A series combination of n_1 capacitors, each | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

In series, the equivalent capacitance is$:$ $C s=\frac{C_{1}}{n_{1}}$ and

Energy stored, $E_{S}=\frac{1}{2} C_{S} V_{S}^{2}=\frac{1}{2}\left(C_{1}

Vedclass · Accepted Answer

$\frac{{16{C_1}}}{{{n_1}{n_2}}}$

Vedclass · Answer

In series, the equivalent capacitance is$:$ $C s=\frac{C_{1}}{n_{1}}$ and

Energy stored, $E_{S}=\frac{1}{2} C_{S} V_{S}^{2}=\frac{1}{2}\left(C_{1} / n_{1}ight)(4 V)^{2}=\frac{8 C_{1} V^{2}}{n_{1}}$

In parallel, the equivalent capacitance is $C_{P}=n_{2} C_{2}$ and

Energy stored $, E_{P}=\frac{1}{2} C_{P} V_{P}^{2}=\frac{1}{2}\left(n_{2} C_{2}ight)(V)^{2}=\frac{n_{2} C_{2} V^{2}}{2}$

Here, $E_{S}=E_{P}$

$\frac{8 C_{1} V^{2}}{n_{1}}=\frac{n_{2} C_{2} V^{2}}{2}$

$	herefore C_{2}=\frac{16 C_{1}}{n_{1} n_{2}}$

Similar Questions

If potential at centre of uniformaly charged ring is $V_0$ then electric field at its centre will be (assume radius $=R$)

There is a square gaussian surface placed in $y-z$ plane. Its axis is along $x-$ axis and centre is at origin. Two identical charges, each $Q$, are placed at point $(a, 0, 0)$ and $(-a, 0, 0)$. Each side length of square is $2a$ then electric flux passing through the square is

Three charges $2q,\, - q,\, - q$ are located at the vertices of an equilateral triangle. At the centre of the triangle

Find the equivalent capacitance across $A$ $\&$ $B$ ........$\mu f$

Two parallel metal plates having charges $+ Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will