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Two parallel plate capacitor\mathrm{A} and \mathrm{B}  having capacitance of  1 \mu \mathrm{F}  and  5 \mu \mathrm{F}  are charged separtely to the same potential of 100 \mathrm{~V}. Now the positive plate of \mathrm{A}  is connected to the negative plate of \mathrm{B} The final charge on each capacitor is-

Option: 1

\mathrm{\frac{200}{3} \mu c, \frac{1000}{3} \mu c }


Option: 2

\mathrm{\frac{100}{3} \mu c, \frac{500}{3} \mu c}


Option: 3

\mathrm{\frac{250}{3} \mu c, \frac{750}{3} \mu c}


Option: 4

None of these.


Answers (1)

best_answer

According to the scheme of connection.

\mathrm{v =\frac{c_1 v_1-c_2 v_2}{c_1+c_2} }
 or    \mathrm{ v =\frac{\left(10^{-6} \times 100\right)-\left(5 \times 10^{-6} \times 100\right)}{1 \times 10^{-6}+5 \times 10^{-6}} \text { volt } }
or   \mathrm{v =\frac{400 \times 10^{-6}}{6 \times 10^{-6}}=\frac{400}{6}=\frac{200}{3} \text { volt }}


\mathrm{\therefore Q_1=c_1 v}  \mathrm{ =\left(1 \times 10^{-6}\right) \times \frac{200}{3}}
 \mathrm{=\frac{200}{3} }  coulomb or   \mathrm{ Q_1=\frac{200}{3} \mathrm{\mu c}}

Similarly, \mathrm{ Q_2=c_2 V}

\mathrm{Q_2=\left(5 \times 10^{-6}\right) \times \frac{200}{3}=\frac{1000}{3} \times 10^{-6} \mathrm{C} }
\mathrm{ Q_2=\frac{1000}{3} \mu \mathrm{c} }
 

Posted by

Irshad Anwar

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