How Does Voltage Variation Affect Power in a 100-W Heater?

[LeakyFrog]

Homework Statement

A 100-W heater is designed to operate with an applied
voltage of 120V.

a) What is the heater's resistance, and what current does the heater carry?

b) Show that if the potential difference V across the heater changes by a small
amount ΔV, the power P changes by a small amount ΔP, where ΔP/P = 2ΔV/V. (Hint:
Approximate the changes by modeling them as differentials, and assume the
resistance is constant.

c) Using the part b result, find the approximate power delivered to the heater
if the potential difference is decreased to 115V. Compare your result to the
exact answer.

Homework Equations

P = IV = V²/R

The Attempt at a Solution

I'm stuck on part b). So far I have
ΔP=ΔV²/R

ΔP/P=(ΔV/V)²

I'm not really sure what to do.

 

[ehild]

The change of a function f(x) can be approximated by Δf=f'(x)Δx, where f' is the derivative of f(x) taken at the original value of x, and Δx is a small change of x.


ehild

 

[LeakyFrog]

Thanks.

Another question, for part c) when you change the voltage on a circuit does the resistance stay the same? Or the current? Although I don't really see how the current could but maybe i suppose.

 

[ehild]

The heater stays the same, isn't it? And it consist of a resistor, and some other parts which do not change either, if you plug it into an other socket or the household voltage decreases because of some overload happening in the network.
Question c. wants you to apply the formula in question b, if V and ΔV are given.

ehild

 

[rdl]

Great job so far! To solve part b), let's start by writing out the equation for power, which is P = IV. We can also substitute in the equation V = IR, which gives us P = I(IR) = I2R. Now, let's take the differential of both sides of this equation:

dP = 2IRdI + I2dR

Since we are assuming that the resistance is constant, dR = 0, so this term disappears. We are left with:

dP = 2IRdI

Now, we can use Ohm's Law (V = IR) to substitute in for I:

dP = 2VdI/R

Finally, we can divide both sides by P to get the desired result:

ΔP/P = 2ΔV/V

This shows that the change in power is directly proportional to the change in voltage, with a factor of 2. This makes sense, since power is also equal to V2/R, and we can see that a small change in V will result in a larger change in V2.

For part c), we can use the equation we just derived to find the approximate power delivered to the heater if V is decreased to 115V. We know that the original power was 100W, so we can set up the following equation:

ΔP/P = 2ΔV/V

We can rearrange this to solve for ΔP:

ΔP = 2ΔV/V * P

Substituting in the values given in the problem, we get:

ΔP = 2(120-115)/120 * 100 = 8.33W

This means that if the potential difference is decreased to 115V, the power delivered to the heater will decrease by approximately 8.33W. To find the exact answer, we can use the original equation for power (P = V2/R) and substitute in the new voltage of 115V:

P = (115)2/R = 100W

We can see that the exact answer matches our approximate result, which confirms the accuracy of our calculation. Great work!