Question

There are 2 resistor and a battery. 8Ω and 16Ω are connected in parallel to a battery of 120V. What is the total power?

Answer 1

please help

Answer 2

Let the resistance of the parallel combination be R ohms.
\[\large R=\frac{8\times16}{8+16}\ ohms\]
The total power P can be found as follows:
\[\large P=\frac{V^{2}}{R}=\frac{120^{2}}{R}=(you\ can\ calculate)\ watts\]

Answer 3

I thought the equation for Power was \[P=VI\] or \[P=I ^{2}R\]?

Answer 4

Thank you for answering by the way!!

Answer 5

are all three equations the same?

Answer 6

You're welcome.
If we take your first equation
\[\large P=VI\ ........(1)\]
we also know from Ohm's Law that
\[\large I=\frac{V}{R}\ ..........(2)\]
If we plug the value of I from equation (2) into equation (1), the result is
\[\large P=V \times \frac{V}{R}=\frac{V^{2}}{R}\]
So we have three equations to find power.

Answer 7

ohhh okay I see. One more equation. While I was doing this problem, I first solved it by doing \[P=I ^{2}R\ for each resistor and add it up together to get the total power. Then I tried to double check by answer by using this equation: P=IV since "The total power is equal to the sum of the power dissipated by the individual resistors. ." However, those value where not the same......But when I used your equation P=V^2/R , the previous statement holds true.

Answer 8

*I first solved the equation by doing P=I^2R

Answer 9

The equation P = VI gives the same answer as the other two equations. Perhaps your calculation needs to be checked.

Answer 10

yes but P=I^2R does not ( after using this equation for each resistor and adding it up)

Answer 11

If you use P = I^2R for each resistor, that is calculate the power dissipated in each resistor separately, the sum of the two values of power is the same as the power calculated by using P = VI.
Using P = VI, the power dissipated in the 8 ohm resistor is given by
\[\large P _{1}=(\frac{120}{8})^{2}\times8\ .........(1)\]
and the power dissipated in the 16 ohm resistor is given by
\[\large P _{2}=(\frac{120}{16})^{2}\times16\ ......(2)\]
If you add the values of power given by (1) and (2) you will get the same answer as calculating using P = VI.
Do you understand?

Answer 12

Using P = I^2R ......*

Answer 13

Okay, I understand. I made the mistake of using the total resistor ( for P=I^2 R) instead of the individual resistor for find current Ex: (120/5.33). Thank you soooooo much for your help!!!

Answer 14

for finding*

Answer 15

Glad to have been of help :)