Two light bulbs have resistances of 400 ω and 800 ωthe two light bulbs are connected in series across a 120- v line. find the current through each bulb. enter your answers numerically separated by a comma. i 400, i 800 = bfind the power dissipated in each bulb. p 400, p 800 = cfind the total power dissipated in both bulbs. p =

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

And so, the current through the circuit is (using Ohm's law):

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

where I is the current and R is the resistance.

For the first bulb:

For the second bulb:

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

3.2x10^7

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .