In an LCR series circuit (of inductance L, capacitance C and resistance R), the impedance is minimum when the angular frequency of the source is given by

Question

In an LCR series circuit (of inductance L, capacitance C and resistance R), the impedance is minimum when the angular frequency of the source is given by (A) √LC (B) (1/√LC) (C) √(L/C) (D) √(C/L)

Tardigrade Admin · Accepted Answer

As at resonance capacitive reactance is equal to inductive reactance, So, ω L=(1/ω C) ⇒ ω2=(1/L C) ⇒ ω=(1/√L C) Where, ω= angular frequency of the source.

Q. In an LCR series circuit (of inductance L, capacitance C and resistance R), the impedance is minimum when the angular frequency of the source is given by

$L C$

$\frac{1}{L C}$

$\frac{L}{C}$

$\frac{C}{L}$

$L CR$

As at resonance capacitive reactance is equal to inductive reactance,
So, $ω L = \frac{1}{ω C}$
$\Rightarrow ω^{2} = \frac{1}{L C}$
$\Rightarrow ω = \frac{1}{L C}$
Where, $ω =$ angular frequency of the source.

Q. In an LCR series circuit (of inductance L, capacitance C and resistance R), the impedance is minimum when the angular frequency of the source is given by

LC​

LC​1​

CL​​

LC​​

LCR​