Two condensers C₁ and C₂ in a circuit are joined as shown in figure. potential of point A is V?

English

Gujarati

Hindi

1. Electric Charges and Fields

normal

Two condensers $C_1$ and $C_2$ in a circuit are joined as shown in figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be

A

$\frac{1}{2}\,\left( {{V_1} + {V_2}} \right)$

B

$\frac{{{C_2}{V_1} + {C_1}{V_2}}}{{{C_1} + {C_2}}}$

C

$\frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$

D

$\frac{{{C_2}{V_1} - {C_1}{V_2}}}{{{C_1} + {C_2}}}$

Solution

As $_1$ and $C _2$ are in series, the charge on $C _1$ is equal to the charge on $C _2$,i.e.,

$Q _1= Q _2$

Let, $V_D$ is the potential at $D$.

therefore, $C_1\left(V_1-V_D\right)=C_2\left(V_D-V_2\right)$

$V _{ D }\left( C _1+ C _2\right)= C _1 V _1+ C _2 V _2$

$V _{ D }=\frac{ C _1 V _1+ C _2 V _2}{ C _1+ C _2}$

Standard 12

Physics

Share

Similar Questions

The plates of a parallel plate capacitor are charged up to $100\, volt$. A $2\, mm$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by $1.6\, mm$. The dielectric constant of the plate is

normal

A point charge $q$ is situated at a distance $d$ from one end of a thin non – conducting rod of length $L$ having a charge $Q$ (uniformly distributed along its length) as shown in fig.Then the magnitude of electric force between them is

normal

Two point charges $+8q$ and $-2q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the $x-$ axis at which the net electric field due to these two point charges is zero is

normal

For shown situation of two dipoles the nature of forces between them are

normal

Two point charges of $ + 2\,\mu C$ and $ + 6\,\mu C$ repel each other with a force of $12\, N$. If each is given an additional charge of $ – 4\,\mu C$, then force will become

normal