Four capacitors are connected as shown in the figure below. ( C = 10.0 F.) 20.0

Question

Four capacitors are connected as shown in the figure below. (C = 10.0 F.)

20.0 F capacitor    
Consider how the charge on the equivalent capacitor, as calculated from the voltage and equivalent capacitance between a and b, is related to the charge on the 20.0-F capacitor. C 6.00 F capacitor    
Find the voltage drop across the 20.0-F capacitor and use that result to find the voltage drop across the 6.00-F capacitor. C 3.00 F capacitor    
Use the previous results you found to calculate the charge on the 3.00-F. C capacitorC    
You appear to have calculated the charge on capacitor C from the charge on the 3.00-F capacitor, but because your answer for the 3.00-F capacitor was incorrect, your answer for the charge on capacitor C is also incorrect. C Four capacitors are connected as shown in the figure below. (C = 10.0 F.) Image for Four capacitors are connected as shown in the figure below. (C = 10.0 F.) (a) Find the equivalent capacitance (a) Find the equivalent capacitance between points a and b. Correct: Your answer is correct. F (b) Calculate the charge on each capacitor, taking ?Vab = 18.0 V. 20.0 F capacitor Incorrect: Your answer is incorrect. Consider how the charge on the equivalent capacitor, as calculated from the voltage and equivalent capacitance between a and b, is related to the charge on the 20.0-F capacitor. C 6.00 F capacitor Incorrect: Your answer is incorrect. Find the voltage drop across the 20.0-F capacitor and use that result to find the voltage drop across the 6.00-F capacitor. C 3.00 F capacitor Incorrect: Your answer is incorrect. Use the previous results you found to calculate the charge on the 3.00-F. C capacitorC Incorrect: Your answer is incorrect. You appear to have calculated the charge on capacitor C from the charge on the 3.00-F capacitor, but because your answer for the 3.00-F capacitor was incorrect, your answer for the charge on capacitor C is also incorrect. C

Explanation / Answer

C and and 3 microF are in series, effective capacitance = 2.3 is in parallel with 6 microF causing effective capacitance = 8.3 microF which is now in series with 20 microF total effective capacitance = 5.86 microF, Vab = 18V

moving from b to a , 20 and 8.3 capacitors are in series , V20 + V8.3 = 18 and V20 = Q / 20 * 10-6 and V8.3= Q/ 8.3 * 10-6 => V20 = 5.27V

12.73 across other capacitors 2.3 and 6 which are in parallel hence V6= 12.73V , V3 = (10/13) * 12.73 = 9.79V and V10 = 2.93 V

Q20 = 105 microC , Q6 = 76.38 microC Q10 = 29.3 microC and Q3 = 29.3 microC

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