JEE Mains · Physics · STD 12 -7. Alternating current
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\(.....s\)
- A \(418\)
- B \(245\)
- C \(348\)
- D \(365\)
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…
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