First slide

Capacitance

Question

A capacitor of $2 μ F$ is charged as shown in the diagram. When the switch $S$ is turned to position 2 , the percentage of its stored energy dissipated is

Difficult

A

$75 %$
B

$80 %$
C

$0 %$
D

$20 %$

Solution

$Initially, the energy stored in 2 μ F capacitor is$
$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (2 \times 10^{- 6}) V^{2} = V^{2} \times 10^{- 6} J$
initially, the charge stored in $2 μ F$ capacitor is
$Q_{i} = C V = (2 \times 10^{- 6}) V = 2 V \times 10^{- 6} coulomb. When switch S is$
$t u r n e d t o p o s i t i o n 2, t h e c h a r g e f l o w s a n d b o t h t h e c a p a c i t o r s s h a r e c h a r g e s t i l l a c o m m o n p o t e n t i a l V_{C} i s r e a c h e d .$
$V_{C} = \frac{total charge}{total capacitance} = \frac{2 V \times 10^{- 6}}{(2 + 8) \times 10^{- 6}} = \frac{V}{5} volt$
$Finally, the energy stored in both the capacitors$
$U_{f} = \frac{1}{2} [(2 + 8) \times 10^{- 6}] {(\frac{V}{5})}^{2} = \frac{V^{2}}{5} \times 10^{- 6} J$
$% loss of energy, Δ U = \frac{U_{i} - U_{f}}{U_{i}} \times 100 %$
$= \frac{(V^{2} - V^{2} / 5) \times 10^{- 6}}{V^{2} \times 10^{- 6}} \times 100 % = 80 %$

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