Six capacitors are connected as shown in the figure. a. If C3 = 2.300 nF, what d
ID: 1645736 • Letter: S
Question
Six capacitors are connected as shown in the figure. a. If C3 = 2.300 nF, what does C2 must 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 C3 as in part (a), what is the value of C1 that will give an equivalent capacitance of 1.914 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 C4 = 1.300 nF, C5 = 1.700 nF, and C6 = 4.700 nF? d) If a battery with a potential difference of 11.70 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) equivalent capacitnace of C2 and C3 = C2+C3 = 5 nF
C3 = 2.3 nF
Therefore, C2 = 5-2.3 = 2.7 nF
b) Equivalent capacitance of C2 and C3 = 5nF
This is in series with C1
equivalent capacitance of 5 nF and C1 = 1.914 nF
1/1.914 = 1/5 + 1/C1
Therefore, C1 = 3.1 nF
c) C4,C5 and C6 are in parallel . Their equivalent capacitance = C4+C5+C6 = 1.3+1.7+4.7=7.7 nF
This is in series with 1.914 nF
Therefore, equivalent capacitance of whole circuit = (1.914*7.7)/(1.914+7.7) = 1.533 nF
d) Q= CV = 11.7*1.533*10-9 = 17.94*10-9 Coulomb
e) Potential drop across C5 = 11.7- (Q/C1) - Q/(C2+C3) = 11.7-(17.94*10-9 / 3.1*10-9) - (17.94*10-9 / 5*10-9)
= 2.325 V