The net impedance of the circuit (as shown in figure) will be: ![image](https://pub-7faa453bd1ec4bf0a04cc031c6f058de.r2.dev/images/837e05f7-b484-47cb- - Physics MCQ NEET 2023 2026 | ExamTest.live | ExamTest.live

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The net impedance of the circuit (as shown in figure) will be:

A $102 Ω$
B $15 Ω$
C $55 Ω$
D $25 Ω$

Solution & Step-by-step Explanation

ShutterstockReactance of inductor X_{L} = 2 π f L = 2 π (50) (\frac{50}{π} \times 1 0^{- 3}) = 5 Ω .Reactance of capacitor X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π ( 50 ) ( \frac{1 0 ^{3}}{π} \times 1 0 ^{- 6} )} = \frac{1}{0.1} = 10 Ω .Impedance Z = R^{2} + (X_{C} - X_{L})^{2} : Z = 1 0^{2} + (10 - 5)^{2} = 100 + 25 = 125 = 55 Ω

Practice this question

Try it yourself before checking the explanation above.

The net impedance of the circuit (as shown in figure) will be:

A

102 Ω

B

15 Ω

C

55 Ω

D

25 Ω

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alternating current impedance

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