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
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?