Change the chapter
Question
For this question, consider the circuit shown in the following figure.
  1. Assuming that none of the three currents ($I_1, I_2, I_3$) are equal to zero, which of the following statements is false?
    1. $I_3 = I_1 + I_2$
    2. $I_2 = I_3 - I_1$
    3. The current through $R_3$ is equal to the current through $R_5$
    4. The current through $R_1$ is equal to the current through $R_5$
  2. Which of the following statements is true?
    1. $\xi_1 + \xi_2 + I_1R_1 - I_2R_2 + I_1r_1 - I_2r_2 + I_1R_5 = 0$
    2. $-\xi_1 + \xi_2 + I_1R_1 - I_2R_2 + I_1r_1 - I_2r_2 - I_1R_5 = 0$
    3. $\xi_1 - \xi_2 - I_1R_1 + I_2R_2 - I_1r_1 + I_2r_2 - I_1R_5 = 0$
    4. $\xi_1 + \xi_2 - I_1R_1 + I_2R_2 - I_1r_1 + I_2r_2 + I_1R_5 = 0$
  3. If $I_1 = 5 \textrm{ A}$ and $I_3 = -2 \textrm{ A}$, which of the following statements is false?
    1. The current through $R_1$ will flow from a to b and well be equal to 5 A.
    2. The current through $R_3$ will flow from a to j and will be equal to 2 A.
    3. The current through $R_5$ will flow from d to e and will be equal to 5 A.
    4. None of the above.
  4. If $I_1 = 5 \textrm{ A}$ and $I_3 = -2 \textrm{ A}$, $I_2$ will be equal to
    1. 3 A
    2. -3 A
    3. 7 A
    4. -7 A
Question Image
<b>Figure 21.67</b>
Figure 21.67
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

(c), (c), (d), (d)

Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 21, Problem 9 (Test Prep for AP® Courses) (6:37)

Sign up to view this solution video!

View sample solution
Video Transcript

This is College Physics Answers with Shaun Dychko. We’re given this circuit here and we have a bunch of questions to answer about it. So the first question is assuming that none of the three currents <i>I1,I2</i> and <i>I3</i> are zero, which of the following statements is false? So, let’s consider this first statement here: <i>I3</i> equals <i>I1</i> plus <i>I2</i> at point a. And this is true because <i>I3</i> is a current going in. And the total current coming out is <i>I1</i> plus <i>I2</i>. And so this is the Junction Rule. At point e, I should say, this is option b, <i>I2</i> equals <i>I3</i> minus <i>I1</i>. Well, we have <i>I2</i> going into this circuit. And we have <i>I3</i> going down from this side. We have <i>I1</i> going down this side, as well. So, <i>I2</i> plus <i>I1</i> equals <i>I3</i>. And can we rearrange this to make it look like that? Indeed we can. So, <i>I2</i> is <i>I3</i> minus <i>I1</i>. And so b is correct. C, the current through <i>R3</i> is current through <i>R5</i>. So here’s <i>R3</i> right here. And here’s <i>R5</i>. <i>R5</i> has a current <i>I1</i> going through it. And resistor three has a current <i>I3</i> going through it. And so those are two different currents. Now, there’s some possibility that they, even though they’re labelled differently they could actually be the same. So we have a strong suspicion that c is false but we can’t be perfectly sure until we consider option d. So the current through <i>R1</i> is the current through <i>R5</i>. Well, that definitely is true because they’re in series. And so we know that d is definitely true. And so c is the only one that has any suspicion of being false. And so that’s going to be our answer. Part B has a whole bunch of gobbily gawk equations here, and the easiest way to figure it out is to just create the loop rule on our own and see which one matches what we created. And there are three different loops to consider. There’s the top loop, the outer loop, and the bottom loop. And the reason that I know to choose the top loop is because all the options here have EMF one and EMF two. And so we should choose a loop which has EMF one and EMF two which is the top loop. So, we begin at point b. And then have a positive EMF one here. So we have EMF one here. And then minus <i>I1</i> times little <i>r1</i> and then minus <i>I1</i> times capital <i>R5</i> because we’re traversing this loop in the same direction as the current through the resistors so it is always minuses across the resistors. And then, however across this branch right here, down the middle, we’re traversing in the opposite direction to the current through it. And so we have a positive <i>I2</i> times <i>R2</i> and then plus <i>I2</i> times little <i>r2</i> and then minus EMF two because we are going from the positive to the negative terminal here. And then up through <i>R1</i> in the same direction as the current. And so we have minus <i>I1</i> times <i>R1</i>. And all that equals zero. And then I just arranged the EMFs beside each other since that’s the way it’s arranged in all this options here so we can easily compare the two. And it looks like c is what we have written here. EMF one minus EMF two minus <i>I1R1</i> plus <i>I2R2</i> minus <i>I1</i> little <i>r1</i> plus <i>I2</i> little <i>r2</i> plus <i>I1R5</i>. And so there good enough. Option c. Part C. If <i>I1</i> is five Amps and <i>I3</i> is negative two Amps, which of the following statements is false? It turns out that none of them is false. So, option a says that the current through <i>R1</i> will flow from <i>a</i> to <i>b</i>. Well, it says <i>I1</i> is positive five Amps. And so that means the drawing in our picture is the correct direction for the current because our value is a positive. And so it does indeed flow from a to b. That is the conventional current flows from a to b. And then equal five Amps ‘coz <i>I1</i> is five Amps. And then option b the current through <i>R3</i> will flow from a to j and will be equal to positive two Amps. So let’s consider capital <i>R3</i>. It will flow from a to j. It will be equal to two Amps. That is correct because we are told that <i>I3</i> is negative two Amps. Which means the way it is drawn on our picture is the opposite direction to what the conventional current actuality is. And so the conventional current is in fact going down with a magnitude of two Amps. And so b is correct. And c, the current through <i>R5</i> will flow from d to e. <i>R5</i> is here will flow from d to e, and be equal to five Amps. Yes indeed because it’s current <i>I1</i> which is positive five Amps. And so this drawing is pointing in the correct direction which is down from d to e through <i>R5</i>. And so c is correct. So that’s why none of them are false. And then part D. If <i>I1</i> is five Amps and <i>I3</i> is negative two Amps, and <i>I2</i> will be equal to what? Well, let’s consider a junction somewhere. I use this junction here a. So <i>I3</i> going into this junction equals the total current coming out which is <i>I1</i> plus <i>I2</i>. And then rearrange this for <i>I2</i> by subtracting <i>I1</i> from both sides. And then switching the sides are on, we got <i>I2</i> is <i>I3</i> minus <i>I1</i>. <i>I3</i> is negative two Amps minus <i>I1</i> which is five Amps giving us negative seven Amps. And so the answer is d.