A capacitor of 0-2- F capacitance is charged to 600 V- After removing the battery- it is connected with a 1-0- Fcapacitor in parallel- then the potential difference across each capacitor will become -

The correct option is C 100 VA capacitor #160; = 2μ F #160; (in series)Charge =600V Parallel capacitor #160; =1.0μ F 0.2× 600= V 0.1 o ...

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Standard X

Physics

Electric Potential and Potential Difference

A capacitor o...

Question

A capacitor of $0.2 μ F$ capacitance is charged to 600 V. After removing the battery, it is connected with a $1.0 μ F$ capacitor in parallel, then the potential difference across each capacitor will become :

300 V

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600 V

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100 V

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120 V

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Solution

The correct option is C 100 V
A capacitor $= - 2 μ F$ (in series)
Charge $= 600 V$
Parallel capacitor $= 1.0 μ F$
$0.2 \times 600 = \frac{V}{0.1}$
or, $V = 0.2 \times 0.1 \times 600$
$= \frac{2}{10} \times \frac{1}{10} \times 600$
$2 \times \frac{600}{12} = 50 \times 2 = 100 V$

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A capacitor of 0.2μF capacitance is charged to 600 V. After removing the battery, it is connected with a 1.0μFcapacitor in parallel, then the potential difference across each capacitor will become :

The correct option is C 100 VA capacitor =−2μF (in series)Charge =600VParallel capacitor =1.0μF0.2×600=V0.1or, V=0.2×0.1×600 =210×110×6002×60012=50×2=100V

A capacitor of $0.2 μ F$ capacitance is charged to 600 V. After removing the battery, it is connected with a $1.0 μ F$ capacitor in parallel, then the potential difference across each capacitor will become :

The correct option is C 100 V
A capacitor $= - 2 μ F$ (in series)
Charge $= 600 V$
Parallel capacitor $= 1.0 μ F$
$0.2 \times 600 = \frac{V}{0.1}$
or, $V = 0.2 \times 0.1 \times 600$
$= \frac{2}{10} \times \frac{1}{10} \times 600$
$2 \times \frac{600}{12} = 50 \times 2 = 100 V$