Mean power dissipated in resistor in ac circuit

[toforfiltum]

Homework Statement

Homework Equations

The Attempt at a Solution

I am surprised that this question came up, I haven't learned ac circuits yet. I don't know how to solve this. I just added both values and current and divide it by two, which gives me 1.5A, since I thought it's the average but I'm wrong. Answer is C. I really don't know how to get that value.

 

[TSny]

The reason your method didn't work is due to the fact that the power is not proportional to the current. It's proportional to the square of the current.

It might help to determine the value of the power when the current is 2 A and the value when the current is -1 A.

 

[toforfiltum]

The reason your method didn't work is due to the fact that the power is not proportional to the current. It's proportional to the square of the current.

It might help to determine the value of the power when the current is 2 A and the value when the current is -1 A.

Oh I see. Thanks. I have to do it this way because power is a scalar, is it? Why can't I square the value of 1.5A instead?

 

[TSny]

Oh I see. Thanks. I have to do it this way because power is a scalar, is it?

It's not because its a scalar quantity, it's because the mean of the square of a quantity is not equal to the square of the mean of the quantity. For example, for the two numbers 2 and 4 the mean of the square of the numbers is (2² + 4²)/2 = 10. But the square of the mean of the numbers is 3² = 9.

Power is proportional to the square of the current. So you need to use the mean of the square of the current to get the mean power.

 

[toforfiltum]

It's not because its a scalar quantity, it's because the mean of the square of a quantity is not equal to the square of the mean of the quantity. For example, for the two numbers 2 and 4 the mean of the square of the numbers is (2² + 4²)/2 = 10. But the square of the mean of the numbers is 3² = 9.

Power is proportional to the square of the current. So you need to use the mean of the square of the current to get the mean power.

Thanks for the really neat explanation.