Question

it is series LCR circuit is made by
taking R = 100ohm , L= 2\π pii henry and
capacitance (= 100 micro ferrad this series
combination is connected to an A.C
source of
220 volte, 50hartz calculate
the impidance of circuit and Peak
value of current​

Answer 1

Given :

• Resistance (R) = 100 Ω
• Inductance (L) = 2/ π H
• Vᵣₘₛ = 220 V
• Frequency = 50 Hz

To find :

• Impedance (z)
• Peak value of current (Iₒ)

Solution :

Step I : To find  Capacitive  reactance

⇒ X꜀ = 1 / ω C

⇒ X꜀ = 1 / 2 x π x f  x 100 x 10⁻⁶

⇒ X꜀ = 1 / 2  x 50 x 3.14  x 100 x 10⁻⁶

⇒ X꜀ = 31.8  Ω

Step II :To find Inductive reactance

⇒ Xₗ = ωL

⇒ Xₗ = 2 x π x f x L

⇒ Xₗ = 2 x π x 50 x 2 / π

⇒ Xₗ = 200 Ω

Step III : Impedance  

⇒ Z = √R² +(  Xₗ - X꜀)²

⇒ Z = √100² +(  Xₗ - X꜀)²

⇒ Z = √100² (  200- 31.8)²

⇒ Z = √100²+ (  168 )²

⇒ Z = √38224

⇒ Z = 195.5 Ω

Step IV :To find peak value of current (Iₒ)

we know that ,

⇒ I ₒ = Vₒ / Z

First we need to find the peak value of voltage ,

⇒ Vᵣₘₛ  = Vₒ/ √2

⇒ Vₒ = 220 x 1.414

⇒ Vₒ = 311 V

Now ,

⇒ I ₒ =  311 / 195.5

⇒ I ₒ = 1.59 A

Answer :

• The impedance of the circuit is 195.5 Ω.
• The peak value of current is 1.59 A .