Verified Solution • 2025–26 Session moderate

QJEEM20Phcc9c: JEE Main 2020 Physics - Class 12 JEEM20Phcc9c Solution: A 750 Hz, 20 V (rms) source is connected to a resistance of 100 …

A 750 Hz, 20 V (rms) source is connected to a resistance of 100 Ω\Omega , an inductance of 0.1803 H and a capacitance of 10 μ\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 :

Multiple Choice Options

  • A 348 s
  • B 418 s
  • C 245 s
  • D 365 s

Step-by-Step Verified Solution

f = 750 Hz, Vrms = 20 V,

R = 100 Ω\Omega , L = 0.1803 H,

C = 10 μ\mu F, S = 2 J/°C

|Z| = R2+(XL−XC)2\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}

= R2+(ωL−1ωC)2\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}}

= R2+(2πfL−12πfC)2\sqrt {{R^2} + {{\left( {2\pi fL - {1 \over {2\pi fC}}} \right)}^2}}

= (100)2+(2×3.14×750×0.1803−12×3.14×750×10−5)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 Ω\Omega

In AC, power (P) = irmsVrms cos ϕ\phi

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

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

∴\therefore P = Vrms∣Z∣.Vrms.R∣Z∣{{{V_{rms}}} \over {\left| Z \right|}}.{V_{rms}}.{R \over {\left| Z \right|}}

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

= (20834)2×100{\left( {{{20} \over {834}}} \right)^2} \times 100

= 0.0575 J/S

Also, H = Pt = S Δ\Delta θ\theta

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

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