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The circuit shown has battery and resistors 6ov (12 pts_ Find current voltage for each resistor (3 pts) Find the power delivered by the battery consumed by the stro...

Question

The circuit shown has battery and resistors 6ov (12 pts_ Find current voltage for each resistor (3 pts) Find the power delivered by the battery consumed by the strongest resistor 528 3150 38 } 52 45

The circuit shown has battery and resistors 6ov (12 pts_ Find current voltage for each resistor (3 pts) Find the power delivered by the battery consumed by the strongest resistor 528 3150 38 } 52 45



Answers

Consider the circuit shown below. The terminal voltages of the batteries are shown. (a) Find the equivalent resistance of the circuit and the current out of the battery. (b) Find the current through each resistor. (c) Find the potential drop across each resistor. (d) Find the power dissipated by each resistor. (e) Find the total power supplied by the batteries.

So in this question we have a complex circuit Uh and we know that the total voltages, nine volts of the battery. And we know that the power in each resistor is 1.5 watts. We want to know what the resistance of each resistor is. So in order to this we need to set up a table voltage, current resistance and power. Okay, I'm gonna need four rows for resistor 123. And our totals, Yeah, I'll throw down a little bit and I'll put in our givens in black. So our total voltage is 9V And each of our powers is 1.5 watts. So now we gotta use some rules and uh Homes Law to figure out what's going on. So we put the rules in red and we'll fill in the blanks as we go. So number one, the first rule that I'm gonna use is the total power has to be equal to the sum of the individual powers. So that means our total power 1.5 plus 1.5 times one plus 1.5 is 4.5 second I'm gonna use power equals voltage times current. To figure out my total current. So .4.5 divided by 9V is .5 eps. Next I'm going to use Holmes law. R equals B over. I. To figure out the total resistance which is 18 oops. Now we have to break this down into individual parts. Okay, so resistor, one Is in series with the battery, which means that the total resistance has to be equal to the sorry, the total current has to be equal to the current. And resistor one. So that means that this is .5 amps. To step five. We're gonna use power equals voltage times current again To get our voltage across the 1st row. So 1.5 divided by .5, gives me three. Again, I'm gonna use R equals the over I for our first row. And three divided by 30.5 is six homes. So our first resistor is six owners. So the next thing that we know is that the total voltage has to be voltage one plus are parallel section and within the parallel section those two voltages have to be the same. So 9 -3 is six. Both of these have six volts. Again, I'm going to use the power equation. Power equals the over I. You could do this slightly differently but I like doing it this way, So 1.5-5 x six. Give me a .25 0.25. And finally, and this number these this should be step seven and step eight And this is step nine. This is our final step across the last two rows, I'm gonna use R equals B over I two more times. So 65.25 is 24 and 24. So resistors two and 3 are 24 homes and just the one is six homes

So for this problem, we're looking forward, the direction and the magnitude of the current going through all these resistors. So first, let's take a guess as to what direction the currents air going in. And then we can write some equations to figure out if our guesses are correct, and so current typically goes out of the positive end of a battery in into the negative end. And so let's take a guess that the current going through our one is going in this direction and we'll call I want, and then we'll make the same guests for our three will say it's going out of the positive end of the battery. Yes, that will be our direction for I three and then for our two. Let's say it's going this way and we'll call it I two and again, these are just guesses. We'll see if they're right when we write our equations out. And so the first equation we're going to write has to do with the junction rule at this junction here. The amount of current going into the junction has to equal the amount of current going out of the junction, So the current going in is going to be I one plus I three and then the current going out is going to be I two. That's going to be our first equation that we're going to use to solve this. And then for the second equation, we're going to look at two of the loops the first loop will look at is going to be this loop, and we're gonna write down all the voltage drops along that loop. And so starting right here we have E m F one and then we're going to subtract the vultures, drop across the resistor r one That's going to be I one times are one and then keep going around clockwise. We're going to subtract the voltage drop across resistor to which is I two times are too and that's all going to equal zero. Then we're gonna do the same thing for this loop here, starting with the first battery battery number two. Then we're going to subtract the voltage drop from resistor three. So we have I three are three, then we're going to subtract the voltage drop from resistor to so I two r two and that completes the loop. And so that's going to equal zero now. There is a lot of different ways to consol of this system of equations. Um, do whichever way works best for you. But if you solve it by plugging in the values of the MF's from the battery and the resisters, you're able to solve it for I one I two and I three. And what we find is that I one is equal 2.133 amps. I to is equal 2.315 AM's and I three is equal 2.181 amps. Now, all of these values are positive, which means that our guesses for the direction of each of these currents is gonna be correct. So for the directions will use the three directions we have in the diagram here and then for the magnitudes. We'll use what we found from the equations

