SATHEE: Electro Magnetic Induction And Alternating Currents Question 49

Electro Magnetic Induction And Alternating Currents Question 49

Question: An LCR circuit contains R = 50 W , L = 1 mH and C = 0.1 mF. The impedance of the circuit will be minimum for a frequency of [Bihar MEE 1995]

Options:

A) $\frac{10^{5}}{2 π} s^{- 1}$

B) $\frac{10^{6}}{2 π} s^{- 1}$

C) $2 π \times 10^{5} s^{- 1}$

D) $2 π \times 10^{6} s^{- 1}$

Show Answer

Answer:

Correct Answer: A

Solution:

Impedance of LCR circuit will be minimum at resonant frequency so $ν_{0} = \frac{1}{2 π \sqrt{L C}}$ $= \frac{1}{2 π \sqrt{1 \times 10^{- 3} \times 0.1 \times 10^{- 6}}}$ $= \frac{10^{5}}{2 π} H z$

Electro Magnetic Induction And Alternating Currents Question 49

Question: An LCR circuit contains R = 50 W , L = 1 mH and C = 0.1 mF. The impedance of the circuit will be minimum for a frequency of [Bihar MEE 1995]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Electro Magnetic Induction And Alternating Currents Question 49

Question: An LCR circuit contains R = 50 W , L = 1 mH and C = 0.1 mF. The impedance of the circuit will be minimum for a frequency of [Bihar MEE 1995]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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