Consider the circuit of Fig. E25.30 (d) Show that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

Video transcript

Hello, fellow physicists today, I want to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. You are provided with a circuit below, prove that the power supplied by the 13.0 volt cell is equal to the total power dissipated in the other components of the circuit. So our little picture here shows our diagram of a circuit we're shown in blue. The jagged line is represents the resistor and for the blue jagged line for 1.25 ohms. And then we also have a battery. It represented with blue for 6.0 volts. We're also given a battery for 13.0 volts, but it's in red. We're also given a resistor and with a jagged red line representing 1.85 ohms. We're also given three other different resistors which are represented by rectangles. So on the far left is a 9.50 ohms resistor. And then on the bottom of our circuit is a 5.00 ohms resistor. And then there's a 4.50 ohms resistor on the right. Ok. We're also given some multiple choice answers. They're all in the same units of watts. Let's read them off to see what our final answer might be or, and I should say to save some time. So the power cell is equal to the power circuit and that's all the same for all the multiple choice and they're all in the same units of watts. So let's read off to see what our final answer might be for the P cell is equal to the P circuit. So A is 4.12 B is 1.91 C is 2.58 D is 3.94 and E is eight 0.89. OK. So first off, let us recall and use the power equation and the power equation states that P power is equal to the current squared multiplied by the resistance. Also, let us recall that the power equation can also be written as P is equal to the voltage multiplied by the current to begin. We need to determine the current flowing through the circuit using Kirchhoff's loop rule. And Kirchhoff's loop rule states that the sum A V is equal to zero. Note that Kirch's loop loop rule states that the potential changes around a circuit are zero. Also, we must recall Ohm's Law which states that the potential difference is equal to the current multiplied by the resistance. OK. So now we can apply and use Kirchhoff's loop roll in the counter clockwise direction from the 13.0 volt battery. So 13.0 volts minus I multiplied by 1.85. So this is going to counter clockwise direction. So clockwise is go is goes this way, but we're going counterclockwise. So we're going this way, we're going the other way. Ok. So going from 13 volt to 1.85 to 9.50 et cetera. Ok. So one 0.85 os minus, I multiplied by 9.50 ohm. Since we're doing resistance minus, I multiplied by 5.00 arms minus, I multiplied by 4.50. Um minus 1.25 ohms or I should say minus I multiplied by minus, I multiplied by 1.25 films minus 6.00 volts is equal to zero. Ok. And this is going counterclockwise. So here to show our route and pink, we're going from the 13 battery and we're going around like that counterclockwise. Ok. And draw a counterclockwise line back in there. There we go. Ok. So we can simplify this. So we can say 13.0 volts minus 6.0 fault minus I multiplied by 1.8 five ohms plus nine point 50 ohms plus 5.00 arms plus 4.50 ohms was 1.25 ohms is equal to zero. So we can then go on even further and simplify and say 7.00 volts minus I multiplied by 22.1 ohms is equal to zero. So now we need to isolate and solve per eye using some algebra. And when we do that, we get I is equal to 7.00 volts divided by 22.1 arms which is equal to 0.317 years when we plug it into a calculator. So to solve for the power supplied by the 13.0 volt cell is given by the power of the 13.0 volt battery is equal to the power supplied represented by epsilon multiplied by I minus I squared multiplied by R. So let's plug in our known values. So the power supplied is 13.0 volts multiplied by our I value which we determine to be 0.317 years minus 0.317 amp here squared multiplied by our resistor which was 1.85 ohms. Let's rewrite that. Oh, ok. So when we plug that into a calculator, we get that piece of script, 13.0 volts is equal to 3.94 watts. And this is the power dissipated in the other components of the circuit can be found by adding the power values across the resistors and the 6.00 volt cell. So considering that we can say that the PC I'm gonna E circuit is equal to P of the 9.50 ohms plus P which or I should say P subscript 9.50 ohms plus P subscript 5.00 ohms plus P subscript 4.50. Oh, let's rewrite that 4.50 alms plus the power of E to the subscript of 6.00 volts. So the power of the circuit is equal to the power of the 9.50 resistor plus the power of the 5.00 resistor plus the power of the 4.50 resistor and the power of the 606.00 volt battery. So now we can plug in our known variables install for the power of the circuit. So the power of the circuit is equal to zero point 317 amperes squared multiplied by 9.50 arms plus 0.317 pi squared multiplied by 5.00 arms plus 0.317 squared, multiplied by 4.50 arms plus zero. Or I should say plus bracket parenthesis, 0.317 amp here squared. A multiplied by 1.25 oh or 125 volts plus 6.00 multiplied by 0.317 and piers bracket. So when we plug in that long equation into a calculator, we will get 3.94 watts and let's box, our correct answer in green. So this is our rounded value. So when you plug it into a calculator, you'll get something like 3.936 but we'll round it up to two decimal places. Uh Since all of our multiple choice answers were to two decimal places. Thus, the power cell equal to the power circuit is equal to 3.94 W watts, 3.94 watts. So looking at going back to look at our multiple choice answers, the correct answer has to be the letter D. The power cell is equal to the power circuit is equal to 3.94 watts. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.