Consider the circuit of Fig. E25.30
. (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery?

Question

Consider the circuit of Fig. E25.30

Circuit diagram showing resistors and voltage sources for calculating power in an 8.0-V battery.

. (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery?

Accepted Answer

Hey, everyone. Welcome back in this problem. We have an electric circuit with two power sources connected as shown in the diagram. Now, the diagram, we have a point S and if we travel counterclockwise from that point S we hit from the negative end to the positive end, a 20 volt power supply With an internal resistance of 2.85. If we continue counterclockwise, we then hit a resistor With resistance 9. homes Continuing counterclockwise. A resistance of 5.50ms, A resistance of 6.5 amps. Then we did an internal resistor of 1.25 homes associated with a six volt power supply going from positive to negative terminal. Okay. And then we're back to that point. We started out and were asked to determine the power At which the electrical energy is converted to other forms in the six volt power supply. Alright, so we're talking about power and the power P it's going to be equal to the voltage V times the current. I recall this formula. And let's start with the current. I, we don't know the current. I. So let's try to find them. How can we find that, well, we're gonna use Kirchoff Slaw, which tells us that the sum of the voltages around this closed loop is going to be zero. Now, we're gonna start at point S okay and moving the counterclockwise direction, we're gonna assume the current goes in that direction. So we're starting with this 20 volt supply going from negative to positive. And so that's going to be a positive 20V. We're gonna keep going counterclockwise and we encounter the 2.85 M internal resistor. Now we have our resistance here, but we don't know the voltage I recall however, that the voltage I is equal to the current or sorry, the voltage V is equal to the current I times of resistance are okay. So we don't know the voltage V but we can find it through I times R. So we can write 2.850m resistor. The voltage is going to be I times 2. homes. Now, I've left a space here so we can fill in the sign. We're assuming that our current is going counterclockwise. So it's going through this resistor. And so this resistance is going to be negative because we're following the direction of the current, same with the next resistor, it's going to be negative value. And again, we have the resistance are so we're going to multiply that by I in order to get our value of VI times 9. we're gonna do the same for the next resistor minus I times 5.5 OEMs. Again, these resistors cause a negative voltage because we're going through them in the direction of the current. I'm gonna add the last two terms below here because we're running out of space. So we have minus I. Okay. Again, we're going through the 6.5 ohm resistor. Now we have minus I times 6.5 homes, the 1.25 oh, my sister is going to be minus I times 1.25 OEMs when we're talking about the voltage And then we're going through this six volt power supply from positive to negative because we're going from the positive terminal to the negative terminal. This is going to be a negative voltage. We have -6V and all of this added together is equal to zero. Alright, so we've got all of this written out the hard part's done. Now, we just need to simplify and again, we're looking for the current I, so we can use that in a power calculation. And you'll notice that's the only unknown in our equation. Now, we have these voltages that we have values for all these resistance is we have values for. Let's find I, so let's move all of the eye terms to the right hand side. We'll keep the voltage on the left hand side, we have 20 volts minus six volts which gives us 14 volts and on the right hand side, we can factor out that current I And then we have 2.85 homes plus 9.5 homes plus 5.5 homes plus 6.5 homes plus 1. homes. And just double check that you get all of the resistors in there. It can be really easy to Miss one. and then the values won't add up properly. Alright. So we want to find we can divide by the value of these resistance is added together. We end up with 14 volts divided by 25.6 OS which gives us a current I of 0. amps. Okay. Alright. We have our current, I we're looking for the power, let's go back up and just look at our equation and we said that the power is equal to V times I, we found the current I V. We have to be a little bit careful because the power supply that we're looking for the voltage on has internal resistance. So it has internal resistance. We can't just write it a V. We have to write it as the following. The power P is equal to the E M F epsilon plus the current I times the internal resistance times the current I. Okay. So we've replaced RV term with epsilon plus I R. Now we have a plus sign here because we're talking about um the power converted in the power supply. Okay. We're not talking about power output. We're talking about the power in the power supply. And so we have a positive value here. So just be careful depending on the situation that might need to be a negative value. Alright, so the EMF for our power supply is 6V the internal or sorry, the current, okay. We're in series of current is the same throughout the resisters. We have 0.546875 amps times the internal resistor. Now, we're talking about the resistor associated with our six volt battery or six volt power supply. And that is the resistor here drawn in blue in our diagram of 1.25. So the internal resistance little r is 1.250ms and all of this is multiplied by that current. Again, that same current we found before the current in series is going to be the same through all of those resistors. So 0.546875. And so this first bracket six volts plus I times R that's gonna give us 6. 3593, Times 0. 6875 Amps. Okay. If we multiply this together on our calculator, we get 3. watts approximately for our power. Alright, so we found the power we were looking for, we go back up to our answer choices. Okay. We can round to three significant digits and we find that the power with which electrical energy is converted to other energy forms in the six volt power supply is going to be answer choice. A it's about 3.66 watts. Thanks everyone for watching. I hope this video helps you in the next one.

Consider the circuit of Fig. E25.30
. (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery?

Verified Solution

Key Concepts

Ohm's Law

Power in Electrical Circuits

Series and Parallel Circuits

Consider the circuit of Fig. E25.30. (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery?

Verified Solution

Key Concepts

Ohm's Law

Power in Electrical Circuits

Series and Parallel Circuits

Consider the circuit of Fig. E25.30
. (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery?