JEE Mains · Physics · STD 12 -7. Alternating current
A series \(AC\) circuit containing an inductor \((20\,mH),\) a capacitor \((120\,\mu F)\) and a resistor \((60\, \Omega )\) is driven by an \(AC\) source of \(24\,V/50\,Hz.\) The energy dissipated in the circuit in \(60\,s\) is
- A \(5.65\times 10^2\,J\)
- B \(2.26\times 10^3\,J\)
- C \(5.17\times 10^2\,J\)
- D \(3.39\times 10^3\,J\)
Answer & Solution
Correct Answer
(C) \(5.17\times 10^2\,J\)
Step-by-step Solution
Detailed explanation
\(E = Pt = \frac{{{E^2}}}{{{Z^2}}}Rt = \frac{{{{(24)}^2}}}{{{{60}^2} + {{(8.33\pi - 2\pi )}^2}}}(60)(60) = 518\,\,J\)
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