Tag: capacitors in parallel and series

Two identical capacitors are connected in series with a source of potential V. If Q is the charge on one of the capacitors, the capacitance of each capacitor is: 

Two capacitors of $4\ \mu F$and $2\ \mu F$ are connected in series with the battery. If total potential difference across the two capacitors is $200$ volts then the ratio  of potential difference across one capacitor to another is

A capacitor of capacitance $ 1 \mu F $ withstands a maximum voltage of 6 kilovolt while another capacitor of $ 2 \mu F $ withstands a maximum voltage 4 kilovolt . if the two capacitor are connected in series, the system will withstand a maximum of:

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

When two condensers of capacitance $1\mu F$ and $2\mu F$ are connected is series then the effective capacitance will be :

Three condensers each of capacitance 2 F, are connected in series. The resultant capacitance will be :

A resistor $ ^{\prime} R^{\prime}  $ and $2  \mu F  $ capacitor in series is connected through a switch to $200  \mathrm{V}  $ direct supply. Across the capacitor is a neon bulb that lights up at $120  \mathrm{V} $ Calculate the value of $  R  $ to make the bulb light up $5  s  $ after the switch has been closed. $ \left(\log _{10} 2.5=0.4\right) $

Two capacitors of capacitances $4\mu F$ and $6\mu F$ are connected across a 120 V battery in series with each other. What is the potential difference across the $4\mu F$ capacitor?

Two capacitor of capacity $C _{1}$ and $C _{2}$ are connected in series. The combined capacity $C$ is given by

Three condenser of capacitance $C(\mu F)$ are connected in parallel to which a condenser of capacitance $C$ is connected in series. Effective capacitance is $3.75$, then capacity of each condenser is