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Ahmedabad Metro – Motera Stadium to GNLU Via Vishwakarma College Route & Time Table

noreply@blogger.com (R D PANCHAL) — Fri, 1 Nov 2024 10:00:00 +0530

Motera Stadium to GNLU Via Vishwakarma College Route

The Ahmedabad metro has started its route from Motera Stadium to GNLU Via Vishwakarma College very soon. The construction of this route is continue and start very soon. The route of the Motera Stadium to GNLU Via Vishwakarma College is already decided.

Motera Stadium to GNLU Via Vishwakarma College Time Table

The time table for Motera Stadium to GNLU Via Vishwakarma College is not decided.

Ahmedabad Metro official site

Prohibited Luggage in Ahmedabad Metro

noreply@blogger.com (R D PANCHAL) — Fri, 1 Nov 2024 09:36:00 +0530

Ahmedabad Metro Do’s, Don’t & Prohibited items.

Ahmedabad metro don’t allowed some luggage. Some Do’s and Don’t condition in the Ahmedabad Metro. Ahmedabad metro don’t allow inflammable solids, manure of any kind, rags, including oily rags, acids and other corrosives, blood, dried, corpses, explosive substances etc.

Ahmedabad metro Do’s

If you hear or see unusual, please inform metro staff
Stand in queqe
Let passengers exit from the train before boarding
Never squat on the train floor
Limit your baggage to 25 kgs
Avoid reading newspaper
Abstain from drunken and unruly behaviour
Watch gap between the train and platform edge
It is illegal to beg, solicit products or conduct group activities in the train
Support activities of station staff
If you lose something, contact GMRC lost and found cell

Ahmedabad metro Don’t

Don’t split or throw garbage within metro premises
Don’t eat or drink inside trains and area of stations
Don’t bring pets inside metro
Don’t smoke inside metro
Don’t travel without tickets
Don’t jump over ticket barrier
Don’t step onto metro track, 3^rd rail is charged with 750 V DC
Don’t open door forcibly
Don’t put posters inside metro
Don’t run or play games on the escalators
Don’t lean or put your hands on the train doors
Don’t rush on platforms

Prohibited Luggage in Ahmedabad Metro

Petroleum and inflammable liquids
Inflammable solids
Manure of any kind
Rags, including oily rags
Oxidising substances
Poisonous substances
Acids and its corrosives
Radioactive substances
Blood, dried
Corpses
Bones, excluding bleached and cleaned bones
Gases, compressed, liquefied or dissolved under pressure
Decayed animal
Human ashes
Parts of human body
Human skeleton
Any other offensive material

Ahmedabad Metro official website

Ahmedabad Metro helpline Number, Customer care

noreply@blogger.com (R D PANCHAL) — Fri, 1 Nov 2024 09:00:00 +0530

Ahmedabad metro customer care

In this blog, Ahmedabad metro helpline number and customer care number is given. The GMRC has developed a mechanism to resolve passenger’s grievance. Passenger may address their Grievances / Query / Suggestion / Appreciation / Feedback to centralized customer care

Ahmedabad metro customer care: 079 – 2296123

Ahmedabad metro general correspondence: 079 – 23248572

Ahmedabad metro Email address: care@gujaratmetrorail.com

Ahmedabad metro official website

Ahmedabad Metro – Sector 1 Gandhinagar to Mahatma Mandir Route & Time Table

noreply@blogger.com (R D PANCHAL) — Fri, 1 Nov 2024 08:17:00 +0530

Sector 1 Gandhinagar to Mahatma Mandir Route

The Ahmedabad metro has started its route from Sector 1 Gandhinagar to Mahatma Mandir. This construction of this route is continue and start very soon. The route of the sector 1 gandhinagar to Mahatma mandir is already decided.

Sector 1 Gandhinagar to Mahatma Mandir Route

Sector 1 Gandhinagar to Mahatma Mandir route takes 6 station between them.

