a) If the capacitance of capacitor 3 is 4.1 nF, what does C 2 have to be to yiel

Question

a) If the capacitance of capacitor 3 is 4.1 nF, what does C2 have to be to yield an equivalent capacitance of 6.0 nF for the combination of capacitors 2 and 3?

b) For the same values of C2 and C3 as in part (a), what is the value of C1 that will give an equivalent capacitance of 0.930 nF for the combination of the three capacitors?

c) For the same values of the capacitances of capacitors 1, 2 and 3 as in part (b), what is the equivalent capacitance (in nF) of the whole system of capacitors, if the values of the remaining capacitances are C4 = 4.3 nF, C5 = 4.1 nF, and C6 = 4.5 nF?

d) If a battery with a potential difference of 12.9 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 capacitor 5 in this case?

Explanation / Answer

a) C2 and C3 in parallel so C eq= C2 + C3 6 = C2 + 4.1 C2 = 1.9 nF b) C1 in series with Ceq in series 1/C = 1/C1 + 1/Ceq 1/0.93 = 1/C1 + 1/1.9 solve for C1 C1 = 1.82 nF c) C4, C5, C6 in parallel so Ceq = C4 + C5 + C6 = 4.3+4.1+4.5=12.9 nF this is in parallel with previous equivalent 1/C = 1/12.9 + 1/0.93 solve for C C=0.867 nF d) Q=CV = 0.867 nF * 12.9=11.18 nC e) thing in parallel share Q so 4,5,6 have 11.18 nC that will be a voltage of V=Q/C=11.18/12.9=0.867 V

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