**What is RL Series Circuit?**

#### A circuit that contains a pure resistance R ohms connected in series with a coil having a pure inductance of L (Henry) is known as** RL Series Circuit**. When an AC supply voltage V is applied, the current, I flows in the circuit.

#### So, I_{R} and I_{L} will be the current flowing in the resistor and inductor respectively, but the amount of current flowing through both the elements will be same as they are connected in series with each other.

#### A RL series circuit having resistance R and inductance L is shown in the figure below.

RL SERIES CIRCUIT |

####
Let, **V** = RMS value of applied voltage.

** I **= RMS value of current

**Vr** = Voltage drop across resistance =** I*R**

**Vₗ** = Voltage drop across inductance = **I*Xₗ**

We know that current in the pure inductance circuit lags behind the voltage by 90 degree. *Therefore, Voltage drop across inductance, Vₗ is ahead of I by 90 degree. *

Phasor diagram of series RL circuit is shown in figure.

* Voltage drop across R is in phase with the current.

* Voltage drop across L leads current by 90.

PHASOR DIAGRAM OF RL SERIES CIRCUIT |

***Note that Applied Voltage leads the current in the RL series Circuit .**

Applied voltage is phasor sum of the voltage across resistance & voltage across inductance.

**V = √{(Vr)² + (Vₗ)²}**

**V = √{(I*R)² + (I*Xₗ)²}**

**V = I * √{(R)² + (Xₗ)²}**

**I = V / √{(R)² + (Xₗ)²}**

The quantity √{(R)² + (Xₗ)²} represents impedance,Z of the RL series circuit.

**Z = √{(R)² + (Xₗ)²}.**

__ Apparent power __-It is the product of the RMS value of applied voltage and RMS value of the current. Its unit is Volt Ampere (VA).

**Apparent power = V*I**

** True power** -It is the product of RMS applied voltage and the component of the current in phase with V. Its unit is in Watts (W).

**True power = V * (I*cosⲫ) ** *{I*cosⲫ is component of current in phase with V}*

__Power factor of RL Series Circuit__

__Power factor of RL Series Circuit__

#### Power factor is defined as the **cosine of the angle between voltage and current. **As seen from the phasor diagram, current is lagging behind the applied voltage by angle ⲫ.

So from phasor diagram,

**Power Factor = cosⲫ = Vr / V.**

**cosⲫ = I*R / I*√{(R)² + (Xₗ)²}.**

**cosⲫ = R / √{(R)² + (Xₗ)²}.**

**cosⲫ = R/Z = Resistance / Impedance.**

Hence,

**– Power factor = Resistance / Impedance**

– Power factor = True Power / Apparent Power