Ahmedabad Metro Official Website

Ahmedabad Metro – Sector 1 Gandhinagar to Motera Stadium Route & Time Table

noreply@blogger.com (R D PANCHAL) — Thu, 31 Oct 2024 08:41:00 +0530

Sector 1 Gandhinagar to Motera Stadium Time Table

The Ahmedabad metro has started its route from Sector 1 Gandhinagar to Motera Stadium. The first train start from 7.20 am and last train at 6.40 pm. There is intermediate change of GNLU if anyone wants to travel for gift city.

Sector 1 Gandhinagar to Motera Stadium
Sector 1 Gandhinagar	GNLU	Motera Stadium
7.20 am	7.36	7.55
8.40	8.56	9.15
10.00	10.16	10.35
11.30	11.46	12.05
12.50 pm	13.06	13.25
14.30	14.46	15.05
16.00	16.16	16.35
17.20	17.36	17.55
18.40	18.56	19.20

Sector 1 Gandhinagar to Motera Stadium Route

Sector 1 Gandhinagar to Motera Stadium route takes 5 station between them and almost takes 35 minutes for travelling.

Ahmedabad Metro official Site

Ahmedabad Metro – Motera Stadium to Sector 1 Gandhinagar Route & Time Table

noreply@blogger.com (R D PANCHAL) — Wed, 30 Oct 2024 18:10:00 +0530

Motera Stadium to Sector 1 Gandhinagar Time Table

The Ahmedabad metro has started its route from motera to Gandhinagar. The first train start from 8.00 am and last train at 6.00 pm. There is intermediate change of GNLU if anyone wants to travel for gift city.

Motera Stadium to Sector 1 Metro Station
Motera Stadium	GNLU	Sector 1 Gandhinagar
8.00 am	8.17	8.35
9.20	9.37	9.55
10.40	10.57	11.15
12.10 pm	12.27	12.45
13.30	13.47	14.05
15.20	15.37	15.55
16.40	16.57	17.15
18.00	18.17	18.35

Motera Stadium, Ahmedabad to Sector 1, Gandhinagar Route

The motera stadium to sector 1 route takes 5 station between them and almost takes 35 minutes for travelling.

How to take ticket of Ahmedabad metro through Vending Machine

noreply@blogger.com (R D PANCHAL) — Wed, 30 Oct 2024 17:44:00 +0530

Ahmedabad metro ticket buy through vending machine

The Ahmedabad metro has facility for take ticket for any route through ticket window or vending machine. It is easy to use vending machine. The following steps full explain for purchase of ticket through vending machine.

Steps for purchase tickets for Ahmedabad metro

Select language
Select place
Click next
Select number of tickets
Scan QR code
Payment done
Take print out of ticket

Government of Gujarat Tighten Rules for Probation Staffs

noreply@blogger.com (R D PANCHAL) — Wed, 30 Oct 2024 13:47:00 +0530

Government of Gujarat Probation Staff

The government of Gujarat tighten rules for newly recruited probational staffs. According to government of Gujarat, Administration department Paripatra : GAD/PRP/e file/1/2024/3607/G2, Date: 25/10/204 rules are given to extent probation period or end of seva or job. According to GR, some guidelines are provided on yearly progress report during probation period. If the probation officer working method is not satisfactory, he / she may be terminated from the job.

Satisfactory performance of Probational staff

He / She should be

Honest
Promptness and responsible
Decisiveness
Obedience
Loyalty

Dissatisfactory performance of Probational staff

The following are some of the parameters to consider for dissatisfactory performance of probation staffs.

Irresponsible
Malpractice
Dishonest
Series lapses in performance
Misconduct in public life and workplace

Download: Paripatra for government of Gujarat probational staff / employee

Also read: Government of Gujarat Vande Bharat Journey LTC GR

Government of Gujarat Employees travel through Vande Bharat in LTC

noreply@blogger.com (R D PANCHAL) — Wed, 30 Oct 2024 08:37:00 +0530

Good news for Government of Gujarat Employees!!

Earlier Gujarat employees may travel through Rajdhani & Shatabdi Express in the LTC. As per GOG notification as on 13/8/2024, GOG gives good news for employees by considering VANDE BHARAT journey in the new LTC rules.

Shatabdi / Vande Bharat Train LTC applicable

Grade pay	Class
Grade pay 7600 and above	Executive class
Grade pay above 2400 and less than 7600	AC Chair class

According to LTC rules, there is direct link between first place and last place of the journey through train. There is no extra charges for Food in the Vande Bharat Train.

