J?

7. Feb 6, 2008

### Tedjn

Sorry, that is my mistake. I'm not thinking very clearly today. Power is iV, so you are right to be confused. You know that she uses 1430 kWh in a month. You can find out how many seconds are in a month, and figure out the average power, in kW or W. Then you divide by voltage to find the current.