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Q43  Option 2
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Asked by manasvijha | 17 Apr, 2019, 07:50: AM
answered-by-expert Expert Answer

First let us calculate the equivalent capacitance across the poits C and D as shown in fig-1 .
Capacitance 3C is connected parallel to a series combination of two capacitance C and 3C.   
Equivalenet capacitance of series combination is 3.75C. Hence equivalent capacitance across points C and D is 4.75 C.
 
if a series combination of two capacitors of respective capacitance C1 and C2 is connected across a poteinial difference V,
then potential difference across capacitance C1 is  begin mathsize 12px style fraction numerator C 2 over denominator C 1 plus C 2 end fraction V end style and potential difference across capacitance C2 is  begin mathsize 12px style fraction numerator C 1 over denominator C 1 plus C 2 end fraction V end style
Hence in fig-1, 190 V is applied across series combination of C and 4.75C.
 
Hence potential difference across C and D is (1/4.75)×190 = 40 V
 
Now in fig-2, we see 40V potential difference across series combination of C and 3C.
 
Hence potential difference across the capacitance 3C is begin mathsize 12px style fraction numerator C over denominator C plus 3 C end fraction cross times 40 space equals space 10 V end style
Answered by Thiyagarajan K | 17 Apr, 2019, 11:47: AM

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