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NEET 2023
Exam
NEET
Year
2023
Subject
Physics
Chapter
Alternating Current

NEET 2023 · Physics · Alternating Current

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

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

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

Options

  1. A25Ω25Ω
  2. B102Ω10\sqrt{2}Ω
  3. C15Ω15Ω
  4. D55Ω5\sqrt{5}Ω

Correct answer

D. 55Ω5\sqrt{5}Ω

Step-by-step solution

The formula to calculate the inductive impedance of the circuit is given by

XL=\omegaL...(1){X}_{L} = \omegaL . . . \left(1\right)

The formula to calculate the capacitive impedance of the circuit is given by

XC=1\omegaC...(2){X}_{C} = \dfrac{1}{\omegaC} . . . \left(2\right)

The formula to calculate the total impedance of the circuit can be written as

{Z=R2+(Xc−XL)2=R2+(1\omegaC−\omegaL)2...(3)\begin{cases}Z & = & \sqrt{{R}^{2} + {\left({X}_{c} - {X}_{L}\right)}^{2}} \\ & = & \sqrt{{R}^{2} + {\left(\dfrac{1}{\omegaC} - \omegaL\right)}^{2}} . . . \left(3\right)\end{cases}

Substitute the values of the given parameters into equation (3) to calculate the required net impedance of the circuit.

{Z=(10Ω)2+(12π×50Hz×103π×10−6F−2π×50Hz×50π×10−3H)2=100+25Ω=125Ω=55Ω\begin{cases}Z & = & \sqrt{{\left(10Ω\right)}^{2} + {\left(\dfrac{1}{2\pi \times 50Hz \times \dfrac{{10}^{3}}{\pi} \times {10}^{- 6}F} - 2\pi \times 50Hz \times \dfrac{50}{\pi} \times {10}^{- 3}H\right)}^{2}} \\ & = & \sqrt{100 + 25}Ω \\ & = & \sqrt{125}Ω \\ & = & 5\sqrt{5}Ω\end{cases}

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