SATHEE: Electro Magnetic Induction And Alternating Currents Question 141

Electro Magnetic Induction And Alternating Currents Question 141

Question: A telephone wire of length 200 km has a capacitance of 0.014 mF per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

Options:

A) 0.35 mH

B) 35 mH

C) 3.5 mH

D) Zero

Show Answer

Answer:

Correct Answer: A

Solution:

Capacitance of wire

$C = 0.014 \times 10^{- 6} \times 200 = 2.8 \times 10^{- 6} F = 2.8 μ F$

For impedance of the circuit to be minimum $X_{L} = X_{C}$

Þ $2 π ν L = \frac{1}{2 π ν C}$

Þ $L = \frac{1}{4 π^{2} ν^{2} C} = \frac{1}{4 {(3.14)}^{2} \times {(5 \times 10^{3})}^{2} \times 2.8 \times 10^{- 6}}$ $= 0.35 \times 10^{- 3} H = 0.35, m H$

Electro Magnetic Induction And Alternating Currents Question 141

Question: A telephone wire of length 200 km has a capacitance of 0.014 mF per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

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 141

Question: A telephone wire of length 200 km has a capacitance of 0.014 mF per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

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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