SERIES L–R CIRCUIT

As we know potential difference across a resistance in ac is in phase with current and it leads in phase by 90° with current across the inductor.

 SERIES C–R CIRCUIT

Potential difference across a capacitor in ac lags in phase by 90° with the current in the circuit.
Suppose in phasor diagram current is taken along positive x-direction. Then VR is also along positive x-direction but VC is along negative y-direction. So we can write

V = V_R – jV_C = iR – j(iX_C)

Here, impedence is,   

The medulus of impedance is,

and the potential difference lags the current by an angle,

 SERIES L–C–R CIRCUIT
Potential difference across an inductor leads the current by 90° in phase while that across a capacitor, it lags in phase by 90°.

Suppose in a phasor diagram current is taken along positive x-direction. Then VR is along positive x-direction, Then VR is along positive x-direction, VL along positive y-direction and VC along negative y-direction.

So, we can write,        V = V_R + jV_L – jV_C = iR + j(iX_L) – j(iX_C)
                 = iR + j [ i (X_L – X_C)] = iZ
Here impedence is,   

The modulus of impedence is,

And the potential difference leads the current by an angle,

Illustration 13:    An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1Ωand inductance 0.01 H. What is the phase difference between the current and the emf in the circuit. Also find the virtual current in the circuit.

Solution:    In case of an ac, the voltage leads the current in phase by an angle,

Here,    X_L = ωL = (2∏fL) = (2∏) (50) (0.01) = ∏Ω
    and    R = 1Ω
  ∴Φ = tan^-1 (∏) ≈ 72.3°                        
    Further,

Substituting the values we have,

Illustration 14:    Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values 1000Ω, 1μF nd 2. henry respectively. Given emf as, V =   √100 sin 1000 t volt

Solution:     The rms value of voltage across the source,

The current will be same every where in the circuit, therefore,
    P.D across resistor V_R = i_rmsR = 0.0707 ∗1000 = 70.7 volt
    P.D across inductor V_L = i_rmsX_L = 0.0707 ∗1000 ∗ 2 = 141.4 volt and
    P.D across capacitor V_C = i_rmsX_C = 0.0707 ∗  = 70.7 volt
Note:
(i)    The rms voltages do not add directly as,
        V_R + V_L + V_C = 282.8 volt
Which is not the source voltage 100 volts. The reason is that these voltages are not in phase and can be added by vector or by phasor algebra.

CIRCUIT ELEMENTS WITH AC