Chapter 3: Circuits Solutions Questions: (4, 5), 14, 7, 8 Exercises & Problems: 5, 11, 19, 3, 6, 41, 49, 61 Q3.4,5: The circuit has two resistors, with 1 >. (a) Which resistor dissipates the larger amount of power if the resistors are in series? (b) In parallel? Explain. P= I. Q3.4. eason: The power dissipated by a resistor is determined by Since the current is the same in both resistors, the larger resistor, 1, will dissipate more power. Assess: If the current is the same in two resistors, the one with the larger resistance will dissipate more power. Q3.5. eason: Since the resistors are connected in parallel across the battery they have the same potential difference (voltage drop) across them. Hence, we use this version P = ( V ). of the power uation: The resistor with the lower resistance will dissipate more power (since is in the denominator); that is. Assess: You might wonder why we don t apply P= Ifor the power instead of P= V ( )/ ; that would seem to give the opposite answer. It is because I is not the same for the two resistors, so P= Idoesn t allow a comparison based on since we don t know P ( V)/ = the currents. We used because we know that V is the same for both resistors. Think about this result in terms of light bulbs. The conclusion is that the resistance of a 100 W light bulb is lower that the resistance of a 60 W light bulb (assuming 10 V in both cases). Measure the resistance of some light bulbs with an ohmmeter. Q3.9: a) If the figure below, what fraction of current I goes through the 3 Ω resistor? b) If the 9 Ω resistor is replaced with a larger resiostor, will the fraction of current going through the 3 Ω resistor increase, decrease, or stay the same?
Q3.9. eason: (a) The current coming from the left into the junction will split according to the ratio of the resistances it sees on the branches of the junctions. The lower the resistance of a branch, the more current will go there. Since the 3 Ω branch has one third the resistance as the other branch, three times as much current will go down the 3 Ω branch. So the fraction of current going through the 3 Ω resistor will be three quarters of the total. (b) If the 9 Ω resistor were replaced with a larger resistor, then even less current would go through it, so the fraction going through the 3 Ω resistor would increase. Assess: Take the question to the extreme: If the 9 Ω resistor were replaced with a much, much larger resistor, then almost no current would go through it, and nearly all the current would go through the 3 Ω resistor. In fact, if you want to you could consider a single resistor to be in parallel with a resistor of infinite resistance. Q3.14: The four bulbs shown are identical. ank the bulbs from brightest to dimmest. Explain. Q3.14. eason: All of the current in the circuit goes through bulb D, so it is the brightest. The same current goes through bulbs A and B, so they are ually bright. Since all the bulbs are identical, the parallel branch with bulbs A and B has twice the resistance as the parallel branch with bulb C and hence half the current. This allows us to rank the brightness of the bulb as follows: D > C > A = B. Assess: The brightness of the bulb depends on the current and the current depends on the resistance. Q3.15: The figure below shows five identical light bulbs connected to a battery. All the bulbs are glowing. ank the bulbs from brightest to dimmest. Explain. Q3.15. eason: The brightness of the bulbs is determined by how much power is dissipated in each bulb, so we use the power uation. When the resistances of the bulbs we are comparing are the same (as is the case here) we compare currents through the bulbs. P= I Bulbs D and E have the same current through them so they are ually bright. We need to examine the resistance of each major branch. They can be compared in one s head to arrive at ABC < DE because of the parallel bulbs B and C. But we can also be more quantitative about it. Call the resistance of each bulb ; then the resistance of the ABC branch is ABC = (3/) and the resistance of the DE branch is DE =.
Because V is the same across each major branch we can use Ohm s law over both branches V = I = I. ABC ABC DE DE I ABC DE 4 = = I (3/) 3 DE ABC The current in bulb A is I ABC and so we see that bulb A is brighter than bulbs D and E. IB = IC = (1/) IA = (1/) IABC = (/3) IDE, so bulbs B and C are not as bright as bulbs D and E. P A > P. D = P P P E > B = C This is the ranking of the brightness. Assess: When you have a thorough understanding and a bit of practice then this question can be answered almost at a glance. All of the details we worked through make sense intuitively. P3.5: The lightbulb in the circuit has a resistance of 1.0Ω. Consider the potential difference between pairs of points in the figure. a) What are the values of V 1, V3, and V34? b) What are the values if the bulb is removed? P3.5. Prepare: Please refer to Figure P3.5. We will use Ohm s law ( V = I) for the resistor and the bulb, use Tactics Box 3.1 for Kirchoff s loop law, and assume that the connecting wires are ideal. Solve: Let us assign clock direction to the current in the circuit. (a): The Kirchoff s loop law is: So, Σ( V) = V + V + V = + ( V V) + ( V V ) = 0 i bat resistor bulb 1 3 3.0 V I(.0 Ω) I(1.0 Ω ) = 0 I = 1.0 A. V = V V = I(.0 Ω ) = (1.0 A)(.0 Ω ) =.0 V 1 1 V = V V = I(1.0 Ω ) = (1.0 A)(1.0 Ω ) = 1.0 V 3 3 V = V V = 0V 34 4 3 (b) When the bulb is removed from the socket, no current flows in the circuit. Thus, (0 A)(.0 Ω ) = 0 V. V1 = V V1 = I(.0 Ω ) = This means that points 1 and are at the same potential as the positive V terminal of the battery. For the same reason, 34 = V4 V3 = 0, implying that points 3 and 4 V = V V = are at the same potential as the negative terminal of the battery. Finally, 3 3 V4 V1 = 3.0 V. Assess: The potential at point 1 is higher than at point 4. This is what you would have expected. P3.11: What is the uivalent resistance of each group of resistors shown?
P3.11. Prepare: Please refer to Figure P3.11. The three resistances in (a), (b), and (c) are parallel resistors. We will thus use Equation 3.1 to find the uivalent resistance. Solve: (a) The uivalent resistance is (b) The uivalent resistance is (c) The uivalent resistance is Assess: 1 1 1 1 = + + = 1.0 Ω.0 Ω 3.0 Ω 6.0 Ω 1 1 1 1 = + + = 1.0 Ω 3.0 Ω 3.0 Ω 3.0 Ω 1 1 1 1 = + + = 0.5 Ω.0 Ω 1.0 Ω.0 Ω We must learn how to combine series and parallel resistors. 3.19: The currents in two resistors in the circuit are shown. What is the value of resistor? P3.19. Prepare: From Kirchhoff s junction law we can easily determine that the current 5 A 15 A 10 A through the unknown resistor must be.. =. down. Also, the potential difference across the unknown resistor is the same as across the 300 Ω resistor. Assume the resistors obey Ohm s law. Solve: The potential difference across the 300 Ω V = I = (1. 5 A)(300 Ω ) = 300 V. resistor is The unknown resistor must have a resistance of V 300 V = = = 300 Ω I 1. 0 A Assess: We used a combination of Kirchhoff s junction law and Ohm s law to solve this one. P3.3: What are the resistance and the emf of the battery in the circuit.
P3.3. Prepare: Please refer to Figure P3.3. Assume that the connecting wire and the battery are ideal. The middle and right branches are in parallel, so the potential difference across these two branches must be the same. Solve: The currents in the middle and right branches are known, so the potential differences across the two branches, using Ohm s law, are Vmiddle = (3.0 A) = Vright = (.0 A)( + 10 Ω). This is easily solved to give = 0 Ω. The middle resistor is connected directly across the battery, thus (for an ideal battery, with no internal resistance) the potential difference V middle uals the emf of the battery. That is (3.0 A)(0 ) = Vmiddle = Ω = 60 V. Assess: Note that the potential difference across parallel resistors is the same. P3.6: Find the current through and the potential difference across each resistor in the circuit. P3.6. Prepare: Assume ideal batteries and wires and ohmic resistors. Label the resistors from left to right 1,, 3, and 4 (so = 10 Ω ). Compute the total resistance by first adding 3 + 4 = 10 Ω. Then that combination in parallel with gives 5. 0 Ω. Lastly, tot = 1 + 5. 0 Ω= 10 Ω. Solve: The current through the battery (and 1 I = V / ) is then tot = (10 V) /(10 Ω ) = 1. 0 A. From the junction law and symmetry that current splits up evenly in the other two I = I = I = 0. 50A. branches, so 3 4 V Now 1 = I 1 1 = (10. A)(50. Ω ) = 50. V. The loop law using the battery, 1and, shows that V = 10 V 5. 0 V = 5. 0 V. V3 = V4 =. 5 V. The potential difference of 5.0 V is split between 3 and 4 I (A) V (V) 1 1.5 5.0 0.50 5.0 3 0.50.5 0.50.5 4 so Assess: From symmetry we expect the results to be the same for 3 and. 4
P3.7: For the circuit shown in the figure below, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading. P3.7. Prepare: The battery and the connecting wires are ideal. The figure shows how to simplify the circuit in Figure P3.7 using the laws of series and parallel resistances. We have labeled the resistors as 1 = 6.0 Ω, = 15 Ω, 3 = 6.0 Ω, and 4 = 4.0 Ω. Having reduced the circuit to a single uivalent resistance, we will reverse the procedure and build up the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: 3 and 4 are combined to get 34 = 10 Ω, and then 34 and are combined to obtain 34 : 1 1 1 1 1 = + = + 15 Ω 10 Ω 34 = 6 Ω 34 34 Next, 34 and 1 are combined to obtain From the final circuit, = 34 = 1 = 6.0 Ω + 6.0 Ω = 1 Ω 4 V I =.0 A = 1 Ω = Thus, the current through the battery and 1 is I 1 =.0 A and the potential difference across 1 is I( 1 ) = (.0 A) (6.0 Ω) = 1 V. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference V. In Step 1 of the previous figure, = 1 Ω is returned to 1 = 6.0 Ω and 34 = 6.0 Ω in series. Both resistors must have the same.0 A current as. We then use Ohm s law to find
V 1 = (.0A)(6.0 Ω) = 1 V V 34 = (.0 A)(6.0 Ω) = 1 V As a check, 1 V + 1 V = 4 V, which was V of the resistor. In Step, the resistance 34 is returned to and 34 in parallel. Both resistors must have the same V = 1 V as the resistor 34. Then from Ohm s law, I = 1 V = 0.8 A I 1 V 34 = = 1. A 15 Ω 10 Ω As a check, I = I 34 =.0 A, which was the current I of the 34 resistor. In Step 3, 34 is returned to 3 and 4 in series. Both resistors must have the same 1. A as the 34 resistor. We then use Ohm s law to find ( V) 3 = (1. A)(6.0 Ω) = 7. V As a check, 7. V + 4.8 V = 1 V, which was V of the resistor 34. ( V) 4 = (1. A)(4.0 Ω) = 4.8 V esistor Potential difference (V) Current (A) 1 3 4 1 1 7. 4.8.0 0.8 1. 1. The three steps as we rebuild our circuit are shown. Assess: This problem ruires a good understanding of how to first reduce a circuit to a single uivalent resistance and then to build up a circuit. P3.5: Two 75 W (10 V) lightbulbs are wired in series, then the combination is connected to a 10 V supply. How much power is dissipated by each bulb? P3.5. Prepare: We will assume ideal connecting wires and an ideal power supply. The two light bulbs are basically two resistors in series. Our strategy is to find the current that flows through the two bulbs and then use Equation.14 to find the power dissipated by each bulb. Solve: A 75 W (10 V) light bulb has a resistance of V (10 V) = = = 19 Ω P 75 W The combined resistance of the two bulbs is = 1 + = 19 Ω + 19 Ω = 384 Ω. The current I flowing through is
V 10 V I = = = 0.315 A 384 Ω Because is a series combination of 1 and, the current 0.315 A flows through 1 and. Thus, P 1 = I 1 = (0.315 A) (19 Ω) = 19 W = P. Assess: Please note that a light bulb rated at 75 W (10 V) delivers 75 W power only when it is connected across a source of 10 V. As we see in this problem, the power delivered by two series bulbs across 10 V is considerably reduced compared to their rated power. P3.61: You have three 1Ω resistors. Draw diagrams showing how you could arrange all three so that their uivalent resistance is (a) 4.0 Ω, (b) 8.0 Ω, (c) 18.0 Ω, and (d) 36.0 Ω. P3.61. Prepare: A visual overview of how to reduce the circuit to an uivalence resistance is shown below. Solve: (a) The three resistors in parallel have an uivalent resistance of 1 1 1 1 = 1 + 1 + 1 Ω Ω Ω = 4.0 Ω (b) One resistor in parallel with two series resistors has an uivalent resistance of 1 1 1 1 1 1 = 1 1 + 1 = 4 + 1 = 8 Ω + Ω Ω Ω Ω Ω = 8.0 Ω. (c) One resistor in series with two parallel resistors has an uivalent resistance of 1 1 1 = 1 Ω+ + = 1 Ω+ 6 Ω= 18.0 Ω 1 Ω 1 Ω (d) The three resistors in series have an uivalent resistance of 1 Ω + 1 Ω + 1 Ω = 36 Ω Assess: Learn how to combine series and parallel resistors. 1