JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 19)

A 750 Hz, 20 V (rms) source is connected to a resistance of 100 $$\Omega $$, an inductance of 0.1803 H and a capacitance of 10 $$\mu $$F all in series. The time in which the resistance (heat capacity 2 J/oC) will get heated by 10oC. (assume no loss of heat to the surroudnings) is close to :
348 s
418 s
245 s
365 s

Explanation

f = 750 Hz, Vrms = 20 V,

R = 100 $$\Omega $$, L = 0.1803 H,

C = 10$$\mu $$ F, S = 2 J/°C

|Z| = $$\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $$

= $$\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}} $$

= $$\sqrt {{R^2} + {{\left( {2\pi fL - {1 \over {2\pi fC}}} \right)}^2}} $$

= $$\sqrt {{{(100)}^2} + {{\left( {2 \times 3.14 \times 750 \times 0.1803 - {1 \over {2 \times 3.14 \times 750 \times {{10}^{ - 5}}}}} \right)}^2}} $$

= 834 $$\Omega $$

In AC, power (P) = irmsVrms cos $$\phi $$

and irms = $${{{V_{rms}}} \over {\left| Z \right|}}$$

Power factor (cos $$\phi $$) = $${R \over {\left| Z \right|}}$$

$$ \therefore $$ P = $${{{V_{rms}}} \over {\left| Z \right|}}.{V_{rms}}.{R \over {\left| Z \right|}}$$

= $${\left( {{{{V_{rms}}} \over {\left| Z \right|}}} \right)^2}R$$

= $${\left( {{{20} \over {834}}} \right)^2} \times 100$$

= 0.0575 J/S

Also, H = Pt = S$$\Delta $$$$\theta $$

$$ \Rightarrow $$ t = $${{2\left( {10} \right)} \over {0.0575}}$$ = 348 sec

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