Download: Government of Gujarat Vande Bharat LTC Rules GR

Important Formula of Design of Transformer - 3

noreply@blogger.com (R D PANCHAL) — Sat, 26 Dec 2020 12:31:00 +0530

Relation of height of window to width of window is 2 to 4

H_w / W_w = 2 to 4

Area of yoke

= ( 1.15 to 1.25 ) A_gi for hot rolled steel

= A_gi for cold rolled steel

The diameter of circumscribing circle is directly proportional to √ A_i

Where A_i = Core Area

The length of mean turns of conductors wound on cold rolled steel is 0.817 times that of hot rolled steel

Volume of conductor in cold rolled steel is 18.3% lesser than that of hot rolled steel = ( 1 – 0.817 ) × 100

MMF of primary winding = N_PI_P

MMF of secondary winding = N_SI_S

Voltage regulation of transformer

= E_r Cos Φ + E_x Sin Φ + ½ ( E_r Sin Φ – E_x Cos Φ )2

Where

E_r = Per unit resistance

= I_PR_P / V_P

E_x = Per unit reactance

= I_PX_P / V_P

The average value of flux density in the duct = B_a / 2

Short circuit fault current = I / E_x

Where

I = RMS fault current

E_x = Per unit leakage reactance

Ratio of maximum radial force under worst fault condition to maximum radial force under normal condition

F_{r ( max ) f} / F_r _{( max )
n} = 2 / E_x²

Where

F_{r ( max ) f}= Radial force under fault condition

F_{r ( max ) n} = Radial force under normal condition

If one winding is only 5% shorter than other winding, it produces large axial force.

No load current

I₀ = √ ( I_w² + I_µ² )

Where

I_w = Loss component of no load current

I_µ = Magnetizing current

RMS value of magnetizing current

I_µ = AT₀ ( N₀ I₀ ) / √ 2 N_P

Where

AT₀ = Total primary turns

N_P = Primary turns

Magnetizing current

I_µ = Magnetizing VA per kg × Weight of core / Number of phases × Voltage per phase

Eddy current loss

We = kf²B_max2

If the supply voltage is kept constant , ( f B_max ) remain constant therefore the eddy current loss remains constant in spite of frequency is changed

Hysteresis loss

W_h = kfB_max2

If the supply voltage is constant, product of f B_max remains constant

W_h = k ( f B_max ) B_max

Let f B_max = k1 constant

B_max = k1 / f

W_h = kk1² / f

If the supply voltage is kept constant, hysteresis loss decreases with increases in the frequency.

The leakage reactance increases with increase in frequency. The effective resistance increases with increases in frequency.

The copper losses don’t depend on supply frequency.

Effect of frequency on flux density if the supply voltage is kept constant

V₂ = 4.44 f₂Φ_max2N₂

V₁ = 4.44f₁Φ_max1N₁

Let us consider that frequency f₁( 50 Hz ) changes to f₂ ( 40 Hz )

Φ_max2f₂ = Φ_max1f₁

Φ_max2 = ( f₁ / f₂ ) Φ_max1

= ( 50 / 60 ) Φ_max1

Φ_max2 = ( 0.82 ) Φ_max1

The eddy current losses do not affect with change in frequency.

Copper loss varies as the square of the output

W_C2 / W_C1 = ( output in 2^nd case )² / ( output in 1^st case )²

The power rating of transformer increases with increase in frequency.

Specific heat dissipation due to convection of oil

λ_conv = 40.3 ( θ / H )^1.4

Where

θ = Temperature difference of surface relative to oil

H = Height of dissipating surface

Effect of W_c / W_i on overload capacity of transformer

Case 1 : W_c / W_i = 1

Case 2 : W_c / W_i = 2

At 20% overlaod

Case 1

Total loss at overload / Total load at full load

[ ( 2 )² × 1 ] + 1 / ( 1 + 1 )

= 2.5

Case 2

Total loss at overload / Total load at full load

[ ( 2 )² × 1 ] + 1 / ( 2 + 1 )

= 3

Therefore the overload losses in the two cases are 2.5 : 3 therefore the transformer with higher W_c / W_i is less as compared to sustain overload

Heating time constant

T_h = Gh / Sλ

Where

G = Weight

h = Specific heat

λ = Specific heat dissipation

T_h = Gh θ / Q ( ⸫ Sλ = Q / θ )

The heating time constant of oil is about 30 times to that of winding.

