Question

1. Three capacitors with individual capacitances of 2 μF, 5 μF, and 10 μ respectively are connected in series with a 12V battery. What are the total capacitance and total charge in the network?
2. Same capacitors in number 1 connected in a parallel. If the combined charge in the network is 50 μC, what are the total voltage and total capacitance in the network?

Answer 1

1. The total capacitance $C$ of series connection is given as follows:

$$\dfrac{1}{C} = \dfrac{1}{2\times 10^{-6}F} + \dfrac{1}{5\times 10^{-6}F} + \dfrac{1}{10\times 10^{-6}F}\\ C = 1.25\times 10^{-6}F$$

The total charge:

$$Q = CV = 1.25\times 10^{-6}F\cdot 12V = 15\times 10^{-6}C$$

2. The total capacitance of parallel connection:

$$C = 2\times 10^{-6}F+5\times 10^{-6}F+10\times 10^{-6}F = 17\times 10^{-6}F$$

The voltage:

$$V = \dfrac{Q}{C} = \dfrac{50\times 10^{-6}C}{17\times 10^{-6}F} \approx 2.9V$$