Question

A seies LCR circuit with R= 25 ohm L =16H and C=25 is connected to. A variable frequency 225V. claculate angular frequency impedance and their resonance

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf \omega=50rad/s}}$

$\red{\boxed{\sf Impedance\:at\: Resonance (Z)=25 ohm}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• R = 25 ohm

• L = 16 H

• C = 25 $\sf{\mu}$ F

━━━━━━━━━━━━━━━━━━━━━━━━━

a) Angular Frequency :

$\large\underline\pink{\sf To\:Find: }$

• Angular Frequency ($\omega$) =?

$\large{♡}$$\large{\boxed{\sf \omega={\frac{1}{\sqrt{LC}}}}}$

$\large\implies{\sf {\frac{1}{\sqrt{16×25×{10}^{-6}}}}}$

$\large\implies{\sf {\frac{1}{\sqrt{400}}}}$

$\large\implies{\sf {\frac{1}{20×{10}^{-3}}}}$

$\huge\red{♡}$$\large\red{\boxed{\sf \omega=50rad/s}}$

________________________________________

b) Impedance at resonance

$\large\underline\pink{\sf To\:Find: }$

• Impedance at resonance (Z) = ?

$\large{♡}$$\large{\boxed{\sf Z={\sqrt{{R}^{2}+{(\omega L -{\frac{1}{\omega C}})}^{2}}}}}$

At resonance ,

$\large{\sf \omega L={\frac{1}{\omega C}}}$

Therefore ,

$\large{\boxed{\sf Z=\sqrt{{R}^{2}}}}$

$\large\implies{\sf Z=R}$

And R value is 25 ohm (given)

Therefore ,

$\huge\red{♡}$$\red{\boxed{\sf Impedance\:at\: Resonance (Z)=25ohm}}$