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Cunent and Potcntial Dixgrams #ith Ammcters and Voltmcicis emmccr Tarmtiet We WilI ile for In 1 cincuit dingrm for a Volumneler MolrmFirt Ammclers Al Volunctets tkz Way We do bclow Complcle investigate why (visually ! Vchonl fint On whiteboard. juId then copy the follou Ing current diagnms (left) and potential diagrms (night). for # reference Kccp the same color scheme for cach diagrm (€ g ifthe cunscnsus Onlo {our Mre color or draw arruxheads Clfcnis uaat an potcutial hc sn1c Iec thcrunc colon Bc Sum not Potentil Diagrms Cumcne Diagrm< Do NOT dnw polential diagrun forthis CC (it $ comiplicted) In which of the circuit dugre aborc through docs the light bulb light? multinieter is likely nayc blown fus ? Which cifcui dinur Bnow amnieter that mcasures the CUrTent through the bulb? Ders cucull dienm voltmcter thit mcasures tha voltage across the bulb? Does circuit digrut cond shov 112-38.22



Answers

A $120-\mathrm{V}$ house circuit has the following light bulbs turned on: $40.0$ $\mathrm{W}, 60.0 \mathrm{~W}$, and $175.0 \mathrm{~W}$. Find the equivalent resistance of these lights.
House circuits are so constructed that each device is connected in parallel with the others. From $\mathrm{P}=V I=V^{2} / R$, for the first bulb $$ R_{1}=\frac{V^{2}}{P_{1}}=\frac{(120 \mathrm{~V})^{2}}{40.0 \mathrm{~W}}=360 \Omega $$ Similarly, $R_{2}=240 \Omega$ and $R_{3}=192 \Omega$. Because devices in a house circuit are in parallel, $$ \frac{1}{R_{\mathrm{eq}}}=\frac{1}{360 \Omega}+\frac{1}{240 \Omega}+\frac{1}{192 \Omega} \quad \text { or } \quad R_{\mathrm{eq}}=82.3 \Omega $$ As a check, note that the total power drawn from the line is $40.0$ $\mathrm{W}+60.0 \mathrm{~W}+75.0 \mathrm{~W}=75.0 \mathrm{~W}$. Then, using $\mathrm{P}=V^{2} / R$ $$ R_{\mathrm{eq}}=\frac{V^{2}}{\text { total power }}=\frac{(120 \mathrm{~V})^{2}}{175.0 \mathrm{~W}}=82.3 \Omega $$

High in the given problem. It is clear that in the first part of the problem, it is clear that in the given arrangement of the circuit Bulbs one and 2 are in series and bulb three is in parallel with both of them. Hence we can see it. The equivalent circuits, E. E and F. Are correct. And these circuits may be shown like this. This is the source providing the potential. Then for the bulbs we can show the resistance is So these are the two bulbs, one and 2 In series, and Bulb three in parallel with them. No, in the second part of the problem, we have to discuss which one of them will be the brightest and which one will be the dimmest. So as bulb three will have maximum current passing through it because it is having The smallest resistance as it is a single one. While the other branch having two resistors, 42 bulbs, the resistance will be doubled so as the resistance is the world. So the current passing through it will be minimum. Hands using expression for the power. All we can see heat dissipated in the bulk. H equals two I square artie. So the bulb in which the current will be maximum. That he developed will be maximum means the bulb will be brightest. That's why we can say that the bulb three maximum current passing through it. So it will be the brightest. Why one and two bulbs one and 2 being in series will be equally bright. Answer for the second part of the problem. Now, in the third part of the problem, resistance of each bill has been given us 24.0 home and IMF blight is given as we is equal to 6.0 Volt. Then for the series combination as the two bulbs are in series. So their resistance net resistance in series combination will be given by 24.0 plus 24.0 home, which will come out to be 48.0 oh no. As potential remains the same in parallel. So both of the branches will be having same potential across them hands. The current passing through Berlin one and to that will be equal means we can see Ivan Is equal to I two is equal to net potential applied divided by the net resistance artists, which is 6.0, divided by artists, which was 48 oh. So this current passing through the two Belts 1 and two comes out to be six x 48 Which is 0.125 mps. And that through I three that will be given by the total potential across It will be equal to the potential plight and divided by its resistance 24 or so, it comes out to be 0.25 MP here, so these two are the currents Passing through these two pumps. These three bulbs. Thank you.

