Question

Arrive at the expression for the impendance of a series LCR circuit using phasor diagram method and hence write the expression for the current through the circuit

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 = v₀ sin ωt ......(1)

where, v is the instantaneous value, v₀ 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 y I₀ the current amplitude. Then V_R = i₀R is (V_L – V_C) in phase with i₀. 

V_L = i₀X_L= i₀(ωL) leads i₀ by 90° 

The current in a pure resistor is phase, with the voltage, whereas the current in a pure 

V_C = i₀X_C = i₀(\(\frac{1}{\omega C}\)) lags behind i₀ by 90° 

The current in a pure resistor is phase with the voltage, whereas the current in a pure inductor lags the voltage by \(\frac{\pi}{2}\) rad. The current in a pure capacitor leads the voltage by \(\frac{\pi}{2}\) 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(QAP = OB)

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