# Solving algebraically for Power, getting two conflicting outcomes

#### groundcontrol

Joined Sep 1, 2015
24
Hello I am trying to solve a problem algebraically rather than trying to use real numbers to try and avoid gaining misleading result due to assuming values incorrectly. In doing this, a problem has arisen that I cant figure out and would appreciate some help please?

The question is:
"If the current flowing through a resistor falls to half its original value, the power will.. :"
a. Double
b. Decrease by 4 times
c. Halve
d. Increase by 4 times

In case my workings are not as clear as I hope, my confusion is that I am getting two different outcomes :/

OUTCOMES
P2 = P1 / 2 (i.e power would halve) AND
P2 = P1 / 4 (i.e. power would decrease by 4 times)

I've been told that the correct answer is B) that the power would decrease by four times. I did get this as one of my outcomes, but I also have got an incorrect outcome, hence there is something going on with my maths (or something???) and I'm not sure what. I cant see what's wrong with the
P2= P1/2 logic - but clearly this is incorrect.

The only thing that I can think of is that with the
P2= P1/2 equation ,
this was formulated involving voltage. There is no mention of voltage in the question, possibly making it an 'unreliable' factor to involve at all?
I know when I originally tackled this problem, I tried to solve it using real numbers and P=VI. As there was no mention of voltage I tried just to kept voltage constant. But doing this, the answer I yielded was that the power halved (when current halved). Again, it was incorrect, again the result was that the power would halve. Same wrong answer. Both approaches to solving the problem involved voltage. Is this what's gone wrong? If so is that a rule?

It just seems like a big margin of error when a 'true' equation can be incorrect. Which makes me think that I am messing it up. Please help if you can see what I'm doing wrong?

Thanks heaps !  • Ramussons

#### JoeJester

Joined Apr 26, 2005
4,390
P = I^2 R

Use 1 ohm and 1^2 Amp = 1 W

decrease the current by half

1 ohm times 0.5 ^2 Amps = 1 ohms times 0.25 Amps = 0.25 W

1 W divided by 0.25W = 4 ... Answer b, decreased by 4 times ....

#### crutschow

Joined Mar 14, 2008
32,840
Power = I²R, thus halving the current will reduce the power to (1/2*I)² R= ¼ I²R, or power is reduced by ¼th.

If you want to get voltage into the act then P = V²/R.
If the current through the resistor is halved, then the voltage drop across the resistor is also halved.
Thus P = (½V)²/R = ¼V²/R, or again the power is reduced by ¼.

• groundcontrol

Joined Dec 4, 2009
328
There is no mention of voltage in the question, possibly making it an 'unreliable' factor to involve at all?
If you have to pass half the current through the same resistance R, then the new voltage across the resistor would be half of what was the initial voltage. To put it differently, applying half the original voltage would make the current half.
So, new power would be,
(V/2) * (I/2) = VI/4, so the power would be a fourth of original power.

#### MrAl

Joined Jun 17, 2014
10,582
Hi,

My take on this is as follows...

The formula for the power sometimes called "Joules Law" is:

P=i^2*R

I used this form because we want to understand what happens when the current "i" changes.

We have the initial condition which is:
P1=i^2*R

and we have the secondary condition which is:
P2=(i/2)^2*R

Without assuming any particular values, we find the ratio of P1 and P2:
P2/P1=(i/2)^2*R/(i^2*R)

and doing the simplification we get:
P2/P1=1/4

and multiplying both sides by P1 we get:
P2=P1/4

so there we have it, the power is one quarter of the original power.

A graphical view is always good too. In the formula:
P=i^2*R
set R=1 and get:
P=i^2
Graph this function and see what it looks like when i=1/2.

#### groundcontrol

Joined Sep 1, 2015
24
Power = I²R, thus halving the current will reduce the power to (1/2*I)² R= ¼ I²R, or power is reduced by ¼th.

If you want to get voltage into the act then P = V²/R.
If the current through the resistor is halved, then the voltage drop across the resistor is also halved.
Thus P = (½V)²/R = ¼V²/R, or again the power is reduced by ¼.
Hi, thanks for helping! I think going from what you have said that the following is incorrect.. Would you mind confirming that P2=P1/2 is incorrect please?

How I came up with P2=P1/2 is this;

P1=V²/R
P2=V²/2R =P1/2
hence P2=P1/2 This is the bit that is confusing me.. Is my maths (or something else) wrong?
Thanks !

#### WBahn

Joined Mar 31, 2012
29,130
You are still making exactly the same mistake that you have been making from the beginning and that everyone has been pointing out to you.

P = V·I

Will give you the power absorbed by a device given the voltage ACROSS THAT device and the current THROUGH THAT device.

If the device is a resistor, then you simply can not take the approach that, "As there was no mention of voltage I tried just to kept voltage constant." The only way to keep the voltage constant across a resistor is to keep the current constant. You CAN NOT just declare that you are going to change the current but not the voltage -- they are related by Ohm's Law such that if you change one the other WILL and MUST change!

• ErnieM

#### tcmtech

Joined Nov 4, 2013
2,867
I am trying to solve a problem algebraically rather than trying to use real numbers to try and avoid gaining misleading result due to assuming values incorrectly. In doing this, a problem has arisen that I cant figure out and would appreciate some help please?

HUH???

How is using the alphabet where every letter or symbol is assumed to represent a number or a correct/accurate sub-equation opposed to real numbers in an equation going to give a more correct resulting answer? Put real numbers in the equations and see what you get and then compare that step by step to you alphabetical gobbledygook and see where what difference. I'm pretty sure your problem is on the alphabetical assumption side not the finite fixed value numerical side.