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A series LCR circuit is connected to an ...

A series LCR circuit is connected to an a.c. source 220 V - 50 Hz as shown in fig. If the reading of the three voltemeters `V_(1), V_(2), V_(3)` are 65 V, 415 V and 204 V respectively. Calculate (i) current in the circuit (ii) value of inductor L (ii) value of capacitor C (iv) value of C for same L required to produce resonance.

Text Solution

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Here, `E_(v) = 220 V, v = 50 Hz`,
`R = 100 Omega`
`V_(R) = 65 V, V_(C) = 415` and `V_(L) = 204 V`
`I_(v) = (V_(R ))/(R ) = (64)/(100) = 0.65 A`
Now `X_(L) = (E_(v))/(I_(v)) = (204)/(0.65) = 313.85 Omega`
As `X_(L) = omega = omega L = 2 pi v L`
`L = (X_(L))/(2 pi v ) = (313.85)/(2 xx 3.14 xx 50) = 1.0 H`.
Again,`X_(C) = (E_(c ))/(I_(v)) = (415)/(0.65) = 638.46 ohm`
Form `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)`
`C = (1)/(2 pi v (X_(C ))) = (1)/(2 xx 3.14 xx 50 xx 638.46)`
`= 5 xx 10^(-6) F = 5 mu F`
Let C' be the capacitance of the capacitor required to produce resonance with `L = 1.0 H`.
From `v = (1)/(2 pi sqrt(LC))`
`C' = (1)/(4 pi^(2) v^(2) L) = (1)/(4 xx (3.14)^(2) xx 50^(2) xx 1.0)`
`= 10.1 xx 10^(-6) F = 10.1 mu F`
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