Five capacitors together with their capacitances are shown in the adjoining figure. The potential difference between the points $A$ and $B$ is $60\, volt.$ The equivalent capacitance between the point $A$ and $B$ and charge on capacitor $5\,\mu F$ will be respectively :-
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At time $\mathrm{t}=0$, a battery of $10 \mathrm{~V}$ is connected across points $\mathrm{A}$ and $\mathrm{B}$ in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them becomes $4$ volt?
The diameter of each plate of an air capacitor is $4\,cm$. To make the capacity of this plate capacitor equal to that of $20\,cm$ diameter sphere, the distance between the plates will be
A uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\frac{{3{V_0}}}{2},\;\frac{{5{V_0}}}{4},\;\frac{{3{V_0}}}{4}$ and $\frac{{{V_0}}}{4}$ have rasius $R_1,R_2,R_3$ and $R_4$ respectively. Then
Electric potential at a point $P$ due to a point charge of $5 \times 10^{-9}\; C$ is $50 \;V$. The distance of $P$ from the point charge is ......... $cm$
(Assume, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^{+9}\; Nm ^2 C ^{-2}$)
There is an electric field $E$ in $X$-direction. If the work done on moving a charge $0.2\,C$ through a distance of $2\,m$ along a line making an angle $60^\circ $ with the $X$-axis is $4.0\;J$, what is the value of $E$........ $N/C$
A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help of insulating handles. Now which of the following quantities will decrease?
The distance between the plates of a parallel plate condenser is $\,4mm$ and potential difference is $60\;volts$. If the distance between the plates is increased to $12\,mm$, then