The frequencies at which the current amplitude in an LCR series circuit becomes 1 2 times its maximum value, are 212 rad s-1 and 232 rad-1. The value of resistance in the circuit is R = 5 Ω . The self inductance in the circuit is_________mH.

i = l 0 2 at ω 1 = 212 rad/s R = 5 Ω ω 2 = 232 rad/s Δ ω = ω 2 − ω 1 = 2 0 r a d / s P = P m a x 2 a t ω 1 a n d ω 2 So, Δ ω = ω 2 − ω 1 = R L ⇒ L = R Δ ω = 5 2 0 = . 2 5 H = 250 mH

The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt[]{2}}$ times its maximum value, are 212 rad s^-¹ and 232 rad^-¹. The value of resistance in the circuit is R = $5 Ω .$ The self inductance in the circuit is_________mH.

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Vishal Baghel | Contributor-Level 10

6 months ago

$\frac{_{}}{}$ $i = \frac{l_{0}}{\sqrt[]{2}}$ at $ω_{1}$ $_{}$ = 212 rad/s $R = 5 Ω$

$_{}$ $ω_{2}$ = 232 rad/s $Δ ω = ω_{2} - ω_{1} = 20 r a d / s$ $_{}_{}$

$P = \frac{P_{m a x}}{2} a t ω_{1} a n d ω_{2}$

So, $Δ ω = ω_{2} - ω_{1} = \frac{R}{L}$ $_{}_{} \frac{}{}$

$\Rightarrow L = \frac{R}{Δ ω} = \frac{5}{20} = . 25 H$

= 250 mH

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