A 20.0 pF capacitor and a 30.0 pF capacitor is connected inseries. This series i

Question

A 20.0 pF capacitor and a 30.0 pF capacitor is connected inseries. This series is then placed in parallel with a series setupcontaining 15.00 pF and 5 pF capacitors. This parallel setup isplaced in series with a 10 pF capacitor. The entire arrangment isconnected to a 24.00 V battery. What is the total capacitance in the system? What is the total charge stored in the circuit? What is the charge stored on each capacitor? What is the voltage drop across each capacitor? How much energy is stored in each capacitor when they arefully charged? A 20.0 pF capacitor and a 30.0 pF capacitor is connected inseries. This series is then placed in parallel with a series setupcontaining 15.00 pF and 5 pF capacitors. This parallel setup isplaced in series with a 10 pF capacitor. The entire arrangment isconnected to a 24.00 V battery. What is the total capacitance in the system? What is the total charge stored in the circuit? What is the charge stored on each capacitor? What is the voltage drop across each capacitor? How much energy is stored in each capacitor when they arefully charged?

Explanation / Answer

Equivalent capacitance of 20 pF and 30pF which are in series =20x30/20+30 =12 pF Equivalent capacitance of 15 pF and 5pF which are in series =15x5/15+5 =15/4 pF Equivalent capacitance of 12pF and15/4 pF which arein parallel = 12+15/4=63/4 pF Equivalent capacitance of 63/4 pF and 10 pF which are inseries = (63/4)x10/(63/4+10 =630/103pF       i) Totalcapacitance of the system,C = 630/103 =6.116 pF       ii) Total chargestored in the circuit,Q = CV = 6.116X10-12 x 24=146.78x10-12 C        iii) Chargestored on each capacitor depends on voltage drop           LetQ1 flows through the branch 20 pF and 30p F which are in series           LetQ2 flows through the branch 15 pF and  5pF which are in series           Let Q flows through the branch 10 pF which are in series thecombination                                         Q1= 12x10-12x24     = 288 x10-12C                                        Q2= (15/4)x10-12x24 = 90x10-12 C                     Q = 146.78x10-12 C     iv) Voltage drop across each capacitor isgiven by:                     V10 = Q/C = 146.78 X10-12 /10X10-12 = 14.67V                        Balance(24-14.67) V will be dropped across two series combinations of20pF,30pF and 15pF and 5pF                       V20 = 30x10-12x9.33/ (20+30)x10-12  =    5.60V                                                  V30 = 20x10-12x9.33/ (20+30)x10-12 =    3.73V                                                   V15 = 5x10-12x9.33/ (15+5)x10-12     =    2.33V                             V5 = 15x10-12x9.33/ (20+30)x10-12 =   7.0V                                       v)Energy stored on each capacitor when they are fully charged may befind out as follows:                Energy stored on each capacitor is given by the formula, E=(1/2) CV2               Use the values of each capacitance and their correspondingvoltage drop for calculating the energy stored on eachcapacitor.                    HOPE THIS WOULD HELP YOU                     Q2= (15/4)x10-12x24 = 90x10-12 C                     Q = 146.78x10-12 C     iv) Voltage drop across each capacitor isgiven by:                     V10 = Q/C = 146.78 X10-12 /10X10-12 = 14.67V                        Balance(24-14.67) V will be dropped across two series combinations of20pF,30pF and 15pF and 5pF                       V20 = 30x10-12x9.33/ (20+30)x10-12  =    5.60V                                                  V30 = 20x10-12x9.33/ (20+30)x10-12 =    3.73V                                                   V15 = 5x10-12x9.33/ (15+5)x10-12     =    2.33V                             V5 = 15x10-12x9.33/ (20+30)x10-12 =   7.0V                                       v)Energy stored on each capacitor when they are fully charged may befind out as follows:                Energy stored on each capacitor is given by the formula, E=(1/2) CV2               Use the values of each capacitance and their correspondingvoltage drop for calculating the energy stored on eachcapacitor.                    HOPE THIS WOULD HELP YOU

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