- Find the equation obtained by applying Kirchhoff’s junction rule at point A.
- Find the equations obtained by applying Kirchhoff’s loop rule in the upper and lower loops.
- Assume $R_1 = R_2 = 6 \Omega$, $R_3 = 12 \Omega$, $r_1 = r_2 = 0 \Omega$, $\xi_1 = 6 \textrm{ V}$ and $\xi_2 = 4 \textrm{ V}$. Calculate $I_1$, $I_2$ and $I_3$.
- For the situation in which $\xi_2$ is replaced by a closed switch, repeat parts (a) and (b). Using the values for $R_1$, $R_2$, $R_3$, $r_1$ and $\xi_1$ from part (c) calculate the currents through the three resistors.
- For the circuit in part (d) calculate the output power of the voltage source and across all the resistors. Examine if energy is conserved in the circuit.
- A student implemented the circuit of part (d) in the lab and measured the current though one of the resistors as 0.19 A. According to the results calculated in part (d) identify the resistor(s). Justify any difference in measured and calculated value.

- $I_1 + I_3 = I_2$
- see video
- $I_1 = \dfrac{8}{15}$, $I_2 = \dfrac{7}{15}$, $I_3 = \dfrac{-1}{15}$
- $I_1 = \dfrac{2}{5}$, $I_2 = \dfrac{3}{5}$, $I_3 = \dfrac{1}{5}$
- $P_{total} = \dfrac{18}{5} \textrm{ W}$, $P_{R_1} = \dfrac{24}{25} \textrm{ W}$, $P_{R_2} = \dfrac{54}{25} \textrm{ W}$, $P_{R_3} = \dfrac{12}{25} \textrm{ W}$

Yes, energy is conserved. - The resistor is $R_3$.

The ammeter introduces additional resistance, $r$, in series with $R_3$, thereby reducing the current.

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View sample solutionThis is College Physics Answers with Shaun Dychko. So fasten your seatbelts ‘coz this question is about seven questions all in one or even more than that, ‘coz some of the questions have sub-questions. But anyway, let’s get her done. So part a says, find the equation obtained by applying Kirchhoff’s Junction Rule at point a. So this junction, we have current three going in and current one going in and current <i>I2</i> coming out. And so the total going in equals the total coming out. So <i>I1</i> plus <i>I3</i> equals <i>I2</i>. And on all of the direction of the currents, by the way. Because we’re told the currents are flowing in the clockwise direction which is this way. And then this way. Then in part b: it says find the equation obtained by applying Kirchhoff’s Loop Rule in the upper and lower loops. So we consider the upper loop first starting at point a. And we’re going to traverse in the clockwise direction. And that means we’re going in the same direction as the current through <i>R2</i>, that means we have a minus <i>I2</i> times <i>R2</i>. And then we’re going from the negative to the positive terminal of this EMF one. So that’s a positive EMF one. And then minus <i>I2</i> times little <i>r1</i>. And then minus <i>I1</i> times little <i>r1</i>. All that equals zero because we end up in point a again. And then for the lower loop, we’ll do point a again. As a starting point we’ll go the clockwise direction in our traversal. But we’re going to begin by going through <i>R1</i> in the opposite direction to the current. And so we have a positive <i>I1</i> times capital <i>R1</i>. And then we’ll go down through <i>R3</i> in the same direction as the currents. So minus <i>I3R3</i> minus <i>I3</i> times little <i>r2</i> minus EMF two because we begin at the positive terminal and end up at the negative terminal. So that makes it a minus there, minus EMF two. We return to point a, and so we say equals zero. And then part c, says assume that we have a bunch of values for all these things in here. And then calculate <i>I1, I2</i> and <i>I3</i>. We have to solve the circuit. So, <i>R1</i> and <i>R2</i> equals zero which is nice ‘coz then they disappear. And EMF one is six. And EMF two is four. And we have <i>R3</i> is 12 ohms. And <i>R1</i> and <i>R2</i> are each six ohms. So we can take this loop rule that we have figured out before. And then get rid of the internal resistance in terms of the little <i>r</i> in them. So EMF one minus <i>I2</i> capital <i>R2</i> minus <i>I1</i> capital <i>R1</i> equals zero. And this loop becomes EMF two plus <i>I1R1</i> minus <i>I3R3</i>. And getting rid of that term because it is zero. And then substituting in numbers: EMF one is six Volts, and then <i>R2</i> is six, and <i>R1</i> is six. And all that equals zero. Then, EMF two is four Volts. But we have a negative EMF two in our equation. And so we have negative four and then plus six times <i>I1</i> because <i>R1</i> is six. Then minus 12 times <i>I3</i> because <i>I3</i> is 12 ohms. And this equation becomes <i>I1</i> plus <i>I2</i> equals one. After you take the six to the right hand side, which makes it a negative six. So subtract six on both sides. Everything is negative. And then divide both sides by negative six and then you get positive <i>I1</i> plus <i>I2</i> equals one. Now, <i>I2</i> we know is <i>I2</i> plus <i>I3</i>, because that’s what we know from this Junction Rule. And so we can substitute to replace <i>I2</i>. And so that becomes two <i>I1</i> plus <i>I3</i> equals one. And then we can work on this equation here. And simplify it a little bit by dividing by the common factor of two. So we get three <i>I1</i> minus six <i>I3</i> minus two equals zero. And now we have two equations with two unknowns <i>I1</i> and <i>I3</i>. And has a variety of ways to solve that system but what I’ve chosen to do is multiply equation a by three and then subtract from that equation b multiplied by two, because that will target the <i>I1</i> term and make it disappear because it’s going to become six <i>I1</i> minus six <i>I1</i>. And we will be having a single equation with only <i>I3</i> as the unknown which we can solve. So we have six <i>I1</i> plus three <i>I3</i> minus two times equation b left side which is three <i>I1</i> minus six <i>I3</i> minus two. And then we have three times the right hand side subtracted. So that’s three times one which is three minus two times zero which is zero. So we have just three there on the right hand side. And then distribute this negative two into the bracket. So we have negative six <i>I1</i> plus 12 <i>I3</i> plus four equals three. And take this four to right hand side by subtracting it from both sides, you get negative one. And then on the left side, the <i>I1</i> is sure enough make zero which was the whole purpose of doing this, getting rid of <i>I1</i>. And then three and 12 makes 15 <i>I3</i>. And so <i>I3</i> then is one over negative 15. Then from equation a, we can solve for <i>I1</i> and say that it’s one minus <i>I3</i>. And then divide both sides by two. We get <i>I1</i> one minus <i>I3</i> over two which is one minus negative one over fifteen times a half which is the same as divided by two multiplying by the reciprocal of the denominator. With fractions within a fraction, I always multiply it with the reciprocal of the denominator whenever it’s about to have a fraction within a fraction. So we have one minus negative one fifteenth times the half so that’s 16 over 15 times a half which is eight over 15. So that’s <i>I1</i>. Then, <i>I2</i> is <i>I1</i> plus <i>I3</i>. So that’s eight fifteenth minus one fifteenth which is seven fifteenth. There, that’s part c then. Now in d, we say replace this EMF two by a close switch or a wire. So essentially you draw a black line through in there and <i>I2</i> is gone. And now do everything again. So we need to find the currents to all three resistors again. So the Junction Rule is still <i>I1</i> plus <i>I3</i> is equal to <i>I2</i> because <i>I1</i> and <i>I3</i> is both going to junction a and <i>I2</i> is the one that comes out of the junction. And then the Loop Rule is going to be for the top loop is going to be negative <i>I2R2</i> plus EMF one. And then this is zero again. And then, let me go through here and in the middle minus <i>I1R1</i>. And then for the bottom loop, we have positive <i>I1</i> plus <i>R1</i> minus <i>I3R3</i>. And there’s nothing on this bottom branch, this <i>R2</i> is zero and this EMF we are told is gone. And so that’s the end of this Loop Rule here, is that equals zero. Then, replacing the resistances and the EMFs with numbers, the EMF one is six and then minus six times <i>I2</i> because is <i>R2</i> is six. And then minus six times <i>I1</i> because <i>R1</i> is six. And divide both sides with six and you get one minus <i>I2</i> minus <i>I1</i> equals zero. And then working on this equation here, this Loop Rule in green we’re replacing <i>R1</i> with six and replacing <i>R3</i> with 12. So it’s six <i>I1</i> minus 12 <i>I3</i> equals zero. And that means <i>I1</i> equals two <i>I3</i> because you can take this to the right hand side by adding 12 <i>I3</i> to the both sides. And then divide both sides by six and you solve for <i>I1</i>, two <i>I3</i>. We need to get two equations with two unknowns. And right now we have three equations with three unknowns. We have this one. And we have this one. So we can revisit this equation in blue. Then we place <i>I2</i> with <i>I1</i> plus <i>I3</i>. That’s using the Junction Rule, so <i>I2</i> is replaced by <i>I1</i> plus <i>I3</i>. And now we have this equation containing the variables <i>I1</i> and <i>I3</i>. And we have this equation containing <i>I1</i> and <i>I3</i>. So we can work with that. And so after <i>I2</i> is replaced with <i>I1</i> plus <i>I3</i>, this negative gets distributed in there. So that makes one <i>I1</i> minus <i>I3</i> minus <i>I1</i>. So the two minus <i>I2</i> become two minus <i>I1</i> minus <i>I3</i> equals zero. Then to get rid of <i>I1</i>, we’re going to multiply equation a by two and add to that equation b. So equation a on the left side multiplied by two and twos become two <i>I1</i>. And on the left side of equation by multiplied by well nothing, just one is going to be one minus two <i>I1</i> minus <i>I3</i> and then equals the right hand sides, add it together. And then A is going to be multiplied by two. That makes two times two <i>I3</i> which is four <i>I3</i>. On the right side of B is just zero, that doesn’t change anything there. Then we have one minus <i>I3</i> on the left because by design these <i>I1</i> terms made zero equals four <i>I3</i>. And then add <i>I3</i> to both sides. And then switch the sides around, you get five <i>I3</i> equals one and then divide both sides by five. And it makes one fifth. So <i>I3</i> is one fifth. Now <i>I1</i> is two times <i>I3</i>, we know from here. And so <i>I1</i> is going to be two times one fifth which is two fifths. And then <i>I2</i> is <i>I1</i> plus <i>I3</i>. So that’s two fifth plus one fifth which is three fifths. Then part e asks us about power dissipation. And so the total power output by the EMF source is going to be six volts times the total current through the EMF. And the total current is this <i>I2</i>. So that’s going to be six Volts times <i>I2</i> which is three fifths and that makes 18 fifths Watts. And the power through resistor one is going to be the current <i>I1</i> squared, multiplied by the resistor’s one resistance. So that’s two fifths. <i>I1</i> is a two fifths here. Two fifths squared times six ohms. And that’s going to be six times four is 24 over 25 Watts. And the power through resistor two is <i>I2</i> squared times <i>R2</i>. So that’s three fifths squared times six ohms. That makes a nine times six which is 54 over 25 Watts. And then the power through <i>R3</i> is <i>I3</i> squared times <i>R3</i> which is one fifth squared times 12 ohms and because <i>R3</i> is 12. And that gives 12 over 25 Watts. And then we are meant to check whether or not energy is conserved. So the total power output of the power source is equal to the total power dissipated in the resistors. So the question is, is 18 fifths equal to the 24 twenty fifths plus 54 twenty fifths plus 12 twenty fifths of each resistor. And that makes 90 twenty fifths. And 90 is five time 18 and the denominator is five times five. Five is cancelled giving us 18 over five. And so indeed yes, we see that this total power output of each resistor is indeed equal to the total power output of the voltage source. So energy is conserved. Then in part f, it says a student created the circuit in picture d here. And measure the currents to one of the resistors as 0.19 Amps. And then to which current here is approximately 0.19 and the answer is <i>I3</i> is one fifth is 0.2. 0.19 is closer to 0.2 than it is to 0.4 or 0.6. And so we conclude that the student is measuring the current <i>I3</i>. So the Ammeter is in here somewhere. Now it’s like less than 0.2 because an Ammeter has internal resistance and by having an additional resistance, it’s going to reduce the current through whatever branch the Ammeter is in. And so we expect it to measure an amount just like and less than with a few radical amount.