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Consider the system of capacitors shown in the figure below (C_1 - 2.00 mu F - 5

ID: 1595114 • Letter: C

Question

Consider the system of capacitors shown in the figure below (C_1 - 2.00 mu F - 5.03 mu F) find the equivalent capacitance of the system Find the charge each capacitor you response is within 10% of the correct value. This may to 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 horror mu C (on C_2) Find the potential difference across each capacitor. Your response differs from the correct answer by more than 10%. Double check your calculator. V (across C_1) Your response differs from the correct answer by more than 10%. Double check your calculator. V (across C_2) Your response differs from the correct answer by more than 100%

Explanation / Answer

Consider from figure C1 = 2 micro farad

C2 = 5 microfarad

C3 = 6 microfarad and C4 = 2 microfarad

a) For equivalent capacitance we should consider two branches separately

C1 and C3 are in series so their equivalent capacitance is C13

C13 = (C1*C3)/(C1 +C3) = (2*6)/(2+6) = 1.5 microfarad

Similarly C2 and C4 are in series so their equivalent capacitance is C24

C24 = (C2*C4)/(C2 +C4) = (5*2)/(5+2) = 1.43 microfarad

Now C13 and C24 are in parallel

So equivalent capacitance of the circuit is

C = C13 + C24 = 1.5 + 1.43 = 2.93 microfarad

b) Now charge flowing through C1 and C3 is same because they are in series

so, C13 * V = Q1

Therefore Q1 = 1.5*90 = 135 micro coulomb

similarly charge flowing through C2 and C4 is same because they are in series

Therefore Q2 = C24*V = 1.43*90 = 128.7 micro coulomb

c) Potential difference across each capacitor :-

As Q1 is the charge throgh C1 and C3

So, V1 across C1 is

V1 =Q1/C1 = 135/2 = 67.5 volts

V2 across C3 is

V2 = Q1/C3 = 135/6 = 22.5 volts

Similarly Q2 is the charge throgh C2 and C4

V3 across C2 is

V3 = Q2/C2 = 128.7/5 = 25.75 volts

V4 across C4 is

V4 = Q2/C4 = 128.7/2 = 64.35 volts

d) Total energy stored by the group is

E = (1/2)CV^2

= (1/2)*(2.93*10^-6)*(90)^2

= 0.01186 J