Question

A series L-C-R circuit with R = 44 ohm, C = 8 uF and
L = 50 H is connected to a variable frequency
220 V ac supply. Calculate angular frequency,
impedance and current at resonance condition.
​

Answer 1

Answer:

The angular frequency is 50 Hz

The circuit impedance 44 ohm

The current in circuit  5 Amp .

Explanation:

Given as :

For series L-C-R circuit

Resistance = R = 44 ohm

Capacitance = C = 8 $\mu$F

Inductance = L = 50 H

The supply voltage = V = 220 volt

Let The Angular frequency = ω Hz

Let The impedance = Z ohm

The Current = I Amp

According to question

At Resonance condition

Inductive impedance = Capacitive impedance

i.e $X_l = X_c$

Or, Lω = $\dfrac{1}{C\omega }$

Or, ω  = $\sqrt{\frac{1}{LC} }$

Or, ω  = $\dfrac{1}{\sqrt{50\times 8\times 10^{-6}}}$

Or, ω  = $\dfrac{1}{\sqrt{4\times 10^{-4}}}$

Or, ω  = $\dfrac{100}{2}$

i.e ω  = 50 Hz

So, The angular frequency =  ω  = 50 Hz

Now, At resonance condition , The circuit is purely resistive , So, At resonance condition .

Impedance = Resistance

i.e X = R

Or, X = 44 ohm

So, The circuit impedance = X = 44 ohm

At resonance , Current = I

i,e I = $\dfrac{voltage}{impedance}$

Or, I = $\dfrac{V}{X}$

Or, I = $\dfrac{220}{44}$

∴   I = 5 Amp

So, The current in circuit = I = 5 Amp

Hence,  The angular frequency is 50 Hz

The circuit impedance 44 ohm

The current in circuit  5 Amp . Answer