A $1 μ F$ capacitor is charged by $100 V$ supply. It is then disconnected from the supply and is connected to another uncharged $1 μ F$ capacitor. The percentage of energy lost in form of heat and electromagnetic radiation is

Option 1 - 25%

Option 2 - 30%

Option 3 - 60%

Option 4 - 50%

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Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 4

Detailed Solution:

$C_{1} = 1 μ F V_{1} = 100 V ? U_{1} = \frac{1}{2} * 1 * 10^{- 6} * {(100)}^{2} = 5 * 10^{- 3} J$

$C_{2} = 1 μ F$ , common potential $V = \frac{C_{1} V_{1}}{C_{1} C_{2}} = \frac{100}{2} = 50 V$

Final electrostatic energy stored in both the capacitors.

$= \frac{1}{2} (C_{1} + C_{2}) V^{2} = \frac{1}{2} * 2 * {(50)}^{2} * 10^{- 6}$

$= 2.5 * 10^{- 3} J$

$E n e r g y l o s t = 2.5 * 10^{- 3}$

$% l o s s o f e n e r g y = \frac{2.5 * 10^{- 3}}{5 * 10^{- 3}} * 100$

$= 50 %$

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A 1 μ F capacitor is charged by 100 V supply. It is then disconnected from the supply and is connected to another uncharged 1 μ F capacitor. The percentage of energy lost in form of heat and electromagnetic radiation is

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