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Important Formula of Design of Transformer - 2

noreply@blogger.com (R D PANCHAL) — Sat, 26 Dec 2020 11:55:00 +0530

Effect of losses and output on linear dimension of transformer

Let us consider that two transformer having linear dimensions in the ratio of x : 1 considering same flux density, current density, frequency and window space factor. The transformer A has dimension x with respect to transformer B.

Output

The output of transformer A = x⁴ times that of transformer B

Losses

The losses in the transformer A = x³ times that of transformer B.

Core area

( a ) Square core

Ratio of net core area / Area of circumscribing circle = 0.58

Ratio of gross core area / Area of circumscribing circle = 0.64

Two stepped core

Ratio of net core area / Area of circumscribing circle = 0.71

Ratio of gross core area / Area of circumscribing circle = 0.79

Relation between transformer rating and voltage per turn

E_t= k√ Q

Where

Q = Transformer kVA rating

E_t = Voltage per turn

Condition for minimum copper losses

Current density in the primary winding = Current density in the secondary winding

Insulation thickness

( a ) Between winding earth and LV and HV winding

Insulation thickness = 5 + 0.9 kV mm

kV = kilo voltage between winding and earth

= kilo voltage between windings

Cross over winding

It should be designed such that 2N_CL E_t > 800 V

Where N_CL = Number of conductor per layer

E_t = Voltage per turn

Continuous disc / Cross over winding

The voltage per coil should not be more than 800 – 1000 V for continuous disc / cross over winding

Number of coils > = V / ( 800 to 1000 V )

The shunt capacitance of winding relative to ground is proportional to its length and series capacitance is inversely proportional to the length.

L’ Hospital’s rule

( dv / dx ) = V / L constant

If the shunt capacitance is zero, the voltage is uniformly distributed over the winding. The voltage gradient at the line end is α times the value of voltage gradient to the uniform voltage distribution.

Where α = √ ( Cg / Cs )

Where Cg = Shunt capacitance Cs = Series capacitance

The number of layers in the cross over winding are calculated on the basis that the maximum voltage between layers should not exceed 300 V.

The voltage distribution across insulation is improved by decreasing shunt capacitance or increasing series capacitance.

Window space factor

Kw = 10 / ( 30 + kV )

kV = High voltage winding in kV

Window area

= Hw × Ww

= Height of window × Width of window

Width of window for optimum output

D = 1.7d

Where

d = Diameter of circumscribing circle

D = Distance between center to center between adjacent limbs

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Important Formula of Design of Transformer - 1

noreply@blogger.com (R D PANCHAL) — Sat, 26 Dec 2020 09:48:00 +0530

EMF induced in the transformer winding

E = 4.44fΦ_mN

Where

E = Induced emf

F = Frequency

Φ_m = Maximum flux density

N = Number of turns

Volts per turn in the transformer

E_S / E_P = 4.44fΦ_mN_S / 4.44fΦ_mN_P

E_S / E_P = N_S / N_P

E_S / N_S = E_P/ N_P

Volts per turn in the primary winding = Volts per turn in the secondary winding

The minimum winding diameter D_m of continuously transposed conductor ( CTC )

D_m = ( Pitch of transposition ) × ( Number of conductors ) / π

Where pitch of transposition is distance between adjacent cross overs measured on one side of conductors and it is normally 15 times strip width.

