Answer to Question #121802 in Electric Circuits for David Isaac

Question

Six capacitors are connected as shown in the figure below.
a) If C3 = 2.3 nF, what does C2 have to be to yield an equivalent capacitance of
5 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 1.194 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 = 1.3 nF, C5 = 1.7 nF, and C6 = 4.7 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?

Answer 1

a) If C2 and C3 are in series, the capacitance of C2 should be

$$C_2=\frac{CC_3}{C-C_3}=\frac{5\cdot2.3}{5-2.3}=4.259\text{ nF}.$$

b) Three capacitors in series will provide equivalent capacitance of

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3},\\\space\\ C_1=5.954\text{ nF}.$$

c) The equivalent capacitance for the combination of capacitors connected in series:

$$C=\bigg(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}+\frac{1}{C_5}+\frac{1}{C_6}\bigg)^{-1}=\\=0.4153\text{ nF}.$$

d) The charge is equal on each of the six capacitors, it is

$$Q=VC=11.7\cdot0.4153\cdot10^{-9}=4.859\cdot10^{-9}\text{ C}.$$

e) The potential drop across C5:

$$V_5=\frac{Q}{C_5}=\frac{4.859\cdot10^{-9}}{1.7\cdot10^{-9}}=2.858\text{ V}.$$