Two capacitors of $1\mu F$ and $2\mu F$ are connected in series and this combination is changed upto a potential difference of $120$ volt. What will be the potential difference across $1 \mu F$ capacitor:
Three long concentric cylindrical shells have radii R, 2R and $2\sqrt{2}R$. Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:
Six identical square metallic plates are arranged as shown in figure length of each plate is l the capacitance of this arrangement should be
Two capacitors of caacity ${ C } _{ 1 }$ and ${ C } _{ 2}$ are connected in series and potential difference V is applied across it. Then the potential difference across${ C } _{ 1 }$ will be
For capacitors in the series combination, the total capacitance C is given by
A series combination of two capacitances of value $0.1\ mu F$ and $1\mu F$ is connected with a source of voltage $500\ volts$. The potential difference in volts across the capacitor of value $0.1\ muF$ will be :
Two parallel plate capacitors are connected in series. Each capacitor has a plate area A and a separation d between the plates. The dielectric constant of the medium between their plates are 2 and 4 . The separation between the plates of a single air capacitors of plate area A which effectively replaces the combination is:
A capacitor comprises of two parallel circular plates. Diameter of each of plates is equal to $6 cm$. If capacitance of above system is equivalent to capacitance of sphere, whose diameter is equal to $200 cm$. Distance between two plates will be:-
A capacitor 1 mF withstands a maximum voltage of 6KV while another capacitor 2 mF withstands a maximum voltage of 4 KV. If the capacitors are connected in series, the system will withstand a maximum voltage of (MNR)
what is the series combination of condenses and $\dfrac { 1 }{ c } =\dfrac { 1 }{ { c } _{ 1 } } +\dfrac { 1 }{ { c } _{ 2 } } +\dfrac { 1 }{ { c } _{ 3 } } $ farad