10/9/2016HW 07 Capacitor Circuits HW 07 Capacitor Circuits Due: 11:59pm on Wednesday, September 28, 2016 To understand how points are awarded, read the Grading Policy for this assignment. Item 1 In the figure are shown three capacitors with capacitances C1 = 6.00 μF, C2 = 3.00 μF, C3 = 5.00 μF . The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge Q2 on the second capacitor is 40.0 μC. Part A What is the charge Q1 on capacitor C1 ? Hint 1. How to approach the problem Consider only the initial section of the network (from a to d). Develop a relation between the capacitance and charge for each capacitor in that section. Use the capacitances C1 and C2 and the final charge Q2 on C2 to calculate the charge Q1 on C1 . Hint 2. Series or parallel? Are capacitors C1 and C2 connected in series or in parallel? ANSWER: series parallel Hint 3. Calculate the potential difference across the second capacitor Calculate the potential difference V2 across the second capacitor. Express your answer in volts to three significant figures. Hint 1. Equation for capacitance Recall that for any capacitor one has the relation C = Q V , where Q is the charge on each plate of the capacitor and V is the potential difference across the capacitor (not the difference across the whole network). https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 1/17 Calculate the potential difference across the first capacitor. Use the result to find the charge on the third capacitor. Parallel capacitors and potential difference If two capacitors are connected in parallel.10/9/2016 HW 07 Capacitor Circuits ANSWER: V2 = 13. for both capacitors.3 V Hint 4. Hint 2.masteringphysics. C3 . Qad . ANSWER: V1 = 13. connected in series or in parallel with the first section of the network (from a to d)? ANSWER: series parallel Hint 3. Series or parallel? Is the third capacitor. How to approach the problem Determine whether the third capacitor. C3 . is connected in series or in parallel with the first section of the network (from a to d). V1 Calculate the potential difference V1 across the first capacitor.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 2/17 . Capacitors in series https://session. respectively. Express your answer in volts to three significant figures.0 μC Correct Part B What is the charge on capacitor C3 ? Express your answer in microcoulombs to three significant figures.3 V ANSWER: Q 1 = 80. Calculate the total charge of the initial section of the network. Hint 1. Q3 . Hint 1. how are the potential differences across each capacitor related? Note that the left plate and right plate are connected to points a and d. Determine the capacitance Cab and charge Qab on the equivalent capacitor. Hint 2. In this problem. Therefore. the positive plate of one capacitor is connected to the negative plate of the other capacitor. in the initial section between points a and d. Hint 4. Since the charge on the two plates must come from separation of charges in the initially neutral plates (charge is never created or destroyed). Capacitance for capacitors in parallel https://session. Express your answer in microfarads to three significant figures. the total charge on that network must equal the charge on C3 . Express your answer in microcoulombs to three significant figures. the charge must be the same for the two capacitors. the total charge stored in the initial section. the network between a and d is in series with capacitor C3 . Hint 1. Calculate the equivalent capacitance of the initial section Calculate the equivalent capacitance Cad of the capacitors in the initial section between points a and d. How to find the charge in the initial section You already know the charge on the first two capacitors. Hint 1. The total charge must therefore be the sum of the charges in each capacitor.masteringphysics. Express your answer in microfarads to three significant figures. How to approach the problem Reduce the network of capacitors to a single equivalent capacitor connecting points a and b directly. Vab? Express your answer in volts to three significant figures. Calculate the total charge in the initial section Calculate Qad . the two plates must have charges of equal magnitude and opposite signs. Hint 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 3/17 . Q1 and Q2 . ANSWER: Qad = 120 μC ANSWER: Q3 = 120 μC Correct Part C What is the applied voltage. Hint 1. Calculate the equivalent capacitance Calculate