question_answer

A series AC circuit containing an inductor\[\left( 20\text{ }mH \right)\], a capacitor \[\left( 120\text{ }\mu F \right)\] and a resistor \[\left( 60\text{ }\Omega  \right)\] is driven by an AC source of\[24\text{ }V/50\text{ }Hz\]. The energy dissipated in the circuit in 60 s is: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) \[5.65\text{ }\times \text{ }{{10}^{2}}J\]                      

B) \[2.26\text{ }\times {{10}^{3}}J\]

C) \[5.17\text{ }\times \text{ }{{10}^{2}}\,J~\]   

D)               \[3.39\text{ }\times \text{ }{{10}^{3}}\,J\]

Correct Answer: C

Solution :

\[{{X}_{L}}=100\pi \,\,\times \,\,20\times {{10}^{-}}^{3}\] \[{{X}_{L}}=2\,\pi \,\,=\,\,6.28\,\Omega \] \[{{X}_{C}}=\frac{1}{100\pi \,\times \,120\,\times \,{{10}^{-6}}}\,\,=\,\,\frac{250}{3\,\pi }\] \[{{X}_{C}}=\frac{{{10}^{6}}}{12000\,\pi }\,\,=\,\,\frac{{{10}^{3}}}{12\,\pi }\] \[\left| {{X}_{L}}-{{X}_{C}} \right|=20.25\] \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\] Energy \[=\,\,\,{{I}^{2}}_{rms}R\,\,\times \,\,t\] \[=\text{ }5.17\times {{10}^{2}}\,J\]