LR Circuit in series with an A.C source

When the inductance L and resistance R are connected in series to an alternating source of voltage, then the circuit is called an LR circuit or LR filter. As they are connected in series, they both will have the same amount of current flowing through them. 

Individual elements of an LR circuit

Let us quickly take a look over individual elements before jumping into combination circuits. 

• If only resistors are connected to an AC source, then it has current in phase with the potential which is represented by the formula I = Iosinwt and voltage is V = Vosinwt, hence the equation for current becomes: 

I = Vo/R

• If only inductors are connected to an AC source, then the current lags the potential by 900 which is represented by the formula I = Iosin(wt- π/2) and voltage is V = Vosin(wt), hence the equation for current becomes I = V0/XLwhere XL is the inductive reactance calculated by:

XL = 2πfL and its unit is ohm (Ω)

• If only capacitors are connected to an AC source, then the current leads the potential by 900 which is represented by the formula I = Iosin(wt+ π/2) and voltage is V = Vosin(wt), hence the equation for current becomes I = V0/XC  where XC is the capacitive reactance calculated by:

XC =1/ 2πfC and its unit is ohm (Ω)

LR Circuit

In a series LR circuit, the resistance, and inductance are connected in series to an AC source. 

Here, the current flowing through the circuit before entering the resistance is i=i0sinwt.

When the current enters the resistance, the potential difference across resistance VR and the potential difference across the inductor VL will be represented with time as 

VR (t) = (V0)R sinwt as current is in phase with the potential in resistance 

And 

VL(t) = (V0)L sin(wt+π/2) as in inductor the current lags the potential by 900

Where (V0)L is the peak value of potential in inductance and (V0)R is the peak value of potential in resistance.

Here, V0 can be calculated by: 

(V0)2 = (V0)R2 + (V0)L2

(V0) = √( (V0)R2 + (V0)L2)

As this is a series circuit, the current through the resistance and the inductance is the same. Hence

 i0 = iL = iR

(V0) = √(i02R2 + i02XL2)

(V0) = √(i02 (R2 + XL2))

(V0) = i0 √ (R2 + XL2)

Here √ (R2 + XL2) is a kind of resistance hence 

(V0) = i0Z

Here, Z is known as the Impedance of the circuit. 

Thus, Impedance is the collective resistance of Resistance R and Inductance L that opposes the flow of current in a series RL circuit. 

According to the phasor diagram, the current is still lagging the potential in the RL circuit, but not by 90° as that in a pure inductor circuit. It is somewhere between 0 and 90°. This angle is known as Φ.

The value of Φ can be found easily by phasor diagram:

tanΦ = XL / R

Power Factor in AC

Usually, the power P is calculated as,

P = VI

But, in an AC circuit, both voltage V and current I vary with time. Hence, we calculate the Instantaneous power P(t) as follows:

P(t) = V(t)I(t)

OR

P(t) = V0sinwt I0(sinwt + Ф)

The RL circuit is always a circuit where current is lagging the potential and the power factor is calculated using the following simple formula

cosФ = R/Z

LR circuit examples

Example 1:

For this circuit let us find the following:

1. Inductive Reactance XL

2. Impedance Z

3. Peak Current i0

4. i(t) 

Solution: 

Inductive Reactance XL

XL = wL = 100ㄫ x 2/ㄫ = 200 Ω

Impedance Z

Z = √ (R2 + XL2) = √((200)2 + (200)2 = 200√2 Ω

Peak Current i0

i0,= V0/Z = 200/200√2 = 1/√2 A

i(t) = i0sin(100ㄫt-Φ)

tanΦ = XL / R = 200/200 = 1

Φ = 450

Hence,

i(t) = 1/√2 sin(100ㄫt-ㄫ/4)

Example 2:

Consider the following circuit:

Find 

1. XL

2. Z

3. I0

4. i(t)

5. VR(t)

6. VL(t)

Solution:

1. XL = wL = 10 * 0.4 = 4Ω

2. Z =  √ (R2 + XL2) =  √ (42+32) = 5Ω

3. I0 = V0/Z = 20/5 = 4A

4. Let us first calculate the value of Φ for this one

tanΦ = XL / R = 4/3 

       Φ = 530 

     i(t) = i0sin(10t + 600 – Φ)

     i(t) = 4sin(10t +600 – 530)

     i(t) = 4sin(10t + 70)

VR(t) = (V0)R sin(10t + 70)

 (V0)R = i0R

     = 4 x 3

     = 12

VR(t) = 12 sin(10t + 70)

VL(t) = (V0)L sin(10t + 70 + 900)

(V0)L = i0XL 

       = 4 x 4 

   = 16 

VL(t) = 16 sin(10t + 970)

Conclusion

With the help of simple formulas and definitions, we can solve an LR circuit easily. The impedance of the LR circuit plays an important role in deciding the total lag of the current from the potential. Various aspects such as power factor, Ф, w work together for the efficient functioning of an LR filter.