An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ____mH.(Take π2=10)
Question
An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ____mH.(Take π2=10)
Solution
In an LCR series circuit, the maximum amplitude of current is achieved when the circuit is in resonance. At resonance, the inductive reactance (XL) equals the capacitive reactance (XC).
The formula for inductive reactance is XL = 2πfL, where f is the frequency and L is the inductance.
The formula for capacitive reactance is XC = 1/(2πfC), where C is the capacitance.
Setting these two equal to each other gives us 2πfL = 1/(2πfC).
Solving for L gives us L = 1/(4π^2f^2C).
Substituting the given values into this equation gives us L = 1/(410(2000)^262.510^-9).
Solving this gives us L = 0.02 H or 20 mH. So, the value of inductance for maximum amplitude of current in the circuit is 20 mH.
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