Six capacitors are connected as shown in the figure C3 C4 C2 C6 C5 (a) lf C3 3.4

Question

Six capacitors are connected as shown in the figure C3 C4 C2 C6 C5 (a) lf C3 3.4 nF, what does C2 have to be to yield an equivalent capacitance of 5.000 nF for the combination of the two capacitors? (b) For the same values of C2 and C as in part (a), what is the value of C1 that will give an equivalent capacitance of 1.560 nF for the combination of the three capacitors? (c) For the same values of C1, C2, and C3 as in part (b), what is the equivalent capacitance of the whole set of capacitors if the values of the other capacitances are C41.2 nF, Cs 2.2 nF and C6 = 5.9 nF? (d) If a battery with a potential difference of 11.7 V is connected to the capacitors as shown in the figure, what is the total charge on the six capacitors? (e) What is the potential drop across C5 in this case?

Explanation / Answer

a. C3 =3.4 nF and C2=?

As C2 and C3 are parallel to each other, the equivalent capacitance = C2 + C3 = 5.00 nF

so, C2 = 5.00 nF - 3.4 nF =1.6 nF

so C2=1.6 nF

b. Now C1 and 5.00 nF are in series to give an equivalent capacitance 1.56 nF

so, 1/1.56 = 1/C1 + 1/5

so, C1 = 2.267 nF

so C1 =2.267 nF

c. C4 =1.2 nF, C5=2.2nF, C6 =5.9 nF

as all three are in parallel, so the equivalent capacitance = C4 + C5 + C6= 1.2 +2.2 +5.9 =9.3 nF

Now this 9.3 nF are in series with 1.56 nF

so, the equivalent capacitance = ( 1.56 x 9.3)/(1.56+9.3) = 1.336 nF

d. The total charge on six capacitors = equivalent capacitance x potential difference of the battery

=1.336 nF x 11.7 V = 15.63 nC

e. potential drop across C5 = total charge/ equivalent capacitance of (C4,C5 & C6) =15.63 /9.3 = 1.68 V

all the best in the course work

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