the equivalent capacitance Cab of the network.10/9/2016 HW 07 Capacitor Circuits For capacitors connected in series. Then calculate the total potential difference Vab. Therefore. https://session. as calculated in Part B. Find the total charge Find the total charge Qab stored in the equivalent capacitor. Remember to solve for Ctotal and not its reciprocal! ANSWER: Cab = 3.00 μF Hint 2.50 mm apart. The magnitude of charge on each plate is 0.21 μF Hint 3. Express your answer in microcoulombs to three significant figures. ANSWER: Cad = 9.0180 μC when the potential difference is 200 V. Capacitance for capacitors in series The total capacitance for two capacitors in series. Using the value of Q3 for the third capacitor in the network. you can show that Qab = Qad = Qdb = Q3 . Hint 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 4/17 .3 V Correct Item 2 An air capacitor is made from two flat parallel plates 1.masteringphysics. ANSWER: Q ab = 120 μC ANSWER: Vab = 37. Finding the charge of the equivalent capacitor Remember that since the section between points a and d and the section between points d and b are in series. is given by 1 Ctotal = 1 Cα + 1 Cβ . Cα and Cβ . the charge on the equivalent capacitor will be the same as the charge stored in each section.10/9/2016 HW 07 Capacitor Circuits The total capacitance for two capacitors with capacitances Cα and Cβ connected in parallel is given by Ctotal = Cα + Cβ . com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 5/17 . what total energy is stored? ANSWER: U = 1.0 × 10 6 V/m.) Express your answer using two significant figures.52×10−2 m2 Correct Part C What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric field strength of 3.0180 μC.80×10−6 J Correct Item 3 Learning Goal: https://session.10/9/2016 HW 07 Capacitor Circuits Part A What is the capacitance? ANSWER: C = 9.masteringphysics.00×10−11 F Correct Part B What is the area of each plate? ANSWER: A = 1. ANSWER: Vmax = 4500 V Correct Part D When the charge is 0. 0 μF and C4 = 36. When capacitors are connected in series. 3. For more complicated combinations. 3. find the parts that are simple series or parallel connections and replace them with their equivalent capacitances. 2. Ceq must be greater than any of the individual capacitances. they always have the same charge. the equivalent capacitance Ceq must be smaller than any of the individual capacitances.10/9/2016 HW 07 Capacitor Circuits To practice Problem Solving Strategy 24. The potential differences are not equal unless the capacitances are equal.0 μF . where C1 =C2 =C3 = 16.1 Equivalent Capacitance.masteringphysics. the potential difference V is always the same for all of the individual capacitors. EXECUTE the solution as follows: 1. By contrast. If the capacitors are connected in series. others in parallel. With more complicated combinations. Identify whether the capacitors are connected in series or in parallel. SET UP the problem using the following steps: 1. you may have to retrace your path to the original capacitors. Keep in mind that when we say a capacitor has charge Q we always mean that the plate at higher potential has charge +Q and the other plate has charge −Q. some of which are connected in series. Consider the capacitor connection shown in the picture below. Determine the equivalent capacitance of the entire combination. you can sometimes identify parts that are simple series or parallel connections. If you then need to find the charge or potential difference for an individual capacitor. The total potential difference across the combination is the sum of the individual potential differences.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 6/17 . The steps outlined in the strategy above will help you determine the equivalent capacitance of the entire combination. assuming that they were uncharged before they were connected.1 Equivalent Capacitance IDENTIFY the relevant concepts: The concept of equivalent capacitance is useful whenever two or more capacitors are connected. SET UP the problem using the following steps https://session. Make a drawing of the capacitor arrangement. ProblemSolving Strategy 24. IDENTIFY the relevant concepts This problem deals with a combination of four capacitors. The charges on the individual capacitors are not equal unless the capacitances are equal. if the capacitors are connected in parallel. The total charge on the combination is the sum of the individual charges. When capacitors are connected in parallel. 