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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2022
  • Hard
  • Numerical

An \(AC\) source is connected to an inductance of \(100\,mH\) a capacitance of \(100\,\mu F\) and a resistance of \(120\,\Omega\) as shown in figure. The time in which the resistance having a thermal capacity \(2\,J^{\circ } C\) will get heated by \(16^{\circ} C\) is ..........\(S\)

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Answer & Solution

Correct Answer

(B) \(15\)

Step-by-step Solution

Detailed explanation

\(\left|\left( X _{ L }- X _{ C }\right)\right|=\left|10-10^{2}\right|=90\,\Omega\)

\(Z =\) Impendance

\(=\sqrt{\left( X _{ L }- X _{ C }\right)^{2}+ R ^{2}}=\sqrt{(90)^{2}+(20)^{2}}=150\,\Omega\)

\(i _{ ms }=\frac{ V _{\text {mas }}}{ z }=\left(\frac{2}{15}\right) A\)

Now \(i _{ mms }^{2} R \Delta t = ms (\Delta T )\)

\(\Rightarrow \Delta t =15\,sec\)

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