JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 21)
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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Explanation
L = 100 $$\times$$ 10$$-$$3 H
C = 100 $$\times$$ 10$$-$$6 F
R = 120 $$\Omega$$
$$\omega$$L = 10 $$\Omega$$
$${1 \over {\omega C}} = {1 \over {{{10}^4} \times {{10}^{ - 6}}}} = 100\,\Omega $$
$$\Rightarrow$$ XC $$-$$ XL = 90 $$\Omega$$
$$ \Rightarrow Z = \sqrt {{{90}^2} + {{120}^2}} = 150\,\Omega $$
$$ \Rightarrow {I_{rms}} = {{20} \over {150}} = {2 \over {15}}A$$
For heat resistance by 16$$^\circ$$C heat required = 32 J
$$ \Rightarrow {\left( {{2 \over {15}}} \right)^2} \times (120) \times t = 32$$
$$t = {{32 \times 15 \times 15} \over {4 \times 120}} = 15$$
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