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Q.

Two condensers of capacities 2C and C are joined in parallel and charged upto potential V. The battery is removed and the condenser of capacity C is filled completely with a medium of dielectric constant K. The p.d. across the capacitors will now be

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a

\frac{{3V}}{K}

b

\frac{V}{{K + 2}}

c

\frac{{3V}}{{K + 2}}

d

\frac{V}{K}

answer is A.

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Detailed Solution

q1 = 2CV, q2 = CV

Now condenser of capacity C is filled with dielectric K, therefore C2 = KC
As charge is conserved

\therefore\,{q_1} + {q_2} = (C_2' + 2C)V'
\Rightarrow V' = \frac{{3CV}}{{(K + 2)C}} = \frac{{3V}}{{K + 2}}

 

 

 

 

 

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