Two condensers, one of capacity C and other of capacity frac{C}{2} , are connected to a V, volt batt

English

Gujarati

Hindi

1. Electric Charges and Fields

normal

Two condensers, one of capacity $C$ and the other of capacity $\frac{C}{2}$ , are connected to a $V\, volt$ battery, as shown. The work done in charging fully both the condensers is

A

$\frac{1}{2}CV^2$

B

$2CV^2$

C

$\frac{1}{4}CV^2$

D

$\frac{3}{4}CV^2$

Solution

Net capacitance $=\frac{3 \mathrm{C}}{2}$

total charge stored $=\frac{3}{2} \mathrm{CV}$

Work done in charging $=\frac{1}{2}\left(\frac{3}{2} \mathrm{CV}\right) \mathrm{V}$

$=\frac{3}{4} \mathrm{CV}^{2}$

Standard 12

Physics

Share

Similar Questions

Figures below show regular hexagons, with charges at the vertices, In which of the following cases the electric field at the centre is not zero.

normal

Three point charges $+q, -2q$ and $+q$ are placed at points $(x = 0, y = a, z = 0), (x = 0, y = 0,z = 0)$ and $(x = a, y = 0, z = 0)$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

normal

Two point charges $+q$ and $-q$ are held fixed at $(-d, 0)$ and $(+d, 0)$ respectively of a $(x, y)$ coordinate system. Then

normal

Two identical charged spheres suspended from a common point by two massless strings of length $l$ are initially a distance $d(d << I) $ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $v$. Then as a function of distance $x$ between them,

normal

A parallel plate capacitor has plates with area $A$ and separation $d$. A battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after slab is introduced, is

normal