Temperature rise

θ = QC / S

Where

Q = Losses

S = Heat dissipating area

C = Cooling co – efficient

Current density

δ = I / a

Where δ = Current density

I = Current

a = Area

Conductor area in window A_C

= 2AT / δ

Where AT = Ampere turns of windings

= I_PN_P + I_SN_S

δ = Current density

Window space factor

K_w = Conductor area in window / Total conductor area of window

= A_C / A_W

kVA rating of single phase transformer

Q = 2.22fB_max δ K_wA_wA_i × 10 ^{– 3} kVA

Where

f = Frequency

Bm = Maximum flux density

δ= Current density

K_w = Window space factor

A_w = Window area

A_i = Net core area

Rating of three phase transformer

Q = 3.33fB_max δ K_wA_wA_i × 10 ^{– 3} kVA

kVA rating of one phase

Q = 4.44fΦ_mN_PI_P × 10 – 3

= 4.44fΦ_m ( AT ) × 10 – 3

Where Φ_m / ( AT ) is constant for given transformer

Voltage per turns E_t= k √ Q

Where k is constant = √ ( 4.44fr × 10³ )

Where r = Φ_m / ( AT ) Constant

Copper loss per meter³

= I²R

= I² ( ρL / a )

= ρ ( I² L / a )

= ρ ( a² δ² L / a ) ( Current density δ = I / a )

= ρ δ² ( aL )

= ρδ²

Where δ = Current density and

ρ = Resistivity of copper

Copper loss per meter³ = ρ δ²

Specific iron loss is equal to iron loss per kg

Ratio of iron loss to copper loss

W_i / W_c = P_i G_i / P_c G_c

Where

P_i = Iron loss per kg

P_c = Copper loss per kg

G_i = Weight of iron

G_c = Weight of copper

Condition for minimum cost of transformer

Weight of iron = Weight of copper

G_i = G_c

Condition for minimum losses ( Maximum efficiency ) in the transformer

Iron losses = Copper losses

W_i = x² W_c

Where x = load at maximum efficiency occurs

W_i / W_c = P_iG_i / P_cG_c

x² = P_iG_i / P_cG_c

G_i / G_c = x² ( P_c / P_i )

It is a condition for maximum efficiency in the transformer.

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Important Formula of Design of Magnetic Circuit

noreply@blogger.com (R D PANCHAL) — Wed, 2 Dec 2020 21:23:00 +0530

Magnetic loading = PΦ

Where

P = Number of poles

Φ = Flux per pole

Electrical loading = I_Z × Z

Where

I_Z = Current in each conductor

Z = Number of stator conductors

Specific magnetic loading

B_AV = PΦ / πDL

= Φ / τL

Where

D = Diameter of stator bore

L = Length of stator core

τ = Pole pitch = P / πD

Output equation of DC Machine

= ( Total magnetic loading )( Total electrical loading )

( Speed in rps )

Output power

= C₀D²Ln_s

Where

C₀ = π² ( B_AV ) ( ac ) × 10 – 3

= Output co-efficient

B_AV = Average magnetic loading

ac = Specific electrical loading

n_s = Speed in revolution per second

Output equation of AC machine

Q = C₀D²Ln_s

Where

C₀ = 11 ( B_AV ) ( ac ) ( K_w ) × 10 – 3

= Output co-efficient

K_w = Winding factor

The volume of active part of DC and AC machines decreases with increases in speed and vice versa.

( π/4 )D²L α ( 1 / N_s )

The volume of active part is inversely proportional to value of output co-efficient. The size and cost of machine increases with increase of specific electrical loading and specific magnetic loading.

The flux density in the teeth is inversely proportional to specific magnetic loading or average flux density in the air. The maximum value of flux density in the teeth occurs where the teeth width is smallest.

The ratio of B_t / B_av is large at the section where the teeth have a smallest width and therefore B_av must be reduced in small machines. The small machines is designed with lower value of specific magnetic loading due to above reason.

The mmf required for air gap is directly proportional to specific magnetic loading. The area of iron part of magnetic circuit α ( 1/ B )

Two machines having linear dimension of x : 1

Core loss x³ : 1

Percentage core loss ( x³ / x⁴ ) = 1 / x

Therefore the percentage core loss decreases with increase in machine size.

The heat dissipation per unit area is proportional to specific electrical loading.

The specific electrical loading is directly proportional to space factor considering that fixed ratio of slot width to slot pitch and fixed value of depth of slot and current density.

( ac ) α S_f ( W_s, y_s, δ are constant )

Where

δ = Current density

W_s = Width of slot

y_s = Slot pitch = πD / S

S = Number of armature slots

D = Diameter of armature

The volume of active part of machine is directly proportional to torque.

