JEE MAIN - Physics (2015 (Offline) - No. 16)
In the given circuit, charges $${Q_2}$$ on the $$2\mu F$$ capacitor changes as $$C$$ is varied from $$1\,\mu F$$ to $$3\mu F.$$ $${Q_2}$$ as a function of $$'C'$$ is given properly by:
$$\left( {figures\,\,are\,\,drawn\,\,schematically\,\,and\,\,are\,\,not\,\,to\,\,scale} \right)$$
_en_16_1.png)
$$\left( {figures\,\,are\,\,drawn\,\,schematically\,\,and\,\,are\,\,not\,\,to\,\,scale} \right)$$
_en_16_2.png)
_en_16_3.png)
_en_16_4.png)
_en_16_5.png)
Explanation
_en_16_6.png)
From figure, $${Q_2} = {2 \over {2 + 1}}Q = {2 \over 3}Q$$
$$Q = E\left( {{{C \times 3} \over {C + 3}}} \right)$$
$$\therefore$$ $${Q_2} = {2 \over 3}\left( {{{3CE} \over {C + 3}}} \right) = {{2CE} \over {C + 3}}$$
Therefore graph $$d$$ correctly dipicts.
_en_16_7.png)
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