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

Find the equivalent capacitance between the points A and B for the following circuit.

After the switch is closed, find potential difference between the plates $Q$ and $S.$

There are six parallel plates of equal area $A$ are arranged as shown in figure $\left(\mathrm{d}<<\mathrm{A}\right).$ The equivalent capacitance between points $2$ and $5,$ is $\alpha \frac{{\epsilon }_0 A}{\mathit{ }d}.$ Then find the value of $\alpha$.

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:

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

Two parallel plate capacitors whose capacities are $C$ and $2 C$ respectively, are joined in parallel. these are charged by $V$ potential difference. if the battery is now removed and a dielectric of dielectric constant $K$ is filled in between the plates of the capacitor $C$, then what will be the potential difference across each capacitor:

A constant potential difference between $A$ and $B$ is applied. When the switch is shifted from position $P$ to position $Q$, then,

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

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 closed. Then the key S₁ is now closed. Then the charges on the capacitors are,

For next three question please follow the same

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 both the keys are open. The the key S₁ is closed. Then the charges on the capacitors are

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 $26.67 \mu \mathrm{F}$.