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Question
An inductance of 2.0 H, a capacitance of 18μF and a resistance of 10 kΩ is connected to an AC source of 20 V with adjustable frequency.
(a) What frequency should be chosen to maximize the current in the circuit?
(b) What is the value of this maximum current?
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Solution
Given:
The inductance of inductor, L = 2.0 H
The capacitance of capacitor, C = 18 μF
The resistance of resistor, R = 10 kΩ
The voltage of AC source, E = 20 V
(a) In an LCR circuit, the current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
XL = XC
`omegaL^2 = 1/{omegaC}`
`rArr omega^2 = 1/{LC} = 1/(2xx18xx10^-6)`
`rArr omega^2 = 10^6/6`
`rArr omega = 10^3/6`
`rArr 2pif = 10^3/6 `
`rArr = f = 1000/(6xx2pi) = 26.539` Hz = 27 Hz
(b) At resonance, reactance is minimum.
Minimum Reactance, Z = R
Maximum current (I) is given by,
`"I" = "E"/"R"`
`rArr "I"=20/(10xx10)`
`rArr "I" = (2"A")/10^3 = 2 "mA"`
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