In the circuit shown in Fig. E26.34, the 6.0-Ω resistor is consum...

Question

In the circuit shown in Fig. E26.34, the 6.0-Ω resistor is consuming energy at a rate of 24 J/s when the current through it flows as shown.  (b) What are the polarity and emf ε of the unknown battery, assuming it has negligible internal resistance?

Accepted Answer

Hello, fellow physicists today, we 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. As shown in the figure a circuit consists of two ideal electromotive force EMF sources and eight resistors. When the current flows through resistor R it produces heat at a rate of 3.0 watts. Find I the value for the emfe and I I the value of the positive terminal of the unknown power source. OK. So that's our end goal is we're trying to find two separate answers. We're trying to find parts I and I, I, so I is asking us to find the value for the emfe and I I is trying to make a or asking us to solve for the value of the positive terminal of the unknown power source. OK. And then there show our figure down below our problem here and it shows the eight resistors represented by J it, you know lines here like little zig zag lines. And we're also given two batteries, one battery that's to the left is 15 volts. And then the battery on the far right is shown, but it doesn't have a voltage value. It's for, for the positive end of the battery is for point M and then the negative side of the battery is for point N. Awesome. So we're given some multiple choice answers and we're given all the answers for part I are given in the units of vaults. And then all the answers for I I is possible answers for where the terminal is connected to what point? So is the positive terminals connected to M or the letter N? So the letter M or N? OK. So let's read them off to see what our final answer pair might be. So A for part I is E is equal to 0.75 and I, I is the positive terminals connected to MB is E is equal to 1.15. And I I is the positive terminals connected to NC is E is equal to 3.25. And I I, the positive terminals connected to M and finally D is E is equal to 12.5 and I I is the positive terminals connected to N. OK. So first off, we need to reduce the parallel combination of the 10.0 ohm resistors. So let's do that really fast, shall we? So one divided by R subscript EQ is equal to one divided by 10.0 ohms plus one divided by 10.0 ohms. So let me isolate and solve for req we get that that is equal to 5.0 ohms. Thus, we obtain the following circuit diagram. So I've gone ahead and drawn up a little version of our circuit considering what we just did. It basically puts in, puts into a di basically, it helps us visualize what we had just did mathematically. So this is what happens when we reduce the parallel combination of the 10.0 ohm resistors, we get the following diagram. OK. So as you can see, we have loop one is going clockwise and we have loop two. So loop ones on the left of our circuit. And then on the right of the circuit is our loop two and they're both going in the clockwise direction. An I two is going to the left and I three is going to the right. OK. So with that in mind, now we must assume that the currents I one I two and I three flow in the clockwise direction as shown in our circuit diagram above the current I one flows through resistor R and the power generated by resistor R is P is equal to R multiplied by I one squared. So now we need to rearrange this equation to solve for I one. So when we do that, we get I one is equal to the square root of P divided by R. So that means when we plug in our values or known variables to solve for I one, we get that I, one is equal to the square root of 3.0 watts divided by 3.0 arms, which means when we plug it into a calculator, I one is equal to 1.0 yours. So now at this stage, we need to apply Kirchhoff's loop rule to loop one. So let's do that. So 3.0 ohms multiplied by I one minus 15.0 volts plots seven point zero arms multiplied by I one plus 5.0 arms multiplied by I two plus 3.0 arms multiplied by I two plus 5.0. Alms multiplied by I two is equal to zero. Oh That was a mouthful. So we get these values from our circuit diagram going in the clockwise direction. So we follow it along that direction and then we just write around like so we fill out our information based on that direction. Awesome. So now we need to take this big old long equation that we wrote and we need to simplify it. So let's start simplifying it. So when we simplify, we get 13.0 ohms multiplied by I square is equal to 15 faults minus 10.0 ohms multiplied by I one. So we need to rearrange this equation to solve for I two. So let's do that. So when we do a little bit of rearranging, use a little algebra we'll get that I two is equal to 15 vaults minus 10 arms multiplied by one here, all divided by 13 arms. Let me plug that into a calculator. We should get 0.385 here. So now we must apply the junction rule and the junction rule states that I one is equal to I two plus I three. So now we need to rearrange this equation to solve for I three. So isolating and solving for I three, we get I three is equal to I one minus I two. So we can now plug in our known variables and solve. So 1.0 pi and that's our I one value minus our I two value which we just determined to be 0.38 five. S is equal to 0.615 an pierce. Fantastic. So now moving right along, we need to apply the loop rule to loop number two. OK. So doing that they have since they were going in the direction it's going through the battery on the negative side, it's negative seven point zero ohms multiplied by I three minus E minus 3.0 ohms multiplied by I three. OK. Mine is 7.0 ohms multiplied by I one plus 15 volts minus 3.0. Arms multiplied by I one is equal to zero. Awesome. So like before we need to simplify, but we also, in this case, we're trying to solve for E So let's simplify and isolate and solve for E. So when we do that, we get that E is equal to 15 volts minus 10 arms multiplied by I three minus 10 ohms multiplied by I one. So we need to simplify this a little bit more. So E is equal to 15.0 bolts minus 10.0 arms or I should say. Um Here, let's move it down. We're getting a little bit, we're running out of room here. Ok. So moving it down below really quick here. So at this point, we can plug in our known variables since we already simplified above. So E is equal to 15 volts minus 10.0 os multiplied by our I three value which was 0.615 and piers minus 10.0 ohms multiplied by our I one value which was 1.0 here. So when we plug that into a calculator, we will get negative 1.5 bolts. Ok? And that is our answer. But before we go and run along with this answer, we need to make the following note. This is very important. So the electric motive force of the power supply, E is 1.15 volts positive. So because there's a minus sign for the electric motive force, since this is why, why we have a negative sign, the polarity of the battery is opposite of what is shown in our circuit diagram provided to us. Therefore the positive terminal is connected to point N. OK. So that means our final answer is for part I is 1.5 volts. And let's scroll up to look at our multiple choice answers to see what our correct answer has to be. So that means our correct answer has to be the letter B. So B is I is equal to E is equal to 1.15 volts. And I, I is the positive terminals connected to N. Awesome. We did it. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

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In the circuit shown in Fig. E26.34, the 6.0-Ω resistor is consuming energy at a rate of 24 J/s when the current through it flows as shown. (b) What are the polarity and emf ε of the unknown battery, assuming it has negligible internal resistance?

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