JEE Mains · Physics · STD 12 -7. Alternating current
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 ..........\(S\)

- A \(14\)
- B \(15\)
- C \(10\)
- D \(13\)
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…
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