Small drop in voltage change in power?

[boderam]

In a brownout the power company's voltage drops. Assuming the voltage drop is small, how would I figure out how much of a voltage drop it takes for a 60W light bulb to act like a 50W bulb? The relevant equation I am guessing would be P=V^2/R. I also know the resistance doesn't change. I tried P=(V-h)^2/R for small h, representing the power in a small voltage drop. thus P is approximately V-2Vh/R but I don't think this helps me any since I haven't eliminated h. What do you think?

 

[Andrew Mason]

In a brownout the power company's voltage drops. Assuming the voltage drop is small, how would I figure out how much of a voltage drop it takes for a 60W light bulb to act like a 50W bulb? The relevant equation I am guessing would be P=V^2/R. I also know the resistance doesn't change. I tried P=(V-h)^2/R for small h, representing the power in a small voltage drop. thus P is approximately V-2Vh/R but I don't think this helps me any since I haven't eliminated h. What do you think?

Your idea is right. Use $P_2 = V_2^2/R$ where $P_2 = 5P_1/6 =\frac{5}{6}V_1^2/R$ and find the ratio of V2 to V1.

AM

 

[kyle8921]

I would approach this problem by first understanding the concept of power and voltage in an electrical circuit. Power (P) is equal to voltage (V) squared divided by resistance (R). This means that as voltage decreases, power also decreases. In the case of a brownout, where the voltage drops, the power supplied to the circuit decreases as well.

To figure out how much of a voltage drop it takes for a 60W light bulb to act like a 50W bulb, we can rearrange the power equation to solve for voltage. This gives us V=sqrt(P*R). Since we know the power (60W) and the resistance (which is constant), we can calculate the voltage needed for the 60W light bulb. Let's call this voltage V1.

Now, to find the voltage drop needed for the light bulb to act like a 50W bulb, we can use the same equation but with the power value of 50W. Let's call this voltage V2. The difference between V1 and V2 will give us the voltage drop needed for the light bulb to act like a 50W bulb.

In conclusion, the equation P=V^2/R is useful in understanding the relationship between power and voltage. By rearranging the equation and using the known values, we can calculate the voltage drop needed for a 60W light bulb to act like a 50W bulb in a brownout situation. However, it is important to note that this is a simplified approach and there may be other factors at play in a real-life scenario.