JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2020
- Hard
- Numerical
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 \(2 J /{ }^{\circ} C\) ) will get heated by \(10^{\circ} C\). (assume no \(105 s\) of heat to the surroundings) is close to\(.....s\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(348\)
Step-by-step Solution
Detailed explanation
\(f =750 Hz , V _{ rms }=2 OV\)
\(R =100 \Omega , L =0.1803 H\)
\(C =10_{\mu} F , S=2 J /{ }^{\circ} C\)
\(Z-\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}-\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}\)
\(=\sqrt{ R ^{2}+\left(2 \pi fL -\frac{1}{2 \pi fC }\right)^{2}}\)
Putting values
\(|Z|=834 \Omega\)
In \(AC\) power \(P = V _{ rms } i _{ rms }\) \(cos\phi\)
\(\operatorname{Cos} \phi=\frac{ R }{| Z |} \quad i _{ rms }=\frac{ V _{ rms }}{| Z |}\)
\(=\frac{V_{\max }^{2} R}{(|Z|)^{2}}\)
\(=\left(\frac{20}{834}\right)^{2} \times 100=0.0575 J / s\)
\(H = Pt = S \cdot \Delta\theta\)
\(t =\frac{2(10)}{0.0575}=348 sec\)
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