Consider the circuit of Fig. E25.30

Question

. (b) What is the power output of the 16.0-V battery?

Accepted Answer

Hey everyone in this problem, we have a section of the circuit of an electrical appliance shown below and were asked to determine the power supplied by the 13.5 volt power supply. Alright, so we have this 13.5 volt power supply shown in purple. Okay. And we see that it has an internal resistor of 2. homes. Now, when we're talking about the power supplied by a particular power supply, recall that this is given by epsilon I minus I squared. Little R. Okay, so we have the electro motive force epsilon, the current I made the current I squared. And then little R. Which is the internal resistance. And our problem, which of these values do we know? Well, we know the electro motive force, the E. M. F. Okay, because we're talking about the 13.5 volt power supply, this is just going to be 13.5V. We also know little are the internal resistance from this diagram, it's going to be the resistance of this purple resistor indicating that it's Part of this power supply and that is going to be 2.2 homes. Alright, so we have epsilon. We have our what about I? Well we don't know our current but we do know that we can find this current using Kirchoff slaw with the summer voltages. So we're gonna use kirchhoff slaw. Where the sum of the voltages around this closed loop circuit is equal to zero. Okay, In order to find I. Alright, so we're gonna pick right here as our starting point and we're going to travel in the counter clockwise direction. So the first thing we're going to encounter is this power supply of 13.5V. We're traveling through it from the negative to the positive end. Okay, so that's going to be a positive voltage. So we get 13.5V there. No we're traveling through these resistors. We're going to assume that we're going in the direction of our current. Okay? And we'll see whether we get a positive or negative at the end. Whether that assumption is true. But for now we're going to assume that this is the direction of our current. So when we travel through these resistors, those are going to give us negative values. Now we're talking about the summer voltages and remember that the voltage v is equal to the current. I. Times of resistance are when we come across a resistor we can write this as and again it's negative because we're in the direction of the current, we can write this as I the current times of value of that resistor or the resistance to point to OEMs. Now we keep moving along and we hit this eight ohm resistor. So we do the same thing. Okay, again we're in the direction of the current. So this is going to be negative. We have I times the resistance eight homes. Okay, I times r gives us a voltage, keep moving along, we hit the six ohm resistor and the exact same thing, it's negative because we're in the direction of current. I times are so I times six ohms Moving along, we hit this resistor of one ohm. And we do the same thing I times one ohm it's negative because we're going in the direction of the current. Now the last thing is this last power supply of 7.5V were traveling through it from the positive to the negative end. Okay, that's going to be a negative voltage. And so we get negative 7.5V there and all of this is going to equal to zero. All right, so on the left hand side we're gonna keep our voltage is we're gonna move these currents and resistance is to the right hand side. So we have 13.5 volts minus 7.5 volts. That's gonna give us six volts. And on the right hand side now we know the current moving through. Each resistor is going to be the same. Okay, so we can factor out this i it's the same for each of these and then we're just gonna have the sum of all of these resistance is 2.2 homes Plus eight homes and these are positive because we moved them over to the right hand side Plus six owns last one. So this is gonna give us six volts is equal to the current I times 17.2 ohms. And now to find I would just need to divide by 17.2. And we're gonna get that I is equal to 0.34884. Okay, voltage divided by OEMs or volts divided by OEM sorry. Or it's gonna give us amperes. So we have amps. Alright, so we found our current I so we can get back to the power. Remember we're trying to find the power and in order to do that we needed epsilon. The current. I and the internal resistance. Are we already knew epsilon and are now we found I. So we can go ahead plug those values into that equation and solve. So the power P is going to be equal to again epsilon I minus I squared r. Okay, this is going to be equal to 13.5 volts times 0.34 84 amps -0.34 Amps squared times of resistance of 2.2. Oops. All right now we have a volt times amp here we're gonna get the unit of what And then here we have amps squared times. Well, an amp times and gives us a volt. So then we get volts times amps as well which gives us the unit of what, which is what we want for power. All right. And if we work this out, this is gonna give us 4. watts for our power supplied by that 13.5 volt power supply. And if we compare this with our answer choices, okay. Um, and we use two significant digits. We see that this is going to correspond with answer choice. A 4.4 watts. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

Consider the circuit of Fig. E25.30
. (b) What is the power output of the 16.0-V battery?

Watch next

Power in Circuits practice set

Struggling with Physics?

Consider the circuit of Fig. E25.30. (b) What is the power output of the 16.0-V battery?

Watch next

Power in Circuits practice set

Consider the circuit of Fig. E25.30
. (b) What is the power output of the 16.0-V battery?