An AC source of 20,V (assuming standard frequ | vedclass

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(B) Given: $L = 100\,mH = 0.1\,H$,$C = 100\,\mu F = 10^{-4}\,F$,$R = 120\,\Omega$,$V = 20\,V$,$f = 50\,Hz$,Thermal capacity $C_{th} = 2\,J/^{\circ}C$,

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Given: $L = 100\,mH = 0.1\,H$,$C = 100\,\mu F = 10^{-4}\,F$,$R = 120\,\Omega$,$V = 20\,V$,$f = 50\,Hz$,Thermal capacity $C_{th} = 2\,J/^{\circ}C$,$\Delta T = 16^{\circ}C$.$1$. Calculate inductive reactance: $X_L = 2\pi fL = 2 	imes 3.14 	imes 50 	imes 0.1 = 31.4\,\Omega$.$2$. Calculate capacitive reactance: $X_C = \frac{1}{2\pi fC} = \frac{1}{2 	imes 3.14 	imes 50 	imes 10^{-4}} \approx 31.8\,\Omega$.*Note: Using standard angular frequency $\omega = 100\,rad/s$ often used in such problems for simplicity:*$X_L = \omega L = 100 	imes 0.1 = 10\,\Omega$.$X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 10^{-4}} = 100\,\Omega$.$3$. Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{120^2 + (10 - 100)^2} = \sqrt{14400 + 8100} = \sqrt{22500} = 150\,\Omega$.$4$. Current $I_{rms} = \frac{V}{Z} = \frac{20}{150} = \frac{2}{15}\,A$.$5$. Heat produced $H = I_{rms}^2 R \Delta t = C_{th} \Delta T$.$(\frac{2}{15})^2 	imes 120 	imes \Delta t = 2 	imes 16$.$\frac{4}{225} 	imes 120 	imes \Delta t = 32$.$\frac{480}{225} 	imes \Delta t = 32 \Rightarrow 2.133 	imes \Delta t = 32 \Rightarrow \Delta t = 15\,s$.

List-$I$	List-$II$
$A$. $AC$ generator	$I$. Presence of both $L$ and $C$
$B$. Transformer	$II$. Electromagnetic Induction
$C$. Resonance phenomenon to occur	$III$. Quality factor
$D$. Sharpness of resonance	$IV$. Mutual Inductance

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