Consider the system of capacitors shown in the figure below (C1 9.00 uF, C2 7.00

Question

Consider the system of capacitors shown in the figure below (C1 9.00 uF, C2 7.00 HF) 600 2.00 AaF 90.0V (a) Find the equivalent capacitance of the system. x Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. HF (b) Find the charge on each capacitor. 324 HC (on C1) 140.4 HC (on C2) 324 HC (on the 6.00 HF capacitor) 140.4 HC (on the 2.00 HF capacitor) (c) Find the potential difference across each capacitor. 36 V (across C1) 20.1 V (across C2) 54 V (across the 6.00 HF capacitor) 70.2 V (across the 2.00 HF capacitor) (d) Find the total energy stored by the group. 202.5 Your response differs from the correct answer by more than 100%. mj

Explanation / Answer

Given
capacitors

a)   C1 = 9*10^-6 F

   C2 = 7*10^-6 F

here the capacitors C1, 6*10^-6 F are in series and

       C2 , 2*10^-6 F are in series

   we know that when the capacitors are in series combination then the resultant capacitance is

   C = C1*C2/(C1+C2)

   Cu = (9*10^-6*6*10^-6)/(9*10^-6+6*10^-6) F = 3.6*10^-6 F

same way
   Cl = (2*10^-6*7*10^-6)/(2*10^-6+7*10^-6) F = 1.56*10^-6 F

now Cu and Cl are in parallel combination and the condition for net capacitance in prallel combination is

   Cn = C1+C2 = Cl+Cu = 1.56*10^-6+3.6*10^-6 F = 5.16*10^-6 F

d)

total energy stored is U = 0.5*c*V^2

           = 0.5*5.16*10^-6*90^2 J

           = 0.020898 J
           = 20.9 mJ

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