A 10μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:
A 10μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:
Option 1 - <p>15μF</p>
Option 2 - <p>20μF<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>10μF<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>30μF</p>
5 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
Answered by
6 months ago
Correct Option - 1
Detailed Solution:
By conservation of charge, (50) (10) = 20 (10+C) ⇒ C=15µF
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