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Old incandescent lightbulbs have purely resistive meta filament, through which flows an alternating current; If an Incandescent bulb uses an average power 0f 75,0 W...

Question

Old incandescent lightbulbs have purely resistive meta filament, through which flows an alternating current; If an Incandescent bulb uses an average power 0f 75,0 W when connected to 60.0 Hz power source having maximum voltage 0f 170 V; what is its resistance (in 0)?Another incandescent bulb, with an average power Usage of 150 W; Is connected to the same AC power source with the same frequency and maximum voltage What I5 Its resistance (In 0)?

Old incandescent lightbulbs have purely resistive meta filament, through which flows an alternating current; If an Incandescent bulb uses an average power 0f 75,0 W when connected to 60.0 Hz power source having maximum voltage 0f 170 V; what is its resistance (in 0)? Another incandescent bulb, with an average power Usage of 150 W; Is connected to the same AC power source with the same frequency and maximum voltage What I5 Its resistance (In 0)?



Answers

(a) What is the resistance of a lightbulb that uses an average power of 75.0 W when connected to a 60.0-Hz power source having a maximum voltage of 170 V? (b) What If? What is the resistance of a 100-W bulb?

Well, let's do part. A resistance R is equal to we. RMS Square, divided by p a. P is the power. And, uh, root means skin won't teach. Um is rate unease 170 divided by route to hold square, multiply by one divided by 7 to 5 where 75 years, the power and therefore are is equal to 193 problems. Now part B again using our is equal to the square Aramis divided by t and we are mrs 1 70 divided vital to home square Multiply by a power with power in this case is 100 therefore resistance park is equal to 144.5 or months.

Again this question we ought to find out the resistance of a. In the part A. We have to find out the resistance of a 75 ward. Well okay so power is given 75. What? Okay. When connected to a 60 horse power source having maximum voltage 1 70 world. Okay so we max is given 1 70 volt. Okay And we have to find out the resistance. So we know the formula of resistance that is we Rms square. We square. Okay. Aramis by power P. Okay. And we have maximum voltage and we know the relation between RMS voltage and maximum voltage that we are M. S. Can be obtained by we max that is peak voltage divided by route to. Okay so here the resistance it will be it will be big voltage divided by Ruutel. It's square. Okay Multiplied by one divided by P. And the resistant. It will be in home understand. Now we will put the values so we mix that is given 1 70. It will be 1 70 divided by road to Squire and multiplied by one divided by 75. And answer well being home so resistance. It will be 1 70. Holy square. That is 289 double zero divided by divided by road to square. That is two multiplied by 75. It will be 1 50. Okay that isn't home. And when we saw this it will be our equals to one. It will be 1 92.66 So it will be approx 1 93 home understand. And this will be the final answer of part one of discussion. And now part B. Okay. Now the bulb is 100 What? Okay so the power of the bulb is given 100 world rest of the dot our same. So we have to find out again. The resistance that will be our that is we Rms Squire divided by P. Okay. And we are M. S. That speak Walters divided by route to peak voltage. It is 1 70 divided by rock and it's a square multiplied by one divided by power is given 100. Okay. And answering will be home. So are it will be to 89 double zero divided by it will be rude to square. That is two multiplied 100. That will be 200 home. And when we saw this this will be 1 44.5. Oh I understand. And this will be the final answer of part vii of discussion. Thank you.

What ails To give a problem, we are asked to find the resistance off the ball. We can write. Resistance is an average power is equal to the square are a mess divided by our from here. Real soulful resistance or resistance will be We are a mess swear divided by V average. Um here will substitute the values for re Artemus so the re Artemus can be written a in terms off equal teach so we peek or a maximum divided by the square root off to square, divided by a P average. Then we can just simply substitute the value for re max, which is 1 70 divided by 0.27 or seven whole square, divided by the average power. Given the 75. This gives us the resistance off 1 93 or home. But be then the resistance for the average power over 100 ports will be are using the same problem simply substituting the values. 17 Did you buy your reporting? 707 Whole square. This whole square divided by 100. This gives us 1 45 homes

Here it is even in the problem that already in power of the light bulb is 60. Word and operated world is of the light bulb is 1 20 world. And in the foster Bardo, this problem you want to call, call it resistance off the light bulb. Not by using their religion that are equal to Secare. A wall dish upon four were here all dishes 12 in the world and bore off. The light bulb is six toward so we have our equal to 1 20 1 14 This is not a distance off a light bulb off our 60 ward. No, the second part again by using the relation that p equal to so good ovary a bomb are a constant. A predator will dish we d'oh for is inversely proportional under your sense every hill, a light bulb off 100 words Then it will less reticence ISS compared to lightbulb off 60 war from the relation it is clear there reasons and Barbara are in worse city proportion. No, by putting the values here in the elation p equal to disappear upon our are we have are equal to recent get a bomb. P were operated wall dishes. 1 20 World and power of the Liable in this case is under war. So we have reasons off underworld libels equal to one foot before. Oh, so this is the but in a sense off 100 word label of. And this is an instance off 60 word Leiber. By seeing these two values, it is clear there a sense off 60 word lightbulb is greater than resistance off under work, lad Buck.


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