The net impedance of circuit (as shown in figure) will be
102 Ω10\sqrt{2} \text{ } \Omega102 Ω
15 Ω15 \text{ } \Omega15 Ω
55 Ω5\sqrt{5} \text{ } \Omega55 Ω
25 Ω25 \text{ } \Omega25 Ω
Given L=50π mH=50π×10−3 HL = \frac{50}{\pi} \text{ mH} = \frac{50}{\pi} \times 10^{-3} \text{ H}L=π50 mH=π50×10−3 H, C=103π μF=10−3π FC = \frac{10^3}{\pi} \text{ } \mu\text{F} = \frac{10^{-3}}{\pi} \text{ F}C=π103 μF=π10−3 F, R=10 ΩR = 10 \text{ } \OmegaR=10 Ω, f=50 Hzf = 50 \text{ Hz}f=50 Hz. Inductive reactance XL=2πfL=2π×50×50π×10−3=5 ΩX_L = 2\pi f L = 2\pi \times 50 \times \frac{50}{\pi} \times 10^{-3} = 5 \text{ } \OmegaXL=2πfL=2π×50×π50×10−3=5 Ω. Capacitive reactance XC=12πfC=12π×50×10−3π=10.1=10 ΩX_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{10^{-3}}{\pi}} = \frac{1}{0.1} = 10 \text{ } \OmegaXC=2πfC1=2π×50×π10−31=0.11=10 Ω. Impedance Z=R2+(XL−XC)2=102+(5−10)2=100+25=125=55 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + (5 - 10)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \text{ } \OmegaZ=R2+(XL−XC)2=102+(5−10)2=100+25=125=55 Ω.
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