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VITEEE 2012 Paper

Section: Physics

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Question : 18 of 120

Marks: +1, -0

Two capacitors of capacities 1 µF and C µF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 µC, the energy stored in the capacitor of capacity C in µJ is

1800
1600
14400
7200

Solution:

Capacitance 1 µF and C µF are connectedin series,

∴

C_{eq}

=

\frac{C}{1+C}

Given, V = 120 V and q = 8 µC

Since q =

C_{eq}

V

80 =

\frac{C}{C+1}

× 20

or C = 2 µF

Energy stored in the capacitor of capicity C

U =

\frac{1}{2}\frac{q^2}{C}

=

\frac{1}{2} \times \frac{(80\times 10^{-6})^2}{2\times 10^{-6}}

=

\frac{1}{2}

×

\frac{80\times 10^{-6}\times 80\times 10^{-6}}{2\times 10^{-6}}

U = 1600 µJ

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