Electric circuit with Bulbs
Solved example: Power dissipated in bulbs
- [Instructor] We have a bulb of 10 volt 50 watt rating, connected across a five volt battery. We are asked to calculate the power dissipated in the bulb. Now, when I saw questions like this for the first time, where we had bulb and some rating was mentioned, I used to always get confused over here. My main confusion was: we're already given power of the bulb as 50 watts, and we're again asked to find the power dissipated in the bulb. So what's going on? So you know what, let's do this step-by-step. The first step to solving any problem is to write down what is asked of us. We are asked to calculate the power dissipated in the bulb. The next step is to write down what is given to us. We are given the rating. The rating is 10 volt, 50 watt. And we're also given the voltage across it is five volt. So this voltage across the bulb, that voltage is going to be five volt, because the voltage across these two points is the same as the voltage across these two points. That is five volts, so the voltage across the bulb is given as five volts. And again, coming back to the question, what's really asked for us? We are already given the power, isn't it? So we need to understand the meaning of the rating. What does this rating mean? You see, whenever rating is mentioned like this, this tells us, that if the voltage across the bulb was 10 volts, then the power dissipated in that bulb would have been 50 watts. Does that make sense? That's what the rating of any device always means. For this much voltage, if this was the voltage across that device, then the power across that, power dissipated would be 50 watts. But clearly, in this example, the voltage across the bulb is not 10 volts, we can see it's 5 volts. So the power dissipated in this bulb won't be 50 watt, in fact we can guess it's gonna be lower than that. And that's exactly what we need to calculate. How much will be the power dissipated in this scenario? So, how do we solve this? If you look at the difference between this question, this problem, and the problems we have solved earlier in circuits, the major difference is that in all the earlier problems, we were given the resistance of the devices which are connected. Most of the time, there were just resistors connected, and we were given the resistance, but over here, we don't know the resistance of the bulb. If we knew the resistance, the problem is done, because if we know the resistance, then we can calculate the current through the circuit using Ohm's Law, we have done that before. And once we know the current and we know the voltage, we can go ahead and calculate the power, because power is just the power of voltage and current. So the big question, or a first big step over here, is going to be to figure out the resistance of this bulb. So how do we do that? The secret lies over here. We already know that the bulb must have some resistance such that when you put 10 volt across it, 50 watt must get dissipated in it. So using this information, we can figure out the resistance of the bulb. And so once we have the resistance, then we can go ahead and solve the circuit, like what we did in all previous cases. So, it would be a great idea to pause this video over here and see if you can try to do this yourself now. All right, let's see. To calculate the resistance, we need a relation connecting power, voltage, and resistance. So, let's write down what we know about power. We know that power is the product of voltage and current. That's the main formula for power. And then if you use Ohm's Law, we can substitute V equals I-R, and so we can also write this as I squared R. This is not a new formula! This is the same thing as before, we just used V is I-R, we just substituted Ohm's Law, that's it. So there's only one formula for power, okay? And of course, we can also substitute I equals V over R. Again, Ohm's Law, and we'll get P as V squared over R. And depending on what we need, we can use any of these to solve a problem. Now, we want power, voltage, and resistance. Power, voltage, and resistance. So let's go ahead and use this formula to figure out the resistance of the bulb. So, we will use P equals V squared over R, and I'm substituting over here, so if the power is 50 watt, so let's substitute that, for 50 watt of power, and when the voltage is 10 volt, so 10 square, what should be the resistance? So divide by R. So from this, R is going to be 10 squared divided by 50, and 10 squared is a hundred, that's a hundred, and so one zero cancels, five goes two times, and from this we know the resistance of this bulb is two ohms. And so we have now what you know the first big step! We have calculated the resistance of this bulb. And once we know the resistance, we can treat this as a simple circuit problem, which we have solved before. So, let me just go ahead and redraw that. So our circuit now has a resistance of two ohms, and it is connected across a five volt, five volt battery, five volt cell. So we know the voltage over here is five volt. And we need to now calculate the power dissipated