The five identical bulbs in FIGURE EX28.11 are all glowing. The b...

Question

The five identical bulbs in FIGURE EX28.11 are all glowing. The battery is ideal. What is the order of brightness of the bulbs, from brightest to dimmest? Some may be equal.

Accepted Answer

Welcome back. Everyone in the figure below all six identical bulbs emit light. The batter used is perfect. What is the sequence of brightness for these bulbs listed? From the brightest to the least bright? Here, we have a set of bulbs, Tuvwxy. And for our answer choices, A says that U has the same brightness and T as T and Y and they are greater than VW and XB says T equals Y which is greater than U, which is greater than V that's equal to W. And XC says U is greater than T, which is greater than Y, which is greater than V that's equal to W and X and D says U is greater than T, which is greater than Y, which is greater than V, which is greater than W, which is greater than X. So from our problem, we need to figure out which of these bulbs are greater, less or equal to each other in terms of the amount of light that they emit. Well, the one that will shine, the brightest will have the most power and recall that power is proportional to current. In other words, that means the ones with the most current will be able to shine the brightest. So if we are able to find a current flowing through each bulb, then we'll be able to order them. So let's see if we can figure that out. So recall that current equals EMF divided by resistance. So let's picture our current here. So it's leaving our voltage source and notice that it goes to why it also goes to t So Kerkhoff junction rule comes into play here because it tells us that the amount of current entering a junction will be the amount of current leaving the Jun Khan. So notice that it gets to our junction at Y and it goes to three places it goes to Y. So let me, let me label our currents here. So we have our original current and one current goes to Y, one current goes to you and one current goes to T OK. But no, OK, let me, let me make sure I represent that properly because it's really important to understand what's going on here. Now, we have a current that's going to you and after it leaves you, it goes to VW and X but VW and X are connected in parallel. So that means our current is going to split equally for V for W and for X. So now the question is which ones have the most power? Well, if we think about it, our current is equally split between Tu and Y. So that means the current going through T will be the same as the current going through. Y. So that's one thing we note so far and those will have the most current because since they have identical uh identical, since the bulbs are identical, that means they have equal resistance. No, what we need to consider is the current that's flowing through the branch in the center. OK. So what's going on in our center? Well, here in our center, the current that's flowing through you is not the same as the current through T. And why? Because we have our resistors connected together in both series and parallel. So we'll need to find the equivalent resistance before we can determine how much current is flowing through the center branch. So let's see if we can figure that out. So in the center, OK, in the center, our resistance will be equal to the equivalent resistance of our resistance in the center branch. OK. And that equivalent resistance will be equal to the resistance of U which is R plus the resistance of all our bulbs VW and X connected in parallel. So that is going to be the reciprocal of R plus the reciprocal of R plus the reciprocal of R are reciprocated no one over R plus one over R plus one over R equals three over R. So equivalent resistance is R plus three divided by R. The inverse of that an inverse of three divided by R equals are divided by three. Now R plus R divided by three, if we think about it, if we write that as a fraction, that would be three, R divided by three plus R over three. So our equivalent resistance is four R divided by three. So that means for our center branch, the current that's flowing through the center, which would have been, you equals our voltage, we know that they are connected in parallel. So it's the same voltage throat. So it's our voltage divided by four are divided by three or or equivalent resistance. And that would be written as three quarters of E divided by R. Now, even if we don't have the values, let's think about it. Conceptually if we have the current flowing through T and Y being E divided by R and the current flowing through U being three quarters of E divided by R. That means the current through U is three quarters, the current flowing through T and Y. So the current flowing through U must be less than the current flowing through T and Y. So that tells us then that the current through T, which is the current through Y is greater than the current flowing through you. So we already have a part of our statement here. We already have a part of our sequence rather, we just need to figure out the rest. In other words, no, we need to compare the current flowing through U to the current that's flowing through VW and X, how can we do that? Well, again, based on Kirk's junction rule, the current that enters a junction will be equal to the current leaving the junction. Since our bulbs have an equal resistance, that means our current is going to split equally into three parts for each bulb and then leave the junction right here. OK. So what does that tell us? Then? That means, so let's write that out. So since the current entering the junction for VW and X is split equally, and that's because the resistance for all are equal, then that current entering or sorry, the current for one of our bulbs, let's say the current four V would be one third of the current that's entering the junction, which is the current flowing through you. And we already know that the current flowing through U is three quarters of the EMF divided by R. So that's gonna be a third of the current flowing through you. And that's a third of 3/4 multiplied by EMF divided by R. If we do some canceling here, three, cancels three and no, we're left with our EMF divided by four, multiplied by I. In other words, if we were to write it as the coefficient of the EMF divided by R, that would be a quarter of EMF divided by R. So no, that means the current flowing through V, which is the same as the coin flowing through dub it would also be a quarter of our EMF divided by R and the current flowing through X would also be a quarter of our EMF divided by R. So now let's think about it. Remember we said the current flowing through you is three quarters of the EF divided by R. So if the current through VW and X is a quarter of the EMF divided by R, then that means the current flowing through VW and X must be less than the current flowing through U. So what does this tell us? Then this tells us or I can write here. Therefore, the current flowing through T which is equal to the current flowing through Y is greater than the current flowing through U. And the current flowing through U is greater than the current flowing through V W and X. Now, remember we said the current is proportional to the power. More current means more power and more power means our bulb is shining brighter. So that means we can assume then that in order of brightness, teeth will have the same brightness as Y. Both of them will be greater than you and you will be greater than the brightness of VW and X. So that will be the sequence for the brightness of our bulb. When we go back to our answer choices that would have been answer choice. B thanks a lot for watching everyone. I hope this video helped

Struggling with Physics?

The five identical bulbs in FIGURE EX28.11 are all glowing. The battery is ideal. What is the order of brightness of the bulbs, from brightest to dimmest? Some may be equal.

Watch next

Power in Circuits practice set