2. in a stepbystep reduction. EVALUATE your answer: Check whether your result makes sense. Thus. however. as described in the following statements. Express your answer in microfarads.0 μF . Which of these statements are correct for this problem? Check all that apply. https://session. Notice that to reduce the latter parallel combination you will first need to reduce the C1 C2 series combination.masteringphysics. ANSWER: C3 is in parallel with C1 and C2 . Hint 1.10/9/2016 HW 07 Capacitor Circuits Part A The four capacitors shown in the diagram are neither all in series nor all in parallel. In Part A you identified that C1 is in series with C2 .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 7/17 .0 μF and C2 = 16. Find the unknown capacitance in the first step of the reduction This diagram shows the first step in the reduction of the capacitor combination. How to approach the problem To find the equivalent capacitance of the combination under consideration you must progressively reduce the capacitor network until only a single capacitor is left. it makes sense to start the reduction by finding the equivalent capacitance of the C1 C2 series combination. You can. C4 is in series with C1 and C2 . Correct EXECUTE the solution as follows Part B What is the equivalent capacitance Ceq of the entire combination? Express your answer in microfarads to three significant figures. Hint 2. What is the value of CeqA ? Recall that C1 = 16. C3 is in series with C4 . identify portions of the arrangement that are either in series or parallel. C1 is in series with C2 . and C3 is in parallel with C1 and C2 . 00 μF Hint 3. Equivalent capacitance in a series combination If Ca and Cb are connected in series.0 μF ANSWER: Ceq = 14. Equivalent capacitance in a parallel combination If Ca and Cb are connected in parallel. Hint 1. Ceq = Ca + Cb ANSWER: CeqB = 24.0 μF . Find the unknown capacitance in the second step of the reduction This diagram shows the second step in the reduction of the capacitor combination. 1 Ceq = 1 Ca + 1 Cb ANSWER: CeqA = 8.10/9/2016 HW 07 Capacitor Circuits Hint 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 8/17 . What is the value of CeqB ? Recall that CeqA = 8 and C3 = 16. the reciprocal of their equivalent capacitance equals the sum of the reciprocals of the individual capacitances.masteringphysics.4 μF Correct https://session. their equivalent capacitance equals the sum of the individual capacitances. Express your answer in microfarads. Your result from Part B satisfies this requirement. Thus. a potential difference of 25. you should expect Ceq < C4 . Hint 2. Since in a series connection the equivalent capacitance is always less than any individual capacitance. For related problemsolving tips and strategies. Relation between the equivalent capacitance and the individual capacitances in a parallel combination The equivalent capacitance of a parallel combination equals the sum of the individual capacitances. Relation between the equivalent capacitance and the individual capacitances in a series combination The reciprocal of the equivalent capacitance of a series combination equals the sum of the reciprocals of the individual capacitances. but with different values for the individual capacitances. you may want to view a Video Tutor Solution of Capacitors in series and in parallel. Correct is the capacitance of a single capacitor equivalent to the series combination between C4 and the branch containing the remaining capacitors. How would you expect the equivalent capacitance Ceq to compare to the values of each individual capacitor in the network? Choose the statement below that is true regardless of the actual values for the individual capacitors.10/9/2016 HW 07 Capacitor Circuits EVALUATE your answer Part C Suppose that you are given another network of the same form as the one analyzed in Part B.masteringphysics. Hint 1. ANSWER: Ceq must be less than C1 + C2 . in a parallel connection the equivalent capacitance is always greater than any individual capacitance.0 V is maintained across ab. You may want to review ( pages 790 793) . Thus. Ceq must be less than C4 . in a series connection the equivalent capacitance is always less than any individual capacitance. Ceq Item 4 For the system of capacitors shown in the the figure below. https://session.