Resistor and Inductor in AC circuit

Basic behaviour of passive components in AC

The three passive components in electricity are resistor, inductor and capacitor. All these components impede (resist) current flow in an ac circuit. Resistor imposes resistance while inductor  and capacitor impose  reactance. Both these quantities are measured  in ohms. Just think of reactance as resistance in inductor and capacitor. 
The current and voltage behaviour of these components are quite interesting. In a pure resistive circuit, the current and voltage are always in phase with each other. While in a pure  inductive circuit, the applied voltage is ahead of current by 90°. 


In pure resistive circuit

In pure inductive circuit 

The applied voltage leads the current by 90°. The applied voltage comes from the AC power supply. 

There is also another type of voltage which produced inside inductor which is created whenever there is change in current and this is called  induced voltage. The induced voltage always opposes the applied( source) voltage. Thus it is  always 180° apart from the applied voltage.

Phasor diagram of V applied, Current I and V induced.

For understanding we can compare pure resistive and pure inductive in one phasor diagram like shown below,

What if we have an AC circuit that has both resistor and inductor in it? What will be the phase difference between the applied voltage and current in the circuit?

In a circuit that has both resistance and inductance, the source voltage and source current (total current) will differ by an angle between 0 and 90°. The angle depends on the ratio of the R and L value.

R-L Series in AC circuit

Lets analyze the circuit below and determine all the circuit parameters such as current, voltage drop across components and power.

Reactance by L1 :  XL = 2πf L = 2 x 3.142 x 50 x 0.01 = 3.142 Ω

Resistance by R1 = 5Ω

The total obstacles to current flow is called impedance ( Z) whereby Z is the vector addition of R and X.

Impedance diagram

In vector addition we cannot simply add the values directly unless there both quantities are in the same direction. In the R-L case we cannot just add the value of R and X even if they are in series.

\[\large Z= \sqrt{R^{2}+X^{2}}\]

Applying the formula in the analysis:

\[\large Z= \sqrt{5^{2}+3.142^{2}} = 5.90\]

I = V/Z

= 100/5.9 = 16.95A

Voltage drop across resistor

VR  = IR = 16.95 x 5 = 84.75V

Voltage drop across inductor

VL = IX = 16.95 x 3.142 = 53.26 V

When both voltages are added directly, the total becomes 84.75 + 53.26 = 138.0 V , which exceeds our supply voltage of 100V.

This is not possible. 

Again, remember that these quantities are not in the same phase. Vector addition must be done to find the total voltage or the supplied voltage and not normal addition. 

\[\large V_{T}= \sqrt{V_{R}^{2}+V_{X}^{2}}\]

\[\large V_{T}= \sqrt{84.75^{2}+ 53.26^{2}} = 100\]

Leading and Lagging self-test

From the waveform below, can you identify which waveform is in phase, which is leading and lagging??