So for this problem, we can go ahead and look at the bullet is drops across each of these registers. So the way you do that is this kind of imagining of a volt ometer here. Ah, vote ometer here and a well, Tom, enter here. All right, so let's look at the first resistor so we can go ahead and quickly calculate the current the current is gonna equal v 90. The equivalent resistance of this whole current is gonna be 30. So our current is gonna be three amps, All right? With that in mind, we know that current stays the same in a serious circuit. So that's helpful because we need to calculate the voltage drop across all of these resistors. So we have hold his drop is we're solving for and we're gonna call this our one r two and r three. So now, in our one the voters drop is gonna be equal to the Resistance Times, the times, the current that flows story, which is gonna be 30 in our two we're gonna have the voltage drop is gonna equal the Resistance times. A current that flows through its kind of equal to 45 and same case in our three. Now, if you look at all these, if we add all these together when we get we get 90 force, which is the total voted supplied by the battery. So if you think about it, as each of these as the current goes through at each of these resistors, the current drops are the sorry that both his drops the voters drops in the voltage drops. Until when it comes back into the negative terminal, you're gonna have a voltage of zero. And that's one of the ways you can verify these types of questions. What?

So in this problem we have t Ray stirs in Syria's is connected by a battery of old is V restaurant one is tomb minister to his tour Warm and rest or three is three on also the maximum power reading for a restroom. One is for what? That for the rest or two is 10. What? And for the rest refire racers created by what So in problem A part? A. We want to find the maximum voltage without burning up one of those wasters. All right, so know that the maximum bold is in the circuit. Depends on the maximum craned each resistance can't taller it. Right. So let's first find out The maximum grand is racers Can't told her it. So to do that, we just use p max equals I score Mexico over our right. So that means Forrester one p one max is going to be Hey, one Max squared over race since one. Great. So this is going to give us I one max. It's going to be square root of p one max over our one. Right. So this is going to be p one maxes for what are one is to what which is gonna be read to and this is gonna give me 1.4 amps now. Similarly, for I to Max, that's going to be P two. Max over are too square root of that. And this is going to give me 0.913 amps. And for I three Max, it is going to be p three max over our three. And take the square root of that and this will give us 1.3 amps. Now, here's the thing, since the resistor too is the one that has less amount of current, right, So and the rest are two is also in the serie circuits, so we don't want the current to be higher than 0.913 Otherwise, the rest er two is gonna bring, and that's what we want, right? We don't want to burn the rest or any of the researchers. So now we have to set the Quran at 0.1913 amps. All right, so that means the current of the eye. The Siri's circuit has to be called thio I to Max. This is three, by the way. Don't be confused. All right? and this is gonna be 0.913 amps. So this is the maximum current that the circuit can hold and fi Max is going to be. I ask Times R s now our s we haven't calculated yet. R s is basically are one plus are two plus our three. Let's go to page One again, If you add this year are one is to Toto plus 12 14 plus 3 17 So this is going to be 17 home. All right, now from here be Max the max is going to be He called his syrup 0.913 amps times 17 um, and this is going to give us 15.5 fold, so that is going to be the maximum voltage. And now, for part B, we just want to find the power. Okay, Power of the circuit. So that's gonna be I times I s times fi, Max, and this is gonna be 0.913 amps, times 15.5 fault. And this will give us 14 points. Two What? This is gonna be 14 point to what


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