A capacitor of capacitance $C_{1} = 1 μ F$ withstand the maximum voltage $V_{1} = 6.0 k V$ while the capacitor of capacitance $C_{2} = 2 μ F$ , the maximum voltage $V_{2} = 4.0 k V$ . What voltage in Kilo Volt will the system of these two capacitors withstand if they are connected in series?

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Answer 1

Maximum charge capacitance $C_{1}$ can withstand is $Q_{1} = C_{1} V_{1}$ while capacitance $C_{2}$ can withstand maximum charge of $Q_{2} = C_{2} V_{2}$

When $C_{1} a n d C_{2}$ are connected in series the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand, $Q_{m a x} = C_{1} V_{1} ∵ C_{1} V_{1} < C_{2} V_{2}$ ... for given values
The resultant capacitance of the combination is $C_{R} = \frac{C _{2} C _{1}}{C _{1} + C _{2}}$
$Q_{m a x} = 6 \times 1 0^{- 3} C$ and $C_{R} = \frac{2}{3} \times 1 0^{- 6} F$

Maximum voltage that the combination of capacitance can withstand is $V_{m a x} = \frac{Q _{m a x}}{C _{R}}$
$V_{m a x} = 9 k V$

Answer 2

Maximum charge capacitance

C_{1}

can withstand is

Q_{1} = C_{1} V_{1}

while capacitance

C_{2}

can withstand maximum charge of

Q_{2} = C_{2} V_{2}

When

C_{1} a n d C_{2}

are connected in series the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand,

Q_{m a x} = C_{1} V_{1} ∵ C_{1} V_{1} < C_{2} V_{2}

... for given values

The resultant capacitance of the combination is

C_{R} = \frac{C _{2} C _{1}}{C _{1} + C _{2}}

Q_{m a x} = 6 \times 1 0^{- 3} C

and

C_{R} = \frac{2}{3} \times 1 0^{- 6} F

Maximum voltage that the combination of capacitance can withstand is

V_{m a x} = \frac{Q _{m a x}}{C _{R}}

V_{m a x} = 9 k V