Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are now connected in parallel across the 120 V line. Find the total power dissipated in both bulbs.

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Hey everyone in this problem we have two electric kettles that have resistance is of 9.6 and 12 homes. They're plugged in parallel to a 110 volt outlet socket. Okay. And were asked to determine the combined power output of the kettles. All right, now, when we're looking for power output, let's recall that the power output P. Is given by the voltage B times the current. I. Now if we look at this problem, we're given information about the voltage V and were given information about the resistance. R. We're not giving information about the current. I recall though that OEMs law relates V. I and R. Which is gonna allow us to write this as V squared over R. Okay. Now the resistance R that we're talking about, we're looking for a combined power output of the kettles. So we're gonna be looking for the resistance sub E que que that equivalent resistance where we have both of our kettles considered. Now in parallel recall that are equivalent resistance is going to be given by the following. We have that the inverse of the equivalent resistance is going to be equal to the sum of the inverse is of the resistance is so we have one over R. E. Q. Is equal to one over R one plus one over R two. Okay, we can find a common denominator simplify this and then take the inverse which gives us a The equivalent resistance is going to be equal to R one. R 2 over R one plus our two. Now, if we use the resistance is we were given this is going to be 9. ohms times 12. OEMs divided by 9.6 S plus 12 ohms. And this is going to give us an equivalent resistance of 5.333 repeated homes. Okay, in the numerator we get home times home. So we have um squared divided by OEM. So we get a unit of home. Alright. So we know our our equivalent are equivalent resistance. We know V squared. Because we were given the voltage V in the problem. So now we can calculate our power P. Okay. And again, power P. The output is going to be V squared for the equivalent resistance. So we have P is equal to V squared over. R equivalent. Okay, the voltage V we were given is 100 and 10 volts All squared divided by 5.333. OEMs repeated. And this is going to give us a power output of 2,268. watts. No. Alright. And if we look at our answer choices, okay? We see that these are rounded to the nearest five or 10. So we have 2268.76. So that's going to give us answer. D. We have approximately 2270 watts as our power output of those kettles. Thanks everyone for watching. I hope this video helped see you in the next one

Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are now connected in parallel across the 120 V line. Find the total power dissipated in both bulbs.

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Key Concepts

Ohm's Law

Parallel Circuit

Power Dissipation