## Current lags and leads voltage in RL series circuit

Table of Contents

**Current lags and leads voltage in RL series circuit**

**i)Â Â Current lags voltage in RL series circuit**

**ii)Â Current leads voltage in RC series circuit**

__Solution:__

**Current lags voltage in RL series circuit**

Consider a circuit consisting of pure Resistance R ohms connected in series with Inductance L henries as shown in fig.

The series combination is connected across ac supply is given by V = Vm Sinwt

The voltage drops in the circuit are,

Drop across pure resistance V_{R} = IR

Drop across pure inductance V_{L} = IX_{L} Where X_{L} = 2Ï€fL

I = rms value of current drawn V_{R}, V_{L} = rms value of pure inductance

By applying KVL,

V = IR + IX _{L}

**Steps to drawn phasor diagram:**

1. I as reference phasor

2. R, V, I are in phase, So V_{R} will be along I phase in case of resistance

3.Â I lags voltage by 90^{0}. But I is reference, V_{L}Â must be shown leading w.r.t I by 90^{0}Â in case of inductance

4. Supply voltage = Vector sum of 2 vectors V_{L} and V_{R} obtained by law of parallelogram

V = âˆš (V_{R}^{2}) + (V_{L}^{2}) = âˆš ((IR) ^{2} + (IX_{L}) ^{2})

= I âˆš (R^{2} + X_{L}^{2})

V = IZ

**Z = âˆš (R ^{2} + X_{L}^{2})**

From voltage triangle we can write,

Tan Ñ„ = V _{L} âˆ•V_{R} = X _{L} âˆ•R Cos Ñ„= V_{R}âˆ•V = Râˆ•Z

Sin Ñ„ = V _{L} âˆ•V =X _{L} âˆ•Z

##### If all the sides of voltage are divided by current, we get triangle called impedance triangle.

Side or triangles are,

1- Resistance R

2- Inductive reactance X _{L}

3- Impedance Z

From this impedance triangle

R = Z Cos Ñ„ X component of Impedance = R

X _{L} = Z sin Ñ„ Y Component of Impedance = X _{L}

In rectangular form the impedance is denoted as,

Z = R + j X _{L}

While in polar form

Z = âˆ£Zâˆ£âˆ Ñ„ Î©

Where,

âˆ£Zâˆ£ = âˆš (R^{2} + X_{L}^{2})

Ñ„ = tan ^{-1}[X_{L}âˆ•R]

**Impedance:**

Impedance is defined as opposition of circuit to flow alternating current. It is denoted by Z and the unit is ohms.

__Power and power triangle__:

__Power and power triangle__

The expression for the current in the series R-L circuit is, i = Im Sin (wt-Ñ„) as current lags voltage

Power = V x i

= Vm Sin (wt) x Im Sin (wt-Ñ„)

= VmIm[Sin(wt).Sin(wt-Ñ„)]

= VmIm [ (CosÑ„ â€“ Cos (2wt-Ñ„))âˆ•2]

= (VmIm âˆ•2) x Cos Ñ„ – (VmIm âˆ•2) x Cos (2wt-Ñ„)

Second term is cosine term whose average value over a cycle is zero.

Average power Pavg = (VmImâˆ•2) x Cos Ñ„

= (Vmâˆ•âˆš2) x (Imâˆ•âˆš2) x Cos Ñ„

P = VI Cos Ñ„ watt

The three side of triangle is,

10.VI

11.VI Cos Ñ„

12.VI Sin Ñ„

These three terms can be defined as below,

**1. Apparent power(S):**

It is defined as the product of rms value of boltage (V) and current (I). it is denoted as â€˜Sâ€™.

