Current lags and leads voltage in RL series circuit
Table of Contents
Current lags and leads voltage in RL series circuit
i) Current lags voltage in RL series circuit
ii) 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 VR = IR
Drop across pure inductance VL = IXL Where XL = 2πfL
I = rms value of current drawn VR, VL = 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 VR will be along I phase in case of resistance
3. I lags voltage by 900. But I is reference, VL must be shown leading w.r.t I by 900 in case of inductance
4. Supply voltage = Vector sum of 2 vectors VL and VR obtained by law of parallelogram
V = √ (VR2) + (VL2) = √ ((IR) 2 + (IXL) 2)
= I √ (R2 + XL2)
V = IZ
Z = √ (R2 + XL2)
From voltage triangle we can write,
Tan ф = V L ∕VR = X L ∕R Cos ф= VR∕V = R∕Z
Sin ф = V L ∕V =X L ∕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 ф X component of Impedance = R
X L = Z sin ф Y Component of Impedance = X L
In rectangular form the impedance is denoted as,
Z = R + j X L
While in polar form
Z = ∣Z∣∠ф Ω
Where,
∣Z∣ = √ (R2 + XL2)
ф = tan -1[XL∕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:
The expression for the current in the series R-L circuit is, i = Im Sin (wt-ф) as current lags voltage
Power = V x i
= Vm Sin (wt) x Im Sin (wt-ф)
= VmIm[Sin(wt).Sin(wt-ф)]
= VmIm [ (Cosф – Cos (2wt-ф))∕2]
= (VmIm ∕2) x Cos ф – (VmIm ∕2) x Cos (2wt-ф)
Second term is cosine term whose average value over a cycle is zero.
Average power Pavg = (VmIm∕2) x Cos ф
= (Vm∕√2) x (Im∕√2) x Cos ф
P = VI Cos ф watt
The three side of triangle is,
10.VI
11.VI Cos ф
12.VI Sin ф
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 ‘S’.
S = VI unit is VA VA – 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 ф 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 ф VAR
Where VAR – 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 VR = IR
6. Drop across pure capacitance Vc = IXc
Where, Xc = 1∕2πfC
Apply KVL we get,
V=VR + VC
V = IxR +IxXC
Steps to draw phasor diagram:
Ø Take current as reference phasor.
Ø In case of resistance, voltage and current are in phase, so, VR will along current phasor.
Ø In case of capacitance, current leads voltage 900 i.e. voltage lags by 900. so, Vc is shown downwards (i.e.) lagging current by 900
Ø The supply voltage being vector sum of these two voltages Vc and VR obtained by completing parallelogram
Form voltage triangle,
V = √ ((VR2) + (Vc2)) = √(( IR2) + (IXc2))
= I √ (R2 + Xc2)
V = IZ
Z = √(R2 + Xc2) is the impedance of circuit
Impedance:
Impedance is nothing but opposition of flow of alternating current. It is measured in ohms given by,
Z = √(R2 + Xc2)
Where, Xc = 1∕1∕2πfC Ω 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 ф The Y component is Xc = Z Sin ф
But direction of Xc is –Ve direction
Z = R- j Xc Ω – Rectangular form Z = ∣Z∣∠-ф Ω – Polar form
Where,
∣Z∣ = √ (R2 + Xc2)
ф = tan -1[-Xc∕R]
Power and power triangle:
The I leads the V by angle ф hence, i = Im Sin (wt + ф)
Power = V x i
= Vm Sin (wt) x Im Sin (wt + ф)
= VmIm[Sin(wt).Sin(wt + ф)]
= VmIm [ (Cos(-ф) – Cos (2wt + ф))∕2]
= (VmIm ∕2) x Cos ф – (VmIm ∕2) x Cos (2wt + ф)
Second term is cosine term whose average value over a cycle is zero.
Average power Pavg = (VmIm∕2) x Cos ф
= (Vm∕√2) x (Im∕√2) x Cos ф P = VI Cos ф 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 = ∣Z∣∠ф Ω ; ф is +Ve for Inductive Z
P = VI Cos ф ;Cos ф is lagging for Inductive circuit
Z = R – j XL = Z = ∣Z∣∠-ф Ω ; ф is -Ve for Capacitive Z
P = VI Cos ф ;Cos ф is leading for Capacitive circuit