1. Tardigrade
  2. Question
  3. Physics
  4. A capacitor of capacitance C1=1 μ F can withstand maximum voltage V 1=6 kV (kilo-volt) and another capacitor of capacitance C2=3 μ F can withstand maximum voltage V 2=4 k V. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

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Q. A capacitor of capacitance $C_1=1\, \mu F$ can withstand maximum voltage $V _1=6 \, kV$ (kilo-volt) and another capacitor of capacitance $C_2=3\, \mu F$ can withstand maximum voltage $V _2=4 \, k V$. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

Electrostatic Potential and Capacitance

Solution:

As $Q=C V,\left(Q_1\right)_{\max }=10^{-6} \times 6 \times 10^3=6\, mC$
While $\left(Q_2\right)_{\max }=3 \times 10^{-6} \times 4 \times 10^3=12 \, mC$
However in series charge is same so maximum charge on $C_2$ will also be $6 \, mC$ (and not $12 mC$ ) and potential difference across it $V_2=6 \, mC / 3\, \mu F =2\, k V$ and as in series $V$
$=V_1+V_2$ so $V_{\max }=6\, kV +2 kV =8 \, kV$