A150-w lamp is placed into a 120-v ac outlet. what is the peak current?

Explanation:

Formula

W = I * E

Givens

W = 150

E = 120

I = ?

Solution

150 = I * 120   Divide by 120

150/120 = I

5/4  = I

I = 1.25

Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)

On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.

However I would go with the other answerer's post and multiply both values by  √2

This question is about as sneaky as they ever get.

First let's do the easy part:

Power = (voltage) x (current)

150 watts = (120 volts) x (current)

current = (150 watts) / (120 volts)

current = 1.25 Amperes but this is NOT the answer to the question.

The voltage at the outlet is a "sinusoidal" wave ... it wiggles up and down 60 times every second.  The number of "120 volts" is NOT the "peak" of the wave.  In fact , the highest it ever gets is  √2  greater than 120 volts.  And all of this applies to the current too.

The RMS current through the lamp is (150/120) = 1.25 Amperes .

The peak current through the lamp is  1.25·√2 = about 1.77 Amperes .

power out to power in less than 100%