If each resistor is rated at 0.50 W (maximum without overheating), what is the maximum voltage that can be applied across the whole network? Express your answer to two significant figures and include the appropriate units.

We found a similar problem:

A 2.8-kΩ and a 2.1-kΩ resistor are connected in parallel; this combination is connected in series with a 1.8-kΩ resistor. If each resistor is rated at 1/2 W (maximum without overheating), what is the maximum voltage that can be applied across the whole network?

The maximum voltage is 57 V

Explanation:

Circuits

The power dissipation of a resistor R and voltage V is:

The maximum power for each resistor must not exceed 0.5 W, this allows us to find the maximum voltage for each one.

For R1=2800

For R2=2100

Since this voltage is less than the first one and they are connected in parallel, we choose this one for both because the greater voltage would cause overheating in the second resistor.

The voltage for R3=1200 is

This maximum voltage will add to the combination of the parallel in the final series connection, thus the total maximum voltage is

The maximum voltage is 57 V

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.