S = VI unit is VA VA â€“ Voltage ampere

**2. Real or true power (P):**

It is defined as the product of applied voltage and active component of the circuit. And It is real component of the apparent power. It is measured in watts (W) or kilowatts (KW)

P = VI Cos Ñ„ watts

**3. Reactive power (Q):**

It is defined as product of applied voltage and reactive component of current. It is also defined as the imaginary of apparent power is represented by Q. Unit is VAR

Q = VI Sin Ñ„ VAR

Where VAR â€“ Volt ampere reactive.

**I lead V in RC series circuit:**

Consider a circuit in which resistance R ohms and capacitance C farads is connected across ac supply is given by,

V = Vm Sin wt

Circuit draws a current I, then there are two voltage drops,

5. Drop across pure resistance V_{R} = IR

6. Drop across pure capacitance Vc = IXc

Where, Xc = 1âˆ•2Ï€fC

Apply KVL we get,

V=V_{R} + V_{C}

V = IxR +IxX_{C}

__Steps to draw phasor diagram:__

__Steps to draw phasor diagram:__

^{Ã˜ }Take current as reference phasor.^{}

^{Ã˜ }In case of resistance, voltage and current are in phase, so, V_{R} will along current phasor.^{}

^{Ã˜Â Â Â Â }In case of capacitance, current leads voltage 90^{0}Â i.e. voltage lags by 90^{0}. so, Vc is shown downwards (i.e.) lagging current by 90^{0}

^{Ã˜ }The supply voltage being vector sum of these two voltages Vc and V_{R} obtained by completing parallelogram^{}

Form voltage triangle,

V = âˆš ((V_{R}^{2}) + (Vc^{2})) = âˆš(( I_{R}^{2}) + (IXc^{2}))

= I âˆš (R^{2} + Xc^{2})

V = IZ

Z = âˆš(R^{2} + Xc^{2}) is the impedance of circuit

**Impedance:**

Impedance is nothing but opposition of flow of alternating current. It is measured in ohms given by,

Z = âˆš(R^{2} + Xc^{2})

Where, Xc = 1âˆ•1âˆ•2Ï€fC Î© called capacitive reactance.

From voltage triangle if all sides of voltage triangle are divided by current, we get impedance triangle.

The sides of triangle are R, Xc, Z

The X component is R = Z Cos Ñ„ The Y component is Xc = Z Sin Ñ„

But direction of Xc is â€“Ve direction

Z = R- j Xc Î© – Rectangular form Z = âˆ£Zâˆ£âˆ -Ñ„ Î© – Polar form

Where,

âˆ£Zâˆ£ = âˆš (R^{2} + Xc^{2})

Ñ„ = tan ^{-1}[-X_{c}âˆ•R]

**Power and power triangle:**

The I leads the V by angle Ñ„ hence, i = Im Sin (wt + Ñ„)

Power = V x i

= Vm Sin (wt) x Im Sin (wt + Ñ„)

= VmIm[Sin(wt).Sin(wt + Ñ„)]

= VmIm [ (Cos(-Ñ„) â€“ Cos (2wt + Ñ„))âˆ•2]

= (VmIm âˆ•2) x Cos Ñ„ – (VmIm âˆ•2) x Cos (2wt + Ñ„)

Second term is cosine term whose average value over a cycle is zero.

Average power Pavg = (VmImâˆ•2) x Cos Ñ„

= (Vmâˆ•âˆš2) x (Imâˆ•âˆš2) x Cos Ñ„ P = VI Cos Ñ„ watts

If we multiply voltage equqtion by current I we get power equation, (RL series circuit)

1. Apparent power (S)

2. Real or true power (P)

3. Reactive power (Q)

Note:

Z = R + j XL = Z = âˆ£Zâˆ£âˆ Ñ„ Î© ; Ñ„ is +Ve for Inductive Z

P = VI Cos Ñ„ ;Cos Ñ„ is lagging for Inductive circuit

Z = R – j XL = Z = âˆ£Zâˆ£âˆ -Ñ„ Î© ; Ñ„ is -Ve for Capacitive Z

P = VI Cos Ñ„ ;Cos Ñ„ is leading for Capacitive circuit