A lightbulb is in series with a 2.0 Ω resistor. The lightbulb ...

Question

A lightbulb is in series with a 2.0 Ω resistor. The lightbulb dissipates 10 W when this series circuit is connected to a 9.0 V battery. What is the current through the lightbulb? There are two possible answers; give both of them.

Accepted Answer

Welcome back. Everyone in this problem. A light bulb with a uniform resistance R is connected in series with two resistors, R one and R two in an electrical circuit. When a 12 volt battery is connected to the circuit, the light bulb dissipates 6.4 watts of power. Given that R one is two ohms and R two is three ohms determine the current passing through the light bulb. For our answer choices. A is 1.6 or 0.8 SB is 1.8 or 1.5 sc is two or three amps and D is 3.2 or 1.6 amps. Now, let's draw what's going on here to try and make sense of it. So, in essence, we have a circuit that's connected to a 12 volt battery that has a light bulb with resistance R. So I just represent it as a resistor and it's connected to two other resistors R one and R two. So let me label those on our diagram. We also know that our first resistor is two ohms. The second resistor is three ohms and we don't know the resistance of a light bulb but we know we know that the power dissipates when connected to the 12 volt battery is 6.4 watts. So the power of our light, but with resistance R equals 6.4 watts. And we're trying to determine the current that's passing through the light bulb. No, because it's connected in series, the current that's passing through the light bulb will be the same current that's passing through our 1st and 2nd resistors. So how can we figure out there is the current for our light bulb? What do we know that relates it to the information we have? Well, recall that power, we know the power, the light bulb dissipates and power equals I squared multiplied by R. Therefore, if we wanted to solve for the current, then we could divide both sides by R OK. And then find the square root of both sides. So I is the square root of P divided by R. Now we have the power but the thing is we don't know the resistance of our light bulb. So if we can figure it out, then we should be able to solve for the current. Now, let's think about it. What else do we know here? Well, we know that it's a closed electrical circuit. So Kirkoff loop rule applies. And Kirkoff loop rule tells us that the sum of changes in potential in a closed circuit must equal zero. So if we can apply Kirkoff loop rule, then we should be able to find the current that's flowing through our circuit. So let's go ahead and do that. So using the car cost little pool. Ok. Then that means starting at our 12 volt battery, assuming the current is going clockwise, 12 volts minus the potential difference across our light bulb, which would be I multiplied by R I think I have enough space to show you the diagram where we work. Yes, minus the potential difference across the two arms is that's two multiplied by I minus three, multiplied by I potential difference across the three ohm resistor. All of that would be equal to zero. No, we have two unknowns here. But we can still solve for I OK. Because that means we'll have 12 volts minus I all multiplied by, let's factor out I as gonna be R negative and negative two, either negative two, sorry, positive two, sorry and negative and negative three. I that's positive three. So all of that equals zero. Let's do some transposition here. We can carry our unknown to the other side of the equation and the two plus three equals five. So that means five plus R I, OK, five plus R equals 12. Therefore, I would be equal to 12 divided by five plus R. Now we still don't have an exact value for I, but we have two equations for I that are equal because if I equals the square of P divided by R and it equals 12 divided by five plus R. Then both of those must be the same. These two equations must be the same. In other words, this implies that the square root of P divided by R equals 12 divided by five plus R. Now we can go ahead and solve because we know what the value of P is. First, let's get rid of our square root. OK. So let me make a little space here. So first we want to get rid of our square root, we can square both sides. So that means P divided by R equals 12 squared, which is 144 divided by five plus R squared. We already know that the power dissipated is 6.4 watts. So we can replace that with 6.4 in our numerator and know to solve let's cross multiply. So this means that 6.4 multiplied by five plus R, all squared equals 144 R. If we expand our equation, then we can square what's going on inside inside the equation five plus R are squared. So it's going to be 6.4 multiplied by 25 plus 10 R plus R squared. None of that would be equal to 144 R. Next we can divide 144 by 6.4 which equals 22.5. So 25 plus 10 R plus R squared equals 22.5. R. Notice that we have like terms and we're also dealing with a squared term R squared. So that means we have a quadratic equation. So let's rewrite it in the general form for a quadratic equation. So we will subtract 22.5 R from both sides and rewriting it in the general form, we'll have R squared minus 12.5 R plus 25 equals zero. Now, as a quadratic equation, we can solve it by using the quadratic formula and recall, OK, recall that the quadratic formula states that for a quadratic formula in the general form A X squared plus BX plus C equals zero, then X will be equal to negative B plus R minus the square root of B squared minus four AC all divided by two A. No. For our problem, our variable is our, our value of A is one that's a coefficient of R squared. That's one, our value of B is negative 12.5 and our value for C equals 25. So subs substituting all these values into our equation, we will get negative negative 12.5 plus R minus the square root of negative 12.5 squared minus four multiplied by A, that's one multiplied by C that's 25. And all of that is being divided by two, multiplied by A, that's two multiplied by one. No, we can solve for our, when you put these values into your calculator, you'll find that we have two possible values. We have either R, one value for R will be equal to 2.5 ohms or another value for R could be 10 ohms. Now that we have these values for R and I don't want to call them R one and R two because we already named those resistors. So instead, let's call them R A and RB. OK. Now that we have those values for R, we can solve or, or find the current in our circuit. Because if we use our first equation, then that tells us that our current, remember it was a square of P divided by R. So that would have been the square root of 6.4 divided by 2.5. And when we put that into our calculator, we get one value for the current being equal to 1.6 pires. When our resistance is 10 ohms, then this tells us that our current would be the square root of 6.4 divided by 10. And when we put that into our calculator, we'll get the other value of the current being equal to 0.8 pires. Therefore, four or light bulbs, the current that's flowing through, it could have been either 1.6 pires or 0.8 pires. When we go back to our answer choices that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped

A lightbulb is in series with a 2.0 Ω resistor. The lightbulb dissipates 10 W when this series circuit is connected to a 9.0 V battery. What is the current through the lightbulb? There are two possible answers; give both of them.

Key Concepts

Ohm's Law

Power in Electrical Circuits

Series Circuits

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