At resonance the peak value of current in L-C-R series circuit is.

Question

At resonance the peak value of current in L-C-R series circuit is. (A) E0/R (B) (E/√R2+(ω L - (1)ω C)2) (C) (E0/√2 √R2+(ω2 L - (1)ω2 C2)2) (D) (EO/√2R)

Tardigrade Admin · Accepted Answer

At resonance. ω L=(1/ω C) (1) So, C =( E 0/√ R 2+(ω L -(1)ω C )2) =( E 0/√ R 2+0) By (1) =( E 0/ R )

Q. At resonance the peak value of current in L-C-R series circuit is.

$E_{0} / R$

$\frac{E}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}$

$\frac{E _{0}}{2 R ^{2} + ( ω ^{2} L - \frac{1}{ω ^{2} C ^{2}} ) ^{2}}$

$\frac{E _{O}}{2 R}$

At resonance.
$ω L = \frac{1}{ω C} (1)$
So, $C = \frac{E _{0}}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}$
$= \frac{E _{0}}{R ^{2} + 0}$ By (1)
$= \frac{E _{0}}{R}$

Q. At resonance the peak value of current in L-C-R series circuit is.

E0​/R

R2+(ωL−ωC1​)2​E​

2​R2+(ω2L−ω2C21​)2​E0​​

2R​EO​​

At resonance. ωL=ωC1​(1) So, C=R2+(ωL−ωC1​)2​E0​​ =R2+0​E0​​ By (1) =RE0​​

$E_{0} / R$

$\frac{E}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}$

$\frac{E _{0}}{2 R ^{2} + ( ω ^{2} L - \frac{1}{ω ^{2} C ^{2}} ) ^{2}}$

$\frac{E _{O}}{2 R}$

At resonance.
$ω L = \frac{1}{ω C} (1)$
So, $C = \frac{E _{0}}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}$
$= \frac{E _{0}}{R ^{2} + 0}$ By (1)
$= \frac{E _{0}}{R}$