QNEET-2023-Ph-246: NEET 2023 Physics - Class 12 NEET-2023-Ph-246 Solution: The net impedance of circuit (as shown in figure) will be: \(15 …

To find the net impedance, Z, of the circuit, we need to calculate the inductive reactance (XL), the capacitive reactance (XC), and use the resistance (R) in the circuit. Since R is given as 10 � $\Omega$ � , we calculate the reactances as follows: Inductive reactance, X L = 2 � f L $X_L = 2\pi fL$ X L _ = 2 � f L , where f $f$ f is the frequency and L $L$ L is the inductance. With f = 50 $f = 50$ f = 50 Hz and L $L$ L given as 50 � $\frac{50}{\pi}$ � 50 _ mH, or 50 _ 10 _ 3 � $\frac{50 \times 10^{-3}}{\pi}$ � 50 _ 1 0 _ 3 _ H, we find X L = 2 � _ 50 _ 50 _ 10 _ 3 � = 5 $X_L = 2 \pi \times 50 \times \frac{50 \times 10^{-3}}{\pi} = 5$ X L _ = 2 � _ 50 _ � 50 _ 1 0 _ 3 _ = 5 � $\Omega$ � . Capacitive reactance, X C = 1 2 � f C $X_C = \frac{1}{2\pi fC}$ X C _ = 2 � f C 1 _ , where C $C$ C is the capacitance. Given C = 10 3 � $C = \frac{10^3}{\pi}$ C = � 1 0 3 _ _F, or 10 _ 3 � $\frac{10^{-3}}{\pi}$ � 1 0 _ 3 _ F, results in X C = 1 2 � _ 50 _ 10 _ 3 � = 10 � $X_C = \frac{1}{2 \pi \times 50 \times \frac{10^{-3}}{\pi}} = 10 \Omega$ X C _ = 2 � _ 50 _ � 1 0 _ 3 _ 1 _ = 10� . With these values, the net impedance Z $Z$ Z of the circuit can be calculated using the formula: Z = R 2 + ( X L _ X C ) 2 $Z = \sqrt{R^2 + (X_L - X_C)^2}$ Z = R 2 + ( X L _ _ X C _ ) 2 _ Substituting R = 10 � $R = 10 \Omega$ R = 10� , X L = 5 � $X_L = 5 \Omega$ X L _ = 5� , and X C = 10 � $X_C = 10 \Omega$ X C _ = 10� , we find: Z = 10 2 + ( 5 _ 10 ) 2 = 100 + 25 = 125 = 5 5 � $Z = \sqrt{10^2 + (5 - 10)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \Omega$ Z = 1 0 2 + ( 5 _ 10 ) 2 _ = 100 + 25 _ = 125 _ = 5 5 _ � Therefore, the correct option is 5 5 � $5\sqrt{5} \Omega$ 5 5 _ � .