Capacitor

Three capacitors each of 4 $\mu$F are to be connected in such a way that the effective capacitance is 6 $\mu$F. This can be done by connecting them :

Effective capacitance of parallel combination of two capacitors C₁ and C₂ is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C₂ is 4 times that of C₁. If these capacitors are connected in series, their effective capacitance will be :

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :

Consider the combination of 2 capacitors C₁ and C₂ with C₂ > C₁, when connected in parallel, the equivalent capacitance is ${{15} \over 4}$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, ${{{C_2}} \over {{C_1}}}$.

The total charge on the system of capacitors $C_{1}=1 \mu \mathrm{F}, C_{2}=2 \mu \mathrm{F}, \mathrm{C}_{3}=4 \mu \mathrm{F}$ and $\mathrm{C}_{4}=3 \mu \mathrm{F}$ connected in parallel is : (Assume a battery of $20 \mathrm{~V}$ is connected to the combination)

If there are $n$ capacitors in parallel connected to $V$ volt source, then the energy stored is equal to

The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 $\mu$C charge, its radius will be : [ Take : ${1 \over {4\,\pi { \in _0}}} =$ 9 $\times$ 10⁹ N $-$ m²/C² ]

A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2μF, how much work is done ?

A capacitor C is fully charged with voltage V₀. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance ${C \over 2}$. The energy loss in the process after the charge is distributed between the two capacitors is :