Two parallel-plate condensers of capacitance of $20 μ F$ and $30 μ F$ are charged to the potential of $30 V$ and $20 V$ respectively. If likely-charged plates are connected together then the common potential difference will be :

10 V

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24 V

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50 V

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100 V

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Solution

The correct option is A $24 V$
The common potential difference is:
$V_{c} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$
$V_{c} = \frac{(20 \times 30) + (30 \times 20)}{20 + 30} = \frac{1200}{50} = 24 V$

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Two condensers of capacities $10 μ F$ and $20 μ F$ have potential differences of 20V and 10V respectively. The total charge is

C

V

2 V

[

]

4 μ F

50 V

2 μ F

100 V

10^{- 2} J

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