a-75-watt-lightbulb-uses-an-average-power-of-75-mathrmw-when-connected-to-an-rms-voltage-of-120-math

Question

A "75 -watt" lightbulb uses an average power of 75 $\mathrm{w}$ when connected to an rms voltage of 120 $\mathrm{V}$ . (a) What is the resistance of the
lightbulb? (b) What is the maximum current in the bulb? (c) What
is the maximum power used by the bulb at any given instant of time?

João Gabriel Alencar Caribé · Accepted Answer

In this question, a lightbulb is connected to an average ROMs voltage off 120 volts and use an average power off 75. What? In the first item we have to calculate what is the resistance off that light bulb to do so we use the following relation. The average power is equals. Two V are a mess squared over our them. The resistance off the light bulb he&#x27;s given by the are a mess squared over the average power consumers. No, we can just plug in the given values and Coakley devalue off the resistance. So 120 square over 75 and easy&#x27;s equals to 192 homes off resistance. In the second item, we have to copulate. What is the maximum correct that flows from the light bulb? So we begin by calculating the Ari mass current by using forms Law. So v are a mess is you goes to our times. I are a mess. This is just on flow. Then we can solve for I r m s to get that these easy goes to V. What a mess over are blocking the given values. We got 120 over 192 and these easy close to zero point 625 and piers. And now, to get the maximum current, we use the relation between the maximum current on their R. M s security, which is the following. All right. R m s is equals two I max over square. I&#x27;m tough too solving. For I Max, we get that the maximum current is he goes to the square root of two times the R M s current, which is equal to the square root of two times zero point 625 which is approximately equals two zero point 88 and piers in the last item. We have to calculate what is the maximum power? Use it by the light bulb at any even instant of time and the maximum power we will be given by be Max musicals to our times. I&#x27;m Max squared by using these version off Max, we get the following be max is equals to 192 times square. Root of two times 0.625 squared and disease equals to 192 times two times 0.625 squared and this is it close to 150. What&#x27;s

Salamat Ali · Answer

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

Mayukh Banik · Answer

we have Aramis. Vulture It waas 1 70 village over route to it was 1 20 Voltage. We have power equals V squared over R So we have our equals. The square overpay. It was 1 20 square over 75. What? It was 1 93 home for part B. You have our It was 1 20 square. Over 100. It was 1. 44.

Zulfiqar Ali · Answer

Well, let&#x27;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&#x27;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.

Averell Hause · Answer

so we can say for part a the peak voltage is going to be equaling Max. Voltage would be equaling radical. Two, multiplied by essentially Delta V sub are our message. The root means squared voltage that is applied. This would be equaling radical two multiplied by the value given 1.20 times 10 to the second bolts, or 120 volts. And this is giving us 1.70 times 10 to the second bolts. This would be the magnitude of the peak voltage four. Part B. We can say that the average power would be equaling. The means squared current squared multiplied by are the resistance. In other words, simply the root means squared, bolted, squared, divided by the resistance are, and this is saying that solving for are there is a sense this would be the change. Delta bi of root means squared, bolted, squared, divided by the average power, and so this would be equaling to simply 100 20 volts divided by 60.0 watts. And so the resistance is found to be 2.4 zero times 10 to the second homes. This would be our final answer for Part B and then for part C, we see that our equaling the root means squared, bolted, squared, divided by the average power. And as you can see here, and as you can see here, if the bulbs or operate are designed to operate at the same voltage than we can say, the 100 watt lightbulb will have the lower resistance. So I would be your answer for part C. That is the end of the solution. Thank you for watching.

Zulfiqar Ali · Answer

Let&#x27;s do part a. Well. The maximum voltage is determined simply from the army&#x27;s will. Teach with the usual relation which is maximum Will teach is equal to route two times we are a mess, right? And here we maximum musical too blue too Multiply be 120 with 100 twenties root mean square Baltes and therefore maximum voltage is equal to 170 walked Now let&#x27;s to part B The resistance is determined from given armies Will Tate in power. So our is equal to we are timis root Mean square Uh, Walt It square divided by p Well, we are mrs 120 square, divided by 60 and therefore are is equal to 240 arms right? And now we do part c Well, a scene from the relation the resistance will decrease for a bulb with higher Well, we&#x27;re so we hire with higher power resistance decreases resistance decreases decreases right

Problem

An array of solar panels produces 9.45 A of direc…

Problem 6 Easy Difficulty

A "75 -watt" lightbulb uses an average power of 75 $\mathrm{w}$ when connected to an rms voltage of 120 $\mathrm{V}$ . (a) What is the resistance of the
lightbulb? (b) What is the maximum current in the bulb? (c) What
is the maximum power used by the bulb at any given instant of time?

Answer

a) 192$\Omega$
b) 0.883 $\mathrm{A}$

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a) 192$\Omega$b) 0.883 $\mathrm{A}$

Physics

Elyse G.

Andy C.

Liev B.

Jared E.

Electric Charge

The Electric Force

James S. WalkerPhysics

Elyse G.

Andy C.

Liev B.

Jared E.

a) 192$\Omega$
b) 0.883 $\mathrm{A}$

James S. Walker

Physics