Two machines have linear dimensions x : 1

Output x⁴ : 1 ( for same speed, current density and flux density )

Losses x³ : 1 ( for same speed, current density and flux density )

Efficiency = Output / ( Output + Losses )

= x⁴ / ( x⁴ + x³ )

= 1 / ( 1 + 1 / x )

Therefore the efficiency increases with increase in the dimension of machines.

Temperature rise α C θ

Where C = Cooling co-efficient

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Important Formula Transformer - 2

noreply@blogger.com (R D PANCHAL) — Sun, 29 Nov 2020 16:55:00 +0530

Voltage regulation of transformer

= [{ I₂ R₀₂ Cos Φ ± I₂X₀₂ Sin Φ } / ₀V₂ ] × 100%

+ Sign for lagging power factor

- Sign for leading power factor

Voltage regulation

= V_r Cos Φ ± V_x Sin Φ

Where

₀V₂ = Secondary voltage at no load

V_r = [ I₂ R₀₂ / ₀V₂ ] × 100%

= Percentage resistive drop

V_x = [ I₂ X₀₂ / ₀V₂ ] × 100%

= Percentage reactance drop

Exact voltage drop in the transformer

= ( V_r Cos Φ ± V_x Sin Φ ) + ( 1 / 200 )( V_x Cos Φ ± V_r Sin Φ )²

Voltage regulation at unity power factor

= [ I₂R₀₂ / ₀V₂ ] × 100%

Condition for maximum voltage regulation

Tan Φ = V_x / V_r

Percentage voltage drop

% drop = ( %R )( kW ) / kVA rating +

( %X )( kVAr ) / kVA rating

Exciting conductance

G₀ = W / V₁²

Exciting susceptance

B₀ = √ ( Y₀² – G₀² )

Hysteresis loss

W_h = ηB_max^1.6fV

Where

V = Volume of material

η = Steinmetz hysteresis coefficient

W_h α B_max^1.6f

Eddy current loss

W_e = kB_max² f²t²

Where

B_max = Maximum flux density

f = Frequency

t = Thickness of laminations

W_e α B_max²f² t²

Total copper loss in the transformer

W_C = I₁² R₁ + I₂² R₂

Where

R₁ = Resistance of primary winding

R₂ = Resistance of secondary winding

W_C = I₁² R₀₁

= I₂² R₀₂

Where

R₀₁ = Total resistance as referred to primary side

R₀₂ = Total resistance as referred to secondary side

Copper loss at half load

W_C α ( I / 2 )² R

= I²R ( At full load ) / 4

Effect on hysteresis loss and eddy current loss if the flux density remains same

W_e α f²

W_h α f

Voltage equation of transformer

E = 4.44fNB_max A

B_max α ( E / f )

Hysteresis loss W_h α B_max^1.6f

α ( E / f )^1.6 f

α ( E )^1.6f ^{– 0.6}

k ( E )^1.6f ^{– 0.6}

Eddy current loss W_e α B_max² f²

α ( E / f )² f²

= kE²

Percentage resistance

%R = [ I₁ R₀₁ / V₁ ] × 100%

= [ I₁² R₀₁ / I₁V₁ ] × 100%

= [ I₂² R₀₂ / I₁V₁ ] × 100%

= % Copper loss at full load

Percentage reactance

%X = [ I₁ X₀₁ / V₁ ] × 100%

= [ I₂ X₀₂ / V₂ ] × 100%

R₀₁ = [ % R × V₁ ] / [ 100 × I₁ ]

= % Copper loss × V₁ ] / [ 100 × I₁ ]

R₀₂ = [ % R × V₂ ] / [ 100 × I₂ ]

= % Copper loss × V₂ ] / [ 100 × I₂ ]

The percentage resistance, reactance and impedance have same value whether it is referred to primary side or secondary side.

