Question

Derive an expression for the impedance of a series LCR circuit, when an AC voltage is applied to it.

Answer 1

Consider a resistance R, an inductor of self inductance L and a capacitor of capacitance C connected in series across an AC source. The applied voltage is given by, 

v = v₀sinωt ……(1)

Where, v is the instantaneous value, v is the peak value and ω = 2πf, f being the frequency of AC.

If i be the instantaneous current at time t, the instantaneous voltages across R, L and C are respectively iR, iX_L and iX_C. The vector sum of the voltage amplitudes across R, L, C equals the amplitude v₀ of the voltage applied.

Let v_R, v_L and V_C be the voltage amplitudes .across R, L and C respectively and I₀ the current amplitude. Then v_R = I₀R is in phase with i₀. 

v_L = i₀X_L = I₀(ωL) leads i₀ by 90° 

v_c = i₀X_c = i₀(\(\frac{1}{\omega C}\)) lags behind io by 90° 

The current in a pure resistor is phase with the voltage, whereas the current in a pure inductor lags the voltage by rad. The current in a pure capacitor leads the voltage by rad. 

For v_L > v_c, phase angle Φ between the voltage and the current is positive. 

From the right angled triangle OAP, 

OP² = OA² + AP² = OA² + OB² (Q AP = OB) 

v² = v²_R + (V_L – V_c)² = (iR)² + (iX_L – iX_c)² 

= i²(R² + (X_L – X_c)²)

Where Z is the impedance of the circuit. Phase angle between v & i.