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​If two capacitors C1 & C2 are connected in parallel then equivalent capacitance is 10μF. If both capacitor are connected across 1V battery then energy stored by C2 is 4 times that of C1. Then the equivalent capacitance if they are connected in series is -
  • a)
    1.6μF
  • b)
    16μF
  • c)
    4μF
  • d)
    1/4μF
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If two capacitors C1 & C2 are connected in parallel then equivalen...

Given C1 + C2 = 10μF ........ (i)

⇒ 4C1 = C2 …(ii)
from equation (i) & (ii)
C1 = 2μF
C2 = 8μF
If they are in series
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Most Upvoted Answer
If two capacitors C1 & C2 are connected in parallel then equivalen...
And C2 are connected in series, the effective capacitance (C) is given by the formula:

1/C = 1/C1 + 1/C2

This means that the reciprocal of the effective capacitance is equal to the sum of the reciprocals of the individual capacitances.

Let's say C1 = 10 μF and C2 = 20 μF. Plugging these values into the formula, we get:

1/C = 1/10 + 1/20

Simplifying this equation:

1/C = 2/20 + 1/20

1/C = 3/20

To find the value of C, we take the reciprocal of both sides:

C = 20/3 μF

Therefore, the effective capacitance when C1 = 10 μF and C2 = 20 μF in series is 20/3 μF.
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If two capacitors C1 & C2 are connected in parallel then equivalent capacitance is 10μF. If both capacitor are connected across 1V battery then energy stored by C2 is 4 times that of C1. Then the equivalent capacitance if they are connected in series is -a)1.6μFb)16μFc)4μFd)1/4μFCorrect answer is option 'A'. Can you explain this answer?
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