Consider the circuit of Fig. E25.30

Question

. (a) What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors?

Accepted Answer

Hey, everyone in this problem, we have a group of students that make the circuit shown below and were asked to determine the combined rate at which the four own and seven ohm resistors convert electrical energy into heat. Now, when we're talking about converting electrical energy into heat, what we really want to find is the power and the power recall is given by the voltage V times. The current I now recall that we also have homes law where V is equal to I times are, the voltage is equal to the current times the resistance. And so we can write the power as I squared times are okay because we can write the as I times R. So we have some information about the resistors. Okay. We know that there are four AM and seven AM. We need to find the current I in order to find our power P. Now, how can we do that? Well, let's recall, Kirchoff Slaw Kirchhoff law tells us that the sum of the voltages around this closed loop is going to be equal to zero. So we assume that the currents traveling in the counterclockwise direction. Okay. And we're gonna see from the sign, whether that's true or not, we're going to assume it's traveling the counter clockwise direction. Now we're gonna start here, We're gonna move counterclockwise. So what we're gonna do first is go through this 12 volt source, We're going from the negative end to the positive end. And so that's going to be a positive voltage. And so our first voltage in our summer voltages is going to be 12V. Now we keep going counterclockwise. We're gonna hit a resistor. Okay. And we don't have information about the voltage of this resistor. We have information about the resistance are, but recall again through OEMs law, we can relate the voltage to the resistance through V is equal to I times are. So we can instead write this as minus I Times 3.2. Now, why did I write that as negative? Well, we're traveling in the direction of the current which means that when we go through a resistor, the voltage is going to be negative. Alright. So we can continue going around, we hit another resistor of seven OEMs again, we're going in the direction of the current. And so this is going to be a negative voltage and it's gonna be I the current times the resistance R which is seven, we keep going, we hit this resistor of forums, same thing in the direction of current. So the voltage is negative in a resistor, we get I times of resistance are which is four times we keep going, we hit this internal resistor here of one own exact same thing direction of the current. And so our voltage is negative, we get I times one own. And finally, we have this 7.5 volt source that we're going through our battery, whatever it may be, and we're going from the positive end to the negative end. And so that is also going to be a negative voltage be of negative 7.5V and all of this is equal to zero. Now, when you look at this equation, you'll see that the eyes, okay. The current is the only thing we don't know and we know that these eyes are the same because this isn't serious. Okay. So the current flowing through each resistor is going to be equal. So all these eyes are the same. Now we can solve for I and then use that in our power equation. So we have 12 volts minus 7.5 volts. That's going to be 4.5 volts moving everything over all these terms with eye to the right hand side and factoring out the eye, we get I times 3.2 ohms plus seven ohms plus four ohms plus one oh And this is going to be equal to i times 15.20ms. And if we divide by 15.2 ohms, we have 4.5 volts divided by 15.2 ohms volts divided by own, the unit is going to be amp like we want for current, we get a value of 0. amps. And that is the current flowing through. And you can see that it's positive and so are assumed direction was correct. Now, let's move to the power and again, our power P is going to be equal to I squared times R and the total power that we're looking for, we're asked for the combined rate at which the four ohm and seven ohm resistors convert electrical energy to heat. And so our total power that we're looking for is going to be the power from the four ohm resistor plus the power from the seven ohm resistor, which is going to be now they both have the same current. So we get I squared times the resistance in four plus I squared times the resistance of seven using this power relationship that we worked out before where P is equal to I squared R. No, we can factor out the I squared. And so we're gonna have 0. 53 amps, all squared times the resistance of four, the four ohm resistors. So, forums plus the resistance of the seven ohm resistor. So seven os And when we work this out on our calculators, we're going to get a power of 0.964- watts. Okay. And not power is The rate at which those two resistors convert electrical energy into heat. And if we go back up to our answer choices, we see that if we go to two significant digits, this is going to be answer choice B 0.96 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
. (a) What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors?

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Power in Circuits practice set

Struggling with Physics?

Consider the circuit of Fig. E25.30. (a) What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors?

Watch next

Power in Circuits practice set

Consider the circuit of Fig. E25.30
. (a) What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors?