Six capacitors are connected as shown in the figure. a) If the capacitance of ca
ID: 2264170 • Letter: S
Question
Six capacitors are connected as shown in the figure.
a) If the capacitance of capacitor 3 is 1.5 nF, what does C2 have to be to yield an equivalent capacitance of 5.0 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 2.126 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 = 1.5 nF, C5 = 1.7 nF, and C6 = 3.3 nF?
d) If a battery with a potential difference of 10.1 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?
Please write answers clearly
Six capacitors are connected as shown in the figure. If the capacitance of capacitor 3 is 1.5 nF, what does C2 have to be to yield an equivalent capacitance of 5.0 nF for the combination of capacitors 2 and 3? 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 2.126 nF for the combination of the three capacitors? 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 = 1.5 nF, C5 = 1.7 nF, and C6 = 3.3 nF? If a battery with a potential difference of 10.1 V is connected to the capacitors as shown in the figure, what is the total charge on the six capacitors? What is the potential drop across capacitor 5 in this case?Explanation / Answer
a) C23 = C2+C3
5 n F = C2 + 1.5 nF
==> C2 = 3.5 n F
b) C123 = C1*(C2+C3)/(C1+C2+C3)
2.126 = C1*5/(C1+5)
(C1+5)*2.126 = C1*5
c1*2.126 + 10.63 = c1*5
C1 = 10.63/(5-2.126)
C1 = 3.7 nF
c)
C456 = C4+C5+C6 = 1.5 + 1.7+3.3 = 6.5 n F
Cnet = C456*C123/(C456+C123) = 2.126*6.5 /(2.126 + 6.5) = 1.602 n F
d)
Qnet = cnet*v = 1.602*10^-9*10.1 = 16.18*10^-9 C
e)
v5 = C123*V/(C123+C456) = 2.126*10.1/(2.126+6.5) = 2.49 volts