In the circuit shown in figure, four capacitors are connected to a battery. The charge on the $5$ $\mu$$F$ capacitor is.......$ \mu C$
Medium
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To form a composite $16\,\mu F,\;1000\,V$ capacitor from a supply of identical capacitors marked $8\,\mu F,\;250\,V$, we require a minimum number of capacitors
The radius of nucleus of silver (atomic number $=$ $47$) is $3.4 \times {10^{ - 14}}\,m$. The electric potential on the surface of nucleus is $(e = 1.6 \times {10^{ - 19}}\,C)$
See the diagram . Area of each plate is $2.0\ m^2$ and $d = 2 \times 10^{-3}\ m$. A charge of $8.85 \times 10^{-8}\ C$ is given to $Q$. Then the potential of $Q$ becomes......$V$
Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20\ V$. (i.e., plate $2$ is at a higher potential). The plates are separated by $d = 0.1\ m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1. $ What is its speed when it hits plate $2?$
$(e = 1.6 \times 10^{-19}\ C, m_e= 9.11 \times 10^{-31}\ kg)$
Two capacitances of capacity ${C_1}$ and ${C_2}$ are connected in series and potential difference $V$ is applied across it. Then the potential difference across ${C_1}$ will be
Assertion : A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant $K$ is introduced between the plates. The energy which is stored becomes $K$ times.
Reason : The surface density of charge onthe plate remains constant or unchanged.
Three charged capacitors, $C_1$ = $17\ μF$, $C_2$ = $34\ μF$, $C_3$ = $41\ μF$and two open switches, $S_1$ and $S_2$ are assembled into a network with initial voltages and polarities, as shown. Final status of the network is attained when the two switches, $S_1$ and $S_2$ are closed. In the figure, the final charge on capacitor $C_3$ in $mC$, is closet to: