For a series LCR circuit, I vs ω curve is shown: (a) To the left of ωr, the circuit is mainly capacitive. (b) To the left of ωr, the circuit is mainly inductive. (c) At ωr, impedance of the circuit is equal to the resistance of the circuit. (d) At ωr, impedance of the circuit is 0 . Choose the most appropriate answer from the options given below :

Question

For a series LCR circuit, I vs ω curve is shown: (a) To the left of ωr, the circuit is mainly capacitive. (b) To the left of ωr, the circuit is mainly inductive. (c) At ωr, impedance of the circuit is equal to the resistance of the circuit. (d) At ωr, impedance of the circuit is 0 . <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/35b1d9e0af67a918debdd553ee074824-.png" /> Choose the most appropriate answer from the options given below : (A) (a) and (d) only (B) (b) and (d) only (C) (a) and (c) only (D) (b) and (c) only

Tardigrade Admin · Accepted Answer

at ω r , X C = X L ⇒ (1/ω r C )=ω r L So if ω < ω r then x C will increase and X L will decrease. Hence to left of ω r circuit is capacitive Z=√R2+(XC-XL)2 at ωr, Z=√R2+O2=R(C)

(a) and (d) only

(b) and (d) only

(a) and (c) only

(b) and (c) only

at ωr​,XC​=XL​ ⇒ωr​C1​=ωr​L So if ω<ωr​ then xC​ will increase and XL​ will decrease. Hence to left of ωr​ circuit is capacitive Z=R2+(XC​−XL​)2​ at ωr​,Z=R2+O2​=R(C)

at $ω_{r}, X_{C} = X_{L}$
$\Rightarrow \frac{1}{ω _{r} C} = ω_{r} L$
So if $ω < ω_{r}$ then $x_{C}$ will increase and $X_{L}$ will decrease.
Hence to left of $ω_{r}$ circuit is capacitive
$Z = R^{2} + (X_{C} - X_{L})^{2}$
at $ω_{r}, Z = R^{2} + O^{2} = R (C)$