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 9/17 . The value of Ceq is not bounded by the value of any individual capacitance in the network. Ceq must be less than C3 . 10/9/2016 HW 07 Capacitor Circuits Part A What is the equivalent capacitance of this system between a and b? ANSWER: C = 19. and this combination is in parallel with the other two capacitors. When these capacitors are replaced by their equivalent we get the network sketched in Figure 24.0 nF and 10.0 nF.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 10/17 .0 nF capacitors in series is 5. SET UP: For capacitors in series the voltages add and the charges are the same; capacitors in parallel the voltages are the same and the charges add; Ceq 1 Ceq = 1 C1 = C1 + C2 + … + 1 C2 . 30.21. = Ceq V = (19.50 nF capacitor store? ANSWER: https://session.29 nF. Part B How much charge is stored by this system? ANSWER: = 482 nC Q Correct Q tot .3 nF. For .masteringphysics.3 nF Correct IDENTIFY: Three of the capacitors are in series. The equivalent capacitance of these three capacitors in parallel is 19.3nF )(25. and this is the equivalent capacitance of the original network. EXECUTE: The equivalent capacitance of the 18. C = +… Q V .0V) = 482 nC Part C How much charge does the 6. 5 V6. Q6. .0 V .50 nF capacitor? ANSWER: V = 25.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 11/17 .masteringphysics. Part A Find the equivalent capacitance of this system between a and b.5 = (6. Item 5 The Figure below shows a system of four capacitors.0 V Correct 25 V .21 is 25 V .5 = C6.10/9/2016 HW 07 Capacitor Circuits = 162 nC Q Correct The potential across each capacitor in the parallel network of Figure 24.50nF)(25. we must go through a series of steps to simplify it as we solve for the unknowns.0V) = 162 nC Part D What is the potential difference across the 7. ANSWER: C = 3. where the potential difference across ab is 50. EVALUATE: As with most circuits.47 μF https://session. 10/9/2016 HW 07 Capacitor Circuits Correct Part B How much charge is stored by this combination of capacitors? ANSWER: = 174 μC Q Correct Part C How much charge is stored in the 10.masteringphysics. and (b) between a and c.0μF capacitor? ANSWER: = 174 μC Q Correct Part D How much charge is stored in the 9. find the equivalent capacitance (a) between b and c.0μF capacitor? ANSWER: = 174 μC Q Correct Item 6 For the system of capacitors shown in .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 12/17 . https://session. com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 13/17 .57 pF Correct Item 7 In the figure . https://session.0 V . each capacitor has 5.00 μF and Vab = 25.0 pF Correct Part B ANSWER: Cac = 8.10/9/2016 HW 07 Capacitor Circuits Part A ANSWER: Cbc = 20.masteringphysics. ANSWER: Q1 = 2.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 14/17 .10/9/2016 HW 07 Capacitor Circuits Part A Calculate the charge on each capacitor.50×10−5 C Correct Part C ANSWER: Q3 = 5.masteringphysics.50×10−5 C Correct Part B ANSWER: Q 2 = 2.00×10−5 C Correct Part D ANSWER: https://session. 0 V Correct Part H ANSWER: V4 = 15.10/9/2016 HW 07 Capacitor Circuits Q4 = 7. ANSWER: https://session.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 15/17 . ANSWER: V1 = 5.0 V Correct Part I Calculate the potential difference between points a and d.50×10−5 C Correct Part E Calculate the potential difference across each capacitor.masteringphysics.00 V Correct Part F ANSWER: V2 = 5.00 V Correct Part G ANSWER: V3 = 10. C1 = 6.00 μF .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 16/17 . and C3 = 5. C2 = 3. ANSWER: Q 1 = 60. The capacitor network is connected to an applied potential Vab. Part A What is the charge on capacitor C1 ? Express your answer with the appropriate units. After the charges on the capacitors have reached their final values.0 μC Correct Part C https://session.00 μF . ANSWER: Q 3 = 90.0 μC Correct Part B What is the charge on capacitor C3 ? Express your answer with the appropriate units.masteringphysics.0 V Correct Item 8 In .10/9/2016 HW 07 Capacitor Circuits Vad = 10. the charge on C2 is 30.0 μC.00 μF . com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4799788 17/17 .9%.masteringphysics.89 out of a possible total of 10 points. ANSWER: Vab = 28. https://session.0 V Correct Score Summary: Your score on this assignment is 98. You received 9.10/9/2016 HW 07 Capacitor Circuits What is the applied voltage Vab? Express your answer with the appropriate units.