Three capacitors of $2\,\mu F,\,3\,\mu F$ and $6\,\mu F$ are joined in series and the combination is charged by means of a $24\, volt$ battery. The potential difference between the plates of the $6\,\mu F$ capacitor is.......$volt$
A$4 $
B$6$
C$8$
D$10 $
Medium
Download our app for free and get started
A$4 $
a (a) $\frac{1}{{{C_{eq}}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \Rightarrow {C_{eq}} = 1\,\mu \,F$
Total charge $Q = C_{eq.}\,\,V = 1 × 24 = 24\, µC$
So $p.d.$ across $6\, µF$ capacitor = $\frac{{24}}{6} = 4\,volt$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A parallel plate capacitor has two layers of dielectrics as shown in fig. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is
Two insulated charged spheres of radii $20\,cm$ and $25\,cm$ respectively and having an equal charge $Q$ are connected by a copper wire, then they are separated
The respective radii of the two spheres of a spherical condenser are $12\;cm$ and $9\;cm$. The dielectric constant of the medium between them is $ 6$. The capacity of the condenser will be
In a region of space, suppose there exists a uniform electric field $\vec{E}=10 i\left(\frac{ v }{ m }\right)$. If a positive charge moves with a velocity $\vec{v}=-2 \hat{j}$, its potential energy
Three concentric metallic spherical shells of radii $R, 2R, 3R$, are given charges $Q_1, Q_2, Q_3$, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, $Q_1 : Q_2 : Q_3$ is
The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be