over here. So again, we can use any of these formulae. We can either calculate the current in the circuit by using Ohm's Law and then power is equal to voltage into current, or we can again directly use this formula, because we know we want to calculate the power across the resistor, right? So we know the voltage across the resistor is five volt, we know the resistance is two ohms, so we can now calculate what the new power is. So let's do that. Power is V squared over R, that's going to be V squared, V squared is five squared, over R, R is two ohms, and that's going to be 25 divided by two, that is 12.5 watts. Tada! There we have it. That is the solution to our problem! So as we predicted, the power dissipated in this bulb will be less than 50 watt, 12.5 watt, and so the bulb won't glow as bright as it should have. And so to summarize, whenever we have problems like this, the first step is to always figure out the resistance of the bulb. The rating is always useful to calculate the resistance of that device, and then once we have the resistance, we'll treat it as a simple circuit problem and solve it. Let's try another problem. This time, we have two bulbs, one with a 20 volt and 200 watt power rating, and another one with 20 volt and 50 watt power rating, connected in series. This whole connection is connected across a 40 volt battery. We're asked to calculate which bulb glows brighter. Basically, we need to figure out in which bulb there will be more power dissipated. So the steps are going to be pretty similar, we're going to use this power rating to figure out the resistance of each bulb, and once we have the resistances, then we can go ahead and treat it as a circuit with resistance in series and try and calculate the power in each resistor. So again, try and solve this yourself. Pause the video and see if you can do this yourself first. All right, I'll skip the initial steps of writing down the data and everything, so that we can save space. Let's directly jump to calculating the resistance. To calculate resistance, we need connection between voltage, power, and resistance, and so just like before, we can use P equals V squared over R for this and for this, and if you do that, I leave that calculation to you, I don't want to waste space over here, but if you do that, we will see that this bulb has a resistance of two ohms. So this is a two ohm resistance. And this bulb has a resistance of eight ohms. Eight ohms. And you can, you know, pause and just verify that. And so we have this circuit now. Forty volts. And now we need to calculate the power across each resistor. How do we do that? Well, we can do it multiple ways. One is we can figure out what is the voltage across each resistor, and then we can use V squared over R and calculate the power, or we can calculate the current through each resistor, and then we can use I squared into R. So either ways we can do it. All right again, if you have not done this before, pause the video and see if you can try this yourself. We have series connection and we have solved this before, so just go ahead and try this yourself. All right, what I will do is I'm going to solve this circuit, this is a series connection, I will try to find out the current in the circuit and then I will use I squared into R. All right, since the resistors are in series, the total resistance adds up, and I get 10 ohms over here. So this would be a 10 ohm resistance. This is 10 ohms. And the power supply is 40 volts, battery is 40 volts, so now we have one resistor, the voltage across that is 40, the resistance is 10, so we can go ahead and use Ohm's Law. So Ohm's Law says V equals I-R, so I is going to be V over R, V is 40, R is 10, and so the current is going to be four amps. Four amps, and therefore, the current in this circuit is also going to be four amps. The same circuit, right? Four amps. And therefore current through each resistor is going to be four amps. And now the power in each can be calculated, I squared into R. So over here, four square into R, I'm just going to try it directly, four square is 16, 16 into two is 32. So the power dissipated over here, let's use green for that, so the power dissipated over here is going to be 32 watts in the first bulb. The second case, again, I squared into R, I is four, 16 into eight, I have to do this, 16 times eight, 48, eight ones are eight, 12, 128. So over here, the power dissipated would be 128 watts. And so since there is a much higher power dissipated in the second bulb, the second bulb glows much brighter. And that's how we solve problems whenever we have bulbs. Again, notice the steps are the same. We first figure out the resistance and then we treat it as a resistance problem. And what's interesting to see is that when these bulbs are in series, the one which had a lower power rating, that's the one which glows brighter. And so can you think of why this is happening? I mean, we understood everything solid, but can you logically try to understand as to why the one which has a lower power rating is glowing brighter than the one which has a higher power rating? Think about this.