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Question
A capacitor of unknown capacitance is connected across a battery of 'V' volt. The charge stored in it is 'Q' coulomb. When the potential across the capacitor is reduced by 'V' volt, the charge stored in it becomes Q' coulomb. The potential V is ______
Options
`(QV^')/((Q - Q^'))`
`(QV^')/((Q + Q^'))`
`((Q - Q^'))/(QV^')`
`((Q + Q^'))/(QV^')`
MCQ
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Solution
A capacitor of unknown capacitance is connected across a battery of 'V' volt. The charge stored in it is 'Q' coulomb. When the potential across the capacitor is reduced by 'V' volt, the charge stored in it becomes Q' coulomb. The potential V is `underline((QV^')/((Q - Q^')))`.
Explanation:
Q = CV
Q' = C(V - V')
∴ C = `Q^'/(V - V^')`
∴ Q = `Q^'/(V - V^') . V`
QV - QV' = Q'V
V(Q - Q') = QV'
V = `(QV^')/(Q - Q^')`
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