An electrical circuit requires a capacitance of \(2~\mu\text{F}\) across a potential difference of \(1.0~\text{kV}.\) A large number of \(1~\mu\text{F}\) capacitors are available, each of which can withstand a maximum voltage of \(300~\text{V}.\) What is the minimum number of these capacitors needed to achieve the required capacitance and voltage rating?
1. \(2\)
2. \(16\)
3. \(24\)
4. \(32\)

The figure shows the charge \((q)\) versus voltage \((V)\) graphs corresponding to two different combinations of the same pair of capacitors:
1. In parallel: \(A\)
2. In series: \(B\)
Based on the graphs (for \(A\) & \(B\)), what are the capacitances of the two capacitors?

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacitances, \(\left ( \dfrac{C_{Series}}{C_{Parallel}} \right )\) in these two cases will be:
1. \(4:1\)
2. \(2:1\)
3. \(1:4\)
4. \(1:2\)

Consider the combination of 2 capacitors \(\mathrm{C}_1\) and \(\mathrm{C}_2\), with \(\mathrm{C}_2>\mathrm{C}_1 \), when connected in parallel, the equivalent capacitance is \(\frac{15}{4}\)times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors,\(\frac{\mathrm{C}_2}{\mathrm{C}_1}\).
1. \(\frac{15}{11}\)
2. \(\frac{111}{80}\)
3. \(\frac{29}{15}\)
4. None of the above