A capacitor has capacitance of $6 n F$ , maximum voltage it can sustain is $40$ volt. It is in series with other capacitor $3 n F$ and this capacitor can sustain $30$ volt, what is maximum voltage that can be applied in series.

70

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50

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60

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10

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Solution

The correct option is C $60$

$q_{s u p p l i e d} = C_{e f f} ϵ = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) ϵ$

$Δ V_{1} = \frac{q}{C_{1}} = \frac{C_{2} ϵ}{C_{1} + C_{2}} = \frac{6 ϵ}{9} = \frac{2 ϵ}{3}$

$(\begin{matrix} C_{1} = 6 C_{2} = 3 \end{matrix})$

$Δ V_{2} = \frac{q}{C_{2}} = \frac{C_{1} ϵ}{C_{1} + C_{2}} = \frac{3 ϵ}{9} = \frac{ϵ}{3}$

$\begin{matrix} Δ V_{1} < 40 & Δ V_{2} < 30 2 \frac{ϵ}{3} < 40 & \frac{ϵ}{3} < 30 ϵ < 60 & ϵ < 90 ϵ < 90 \end{matrix}$

$ϵ_{m a x} = 60$

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Similar questions

Q. A capacitor has capacitance of

6 n F

, maximum voltage it can sustain is

40

volt. It is in series with other capacitor

3 n F

and this capacitor can sustain

30

volt, what is maximum voltage that can be applied in series.

A capacitor of capacitance $C_{1} = 1.0 μ F$ can with stand a maximum voltage $V_{1} = 6.0 k V$ . Anothercapacitor of capacitance $C_{2} = 2.0 μ F$ can withstand a maximum voltage $V_{2} = 4.0 k V .$ If the capacitors are connected in series, the combination can withstand a maximum voltage of