How Does a Small Change in Voltage Affect the Power of a Heater?

[cseet]

Hi all,

I really need somebody to explain to me how to work the following problem...

Q:
A 1.50E2 - W heater is designed to operate with an EMF of 120V. If the potential difference across the heater changes by a small amount such as 2.6%, approximately by what factor does the power change? (Hint: approximately the changes with differentials) (unit %).

can somebody kindly explain to to go about it?? thanks
cseet

 

[Doc Al]

The first step is to express the power as a function of voltage. That should be easy. Then you have P = f(V).

Then you approximate the changes as follows:

ΔP = f'(V)ΔV

You can then manipulate this equation to give you the relationship between ΔP/P and ΔV/V. Give it a try.

 

[M98Ranger]

Hi cseet,

I'd be happy to help you with this problem! Let's break it down step by step:

1. First, let's define some terms. EMF stands for electromotive force, which is essentially the potential difference or voltage across a circuit. In this problem, the EMF is given as 120V.

2. Next, we need to understand the concept of power. Power is the rate at which energy is transferred or used, and it is measured in watts (W). In this case, we are dealing with a 1.50E2 - W heater, which means it has a power of 150 watts.

3. Now, let's look at the question. It asks for the change in power when the potential difference changes by 2.6%. The key here is to understand that power is directly proportional to the potential difference. This means that as the potential difference changes, the power will change by the same factor.

4. To find the change in power, we can use the formula P2/P1 = (V2/V1)^2, where P2 is the new power, P1 is the original power, V2 is the new potential difference, and V1 is the original potential difference. Plugging in the values we have, we get (P2/150) = (122.6/120)^2.

5. Solving for P2, we get P2 = 156.72 watts. This means that the power has increased by a factor of 156.72/150 = 1.045 or 4.5%.

So, to answer the question, when the potential difference changes by 2.6%, the power changes by approximately 4.5%. I hope this explanation helps you understand the problem better. If you have any further questions, please don't hesitate to ask. Good luck!