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i) Current lags voltage in R-L series circuit
ii) Current leads voltage in R-C series circuit

**i) ****Current
lags voltage in R-L series circuit**

**ii) ****Current
leads voltage in R-C series circuit**

__Solution:__

**Current lags voltage in R-L 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} = 2πfL

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

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

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

= I √ (R^{2} + X_{L}^{2})

V = IZ

**Z = √ (R ^{2}
+ X_{L}^{2})**

From
voltage triangle we can write,

Tan ф = V _{L} ∕V_{R}
= X _{L} ∕R Cos ф= V_{R}∕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∣ = √ (R^{2} + X_{L}^{2})

ф = tan ^{-1}[X_{L}∕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. 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 R-C
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 = 1∕2πfC

Apply
KVL we get,

V=V_{R}
+ V_{C}

V = IxR
+IxX_{C}

__Steps to draw phasor diagram:__

^{Ø }Take current as reference phasor.^{}

^{ }

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

^{ }

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

^{ }

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

Form
voltage triangle,

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

= I √ (R^{2} + Xc^{2})

V = IZ

Z = √(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 = √(R^{2} + Xc^{2})

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∣ = √ (R^{2} + Xc^{2})

ф = tan ^{-1}[-X_{c}∕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,

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

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