A condenser having a capacity of 6,mu F is charged to 100, V and is then joined to an uncharged condenser of 14,mu F and

Q: A condenser having a capacity of $6\,\mu F$ is charged to $100\, V$ and is then joined to an uncharged condenser of $14\,\mu F$ and then removed. The ratio of the charges on $6\,\mu F$ and $14\,\mu F$ and the potential of $6\,\mu F$ will be

c (c) Let ${q_1},\,{q_2}$ be the charges on two condensers $V = \frac{{{q_1}}}{6} = \frac{{{q_2}}}{{14}}$ $==>$ $\frac{{{q_1}}}{{{q_2}}} = \frac{6}{{14}} = \frac{3}{7}$ Also ${q_1} + {q_2} = 600$ $==>$ ${q_1} + \frac{{14}}{6}{q_1} = 600$ $==>$ ${q_1} = \frac{{600}}{{20}} \times 6$ $V = \frac{{{q_1}}}{6} = \frac{{600}}{{20}} = 30\,volt$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A condenser having a capacity of $6\,\mu F$ is charged to $100\, V$ and is then joined to an uncharged condenser of $14\,\mu F$ and then removed. The ratio of the charges on $6\,\mu F$ and $14\,\mu F$ and the potential of $6\,\mu F$ will be

A$\frac{6}{{14}}$ and $50\;volt$
B$\frac{{14}}{6}$ and $30\;volt$
C$\frac{6}{{14}}$ and $30\;volt$
D$\frac{{14}}{6}$ and $0$ $volt$

Medium

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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