Condition for maximum efficiency

Iron loss = Copper loss

Wi = I₁² R₀₁

= I₂² R₀₂

Load corresponding to maximum efficiency

I2 = √ ( W_i / R₀₂ )

= I_2full / I_2full [ √ ( W_i / R₀₂ ) ]

= I_2full √ ( W_i / W_cu )

= I_2full √ ( Iron loss / Full load copper loss )

Efficiency at any load

η = ( output power / input power) × 100%

Where

Output power = ( x ) ( kVA ) ( Power factor )

Input power = ( x ) ( kVA ) ( Power factor ) + ( W_i ) + ( W_cu )

x = Ratio of actual to full load kVA

kVA for maximum efficiency

kVA = kVA_full ( √ W_i/ W_cu )

Transformer efficiency

η = 1 – ( x / Cos Φ + x )

= 1 – { ( x / Cos Φ ) / ( 1 + x / Cos Φ ) }

Where x = Losses / V₂I₂

All day efficiency

η_Allday = { Output in kWh / Input in kWh } × 100%

Saving in copper in auto transformer as step down transformer

Saving = k × Weight of copper in ordinary transformer

Where k = Voltage transformation ratio

Power transfer inductively = ( 1 – k ) Input

Power transfer conductively = k × Input

Ratio of V – V connection to Delta – Delta connection

V – V Capacity / Delta – Delta Capacity = 57.7%

V – V connected two transformer supply power at Cos Φ power factor

P₁ = kVA Cos ( 30 – Φ )

P₂ = kVA Cos ( 30 + Φ )

Where Φ = Power factor of load

Rating of teaser transformer

= 86.6% of Main transformer rating

Sharing of load when transformer P and transformer Q are connected in parallel

S_P = S ( Z_Q / Z_P + Z_Q )

S_Q = S ( Z_P / Z_P + Z_Q )

Where

S_P = kVA rating of transformer P

S_Q = kVA rating of transformer Q

Z_P = Impedance of transformer P

Z_Q = Impedance of transformer Q

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SSC Junior Engineer MCQs 2017 – 4

Important Formula Transformer - 1

noreply@blogger.com (R D PANCHAL) — Sun, 29 Nov 2020 11:19:00 +0530

EMF Equation of transformer

E = 4.44fNB_max A

Where

f = Supply frequency in Hz

N = Number of winding turns

B_max = Maximum flux density in weber per meter square

A = Cross section area of core in meter square

Primary induced emf

E1 = 4.44fN₁B_max A

Secondary induced emf

E1 = 4.44fN₂B_max A

Voltage transformation ratio ( k )

K = E₂ / E₁ = V₂ / V₁ = N₂ / N₁ = I₁ / I₂

Turns ratio

N₁ / N₂ = E₁ / E₂ = V₁ / V₂ = I₂ / I₁ = 1 / k

Volts per turns

E₂ / N₂= E₁ / N₁ = 4.44fΦ_max
The volts per turn ratio remains same for primary winding and secondary winding.

No load input power

W₀ = V₁I₀ Cos Φ₀

Where

W₀ = No load input power

V₁ = Supply voltage

I₀ = No load current

Cos Φ₀ = No load input power factor

Active / working iron loss component

I_w = I₀ Cos Φ₀

Magnetizing current

I_µ = I₀ Sin Φ₀

No load current I₀ = √ ( I_w² + I_µ² )

No load power factor angle

tan Φ₀ = I_µ / I_w

Φ₀ = tan ^{– 1} ( I_µ / I_w )

Primary current ( vector addition )

I₁ = I₀ + I₂’

Where

I₀ = No load current

I₂’ = Reflected secondary current

Relation between load flux ( Φ₂ ) and reflected load component flux ( Φ₂’ )

( Φ₂ ) = ( Φ₂’ )

N₂I₂ = N₁I₂’

I₂’ = ( N₂ / N₁ ) I₂

I₂’ = kI₂

k = I₂’ / I₂ = I₁ / I₂ = N₂ / N₁ = V₂ / V₁

Where

I₂’ = Reflected load current

I₂ = Load current

Transformer secondary resistance, reactance and impedance as referred to primary side

R₂’ = R₂ / k²

X₂’ = X₂ / k²

Z₂’ = Z₂ / k²

Where R_2, X₂ and Z₂ are secondary resistance, reactance and impedance respectively and k is voltage transformation ratio