We have to find a figure for solution. A park here. Val would two and three are connected. Are connected in parallel in parallel to the hypertension. So high potential side of diverting hide potential side of the battery. Thus, elements that eliminates C. And D. So any minute that is C. N. D. So we can see that Bulb one and 2 well one end who are connected and see it connected in city and E mm and F. R. A cube lint. And they all represent the circuit. So here are the answers. You represent the circuit dirties. E N. F. These are the answers. Stop. This is E. N. F. And the B. Part is there since the world since Well with three is in Peril Vida. C. Rich combination of worlds one and two. So the bold age drop across voltage drop across. But that is well Victory is the same as that or for that see rich combination. So here the power is equal to be square upon our and the light well the light bulb odd identical that is R one, R 2, R three as opposed to us. If the potential across if the port in here, across the battery. H. E. The power dissipated the power desecrated In ball with three. It's be critical to is square upon us. So that voltage dropped by the half. So the voltage drop. The bold age drop here. We can write the boldest drop by half across across each other each other. So Egypt. So now this is these are in the series combination in the series combination. Since there identical. Soap even is equal to P two is equal to e upon tools Holy square upon are so this is is squared up on Fora so this is be pity upon foot. So now ball with three. Ball with three is the brightest and one and 2 ball odd seen. So the answer is but who is the bright herbal? There's 33 is the brightest ball and velvet to Andrea same. And the next one is there see part. So for the valves Tracy through a vote Ball with three so I three. That was to be upon our that is six upon 24 is equal to 0.25 mps. So and the next one is trouble so and the one so this is valued three and this is value one and two. So if you call to Ivan is equal to I two is equal to E. Upon to us. 36.2 is multiplied by 24 is equal to 0.13 Mp. So the answers is the 0.25 M. P. A. And 0.13 mps.

Here we have psychic A e and F are equivalent. And they are the represent a circuit required. Here we will have. Yeah, but three Toby, brightest and by one and by two off equals the brightness, but less than bite us off. About three. We have I three equals 6 ft over 24 home. It was 0.25 Dampier and I want it. Was I to it was six folds over two, divided by 24 arm. It was 0.125 And the

What a off the given problem. The resistance off a bulb is given by potential square, divided by the power off the bulb. Eh? Potential. We have won 20 wards squared, divided by 25. This gives us the resistance off obeys 5 76 home. The resistance of a bull be is similarly given by the same for Mama. So we'll just substituted Chinese who won 20 square divided by 100 Excuse us 1 44 homes. But to be the direction for bulb they will be Don't a t A. Is called to CUNY body by a from the equation off the current that is charged by the current. From here, we could write I A is a Q is equal to off a resistance of 80 right by potential difference substituting your values. So one into 5 70 5 76 divided by 1 20 This gives us a duration for 12 days. 4.80 seconds. What? See, um, the charges seem however, yes, it leaves the bulb. So let's say this is a Baldwin. This is charged is it leaves the bulb. It is a glory potential. More potential. Then when it entered the bulb. Oh, buddy. We can find duration for Delta T A from the focal mother is work divided by fall off A. So work is one Joel, divide up, all 25. This gives a 0.0 for 00 seconds. What he energy entered the bold by electrical transmission, electrical transmission, transmission and leaves used by heat. He'd and radiation radiation. But if we can find the coast, my writing equation that is coast is equal to energy times the rate. Uh, well, first, find energy so energy will be energies. Uh, power times, Delta t power off a 25 times converting to kill awards. Be writing by 1000 words when playing with the time. That is 30 days times 24 hours. This gives us the energy off. 18 killed. Want you are. So what's up? Students value here. So 18 killer walk are times of rate is 0.11 $0 for Kimble. Want our So this gives us the cost off whole store for $1.98.


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