Combination of Capacitors

A network of four capacitors each of $15 \mu F$capacitance is connected to a $500 V$ supply as shown in the figure. Determine $\left(a\right)$ equivalent capacitance of the network and $\left(b\right)$ charge on $C_{4 }$capacitor.

Two capacitors of capacitance $6 \mathrm{\mu F}$ and $12 \mathrm{\mu F}$ are connected in series with a battery. The voltage across the $6 \mathrm{\mu F}$ capacitor is $2 \mathrm{V}$. The total battery voltage is

Three capacitors $2 \mathrm{\mu F}, 3 \mathrm{\mu F}$ and $6 \mathrm{\mu }\mathrm{F}$ are joined with each other. What is the minimum effective capacitance (in $\mathrm{\mu F}$)?

Four plates of equal area $A$ are separated by equal distance $d$ and are arranged as shown in the figure. The equivalent capacity is:

Three identical capacitor are given a charge Q each and they are allowed to discharge through resistor R₁, R₂ and R₃. Their charges as a function of time are shown in graph below. The smallest of the three resistance is:

Find the equivalent capacitance between the points $A$ and $B$ in micro Farads.

Find the equivalent capacitance of the following connection in micro Farads. Round the answer to two digits after the decimal point.

Capacitor $C_1$, $C_2$ and $C_3$ are connected in series. Their equivalent capacitance $C_{eq}$ is given as:

A network of four capacitors $5 \mathrm{\mu F}$ each are connected to a $240 \mathrm{V}$ supply. Determine the equivalent capacitance of the network.

The equivalent capacitance between points $P$ and $Q$ in the following circuit is

Five capacitors each of capacitance value $C$ are connected as shown in the figure. The ratio of capacitance between $P$ and $R$ and the capacitance between $P$ and $Q$ is:-

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

Four capacitors ${\text{C}}_1\left(=1\text{ μ}\text{F}\right)$, ${\text{C}}_2\left(=2\text{ μ}\text{F}\right)$, ${\text{C}}_3\left(=3\text{ μ}\text{F}\right)$, and ${\text{C}}_4\left(=4\text{ μ}\text{F}\right)$, are connected in a network as shown in the diagram. The e.m.f of the battery is E = 12 V and its internal resistance is negligible. The keys S₁ and S₂ can be independently put on or off. Indicate the charge on the capacitors C₁, C₂, C₃ and C₄ by q₁, q₂, q₃ and q₄ respectively and the potential drops across them by V₁, V₂, V₃ and V₄ respectively.

 Initially key S₂ is open. The key S₁ is closed and the capacitors are charged. If now the key S₂ be closed, the charged that will flow across this key is

A network of four $20\mathrm{ }\mathrm{\mu }\mathrm{F}$ capacitors is connected to a $600 \mathrm{V}$ supply as shown in the figure.
Then, the equivalent capacitance of the network is $26.67 \mu \mathrm{F}$.

A parallel-plate capacitor consists of two identical metal plates. Two dielectric slabs having dielectric constants $K_1$ and $K_2$ are introduced in the space between two plates as shown below. Show that capacity of capacitor in the arrangement is given by $C=\frac{{\epsilon }_0\left(K_1+K_2\right)A}{2d}$

A network of four capacitors $5 \mathrm{\mu F}$ each are connected to a $240 \mathrm{V}$ supply. Determine the charge on each capacitor.

Derive an expression for effective capacitance of three capacitors connected in parallel.

What is the equivalent capacitance of two $2 \mathrm{\mu F}$ capacitors connected in series.

How will you obtain a capacitance of higher value if you have been provided with capacitors of relatively low capacitance? 

If $n$ capacitors of equal capacitance $C$ are connected in series what will be the equivalent capacitance.