Question

Establish the expression for impedance of circuit when elements X, Y and Z are connected in series to an ac source. Show the variation of current in the circuit with the frequency of the applied ac source.

Answer 1

Circuit diagram

Phasor diagram

Now,

V_R = i₀R

V_L = i₀XL

V_C = i₀X_C

⇒ V² = (V_R)² + (V_L − V_C)2

V² = i²(R² + (X_L − X_C)²)

`i = V/sqrt(R^2 + (X_L - X_C)^2) = V/Z`

⇒ `Z = sqrt(R^2 + (X_L - X_C)^2`

When X_L = X_C

`omega_rL = 1/(omega_rC)`

⇒ `omega_r = 1/sqrt(LC)`

`2 piv_r = 1/sqrt(LC)`

`v_r = 1/(2pisqrt(LC))`

At this frequency v_r as X_L =  X_C

⇒ Z = R → minimum

The above connection indicates that at frequencies larger than or less than ω_r, the current values are smaller than the maximum value (I₀).