Equivalent resistance as refereed to primary side

R₀₁ = R₁ + R₂’

= R₁ + R₂ / k²

Transformer primary resistance, reactance and impedance as referred to secondary side

R₁’ = k²R₁

X₁’ = k²X₁

Z₁’ = k²Z₁

Where R₁, X₁and Z₁ are primary resistance, reactance and impedance

Equivalent resistance as refereed to secondary side

R₀₂ = R₂ + R₁’

= R₂ + k²R₁

Shifting of secondary voltage / current from secondary side to primary side , only transformation ratio k is used

Secondary voltage as refer to primary side

V₂’ = V₂ / k

Similarly, secondary current as referred to primary side

I₂’ = kI₂

Leakage reactance

X_L = e_L / I

No load resistance R₀ and reactance X₀

R₀ = E₁ / I_µ

X₀ = E₁ / I_w

Where E₁ = Primary induced emf

Magnetizing ampere turns = N₁I₁

Demagnetizing ampere turns = N₂I₂

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Important Formula of Transmission Line Parameters

noreply@blogger.com (R D PANCHAL) — Sun, 22 Nov 2020 19:59:00 +0530

( 1 ) Resistance of the conductor
R = ρL / a

Where

R = Resistance

L = Length of conductor

a = Area of conductor

ρ = Resistivity of conductor

( 2 ) Inductance of conductor

L = Nφ / I

= Flux Linkage / Ampere

= ψ ( Symbol Psi ) / I ( Where ψ = Flux linkage )

( 3 ) Inductance of conductor due to internal flux

Internal flux linkage ψ_int = µI / 8π weber – turns / meter

Where µ = Permeability

= µ₀ µ_r

µ₀= Absolute permeability = 4 π × 10 ^{– 7}

µ_r = Relative permeability

ψ_int = (4 π × 10 ^{– 7} ) × I / 8π

= ( I / 2 ) 10 ^{– 7} ……………. ( 1 )

L_int = ( 1 / 2 ) 10 ^{– 7}

( 4 ) Inductance of conductor due to external flux linkage

ψ_out = ( 2I × 10 ^{– 7} ) Ln {( D – r ) / r } weber – turns / meter

Where

D = Diameter of conductor

r = Radius of conductor

Ln = Log natural

I = Current

If D >> r

Flux linkage ψ_external = ( 2I × 10 ^{– 7} ) Ln { D / r } .. ( 2 )

Inductance = ( 2 × 10 ^{– 7} ) Ln { D / r }

( 5 ) Inductance of single conductor

ψ = ψ_internal + ψ_external

= equation ( 1 ) + equation ( 2 )

= ( I / 2 ) 10 ^{– 7} + ( 2 × 10 ^{– 7} ) Ln { D / r }

= ( 2 × 10 ^{– 7} ) × I [ Ln { D / r’ } ]

Where r’ = re ^{– ¼}= 0.7788r

= GMR of conductor

L = ( 2 × 10 ^{– 7} ) [ Ln { D / r’ } ]

Inductance of single phase two wire circuit

L = ( 4 × 10 ^{– 7} ) Ln { D / r’ }

Inductance per conductor

L = ( 2 × 10 ^{– 7} ) Ln { D / r’ }

( 6 ) Inductance single phase composite conductors

L = ( 2 × 10 ^{– 7} ) Ln { D_m / D_s } H / m

= ( 2 × 10 ^{– 7} ) Ln { GMD / GMR } H / m

= 0.2 Ln { GMD / GMR } mH / km

The GMD of bundle conductor can be found by taking root of product of distances from each conductor of a bundle to every other conductor of bundle.

GMR = Geometrical mean radius

= √ ( r’D ) ( for two conductors )

= [ ( r’D² ) ]^1/3 ( for three conductors )

= 1.09 × [ ( r’D³ ) ]^1/4 ( for four conductors )

( 7 ) Inductance of three phase unsymmetrical spacing three

conductor ( transposed )

L = ( 2 × 10 ^{– 7} ) Ln { D_eq / Ds } H / m

Where

D_eq = ( D12 × D23 × D32 )^1/3

D = Distance between conductors

D12 ≠ D23 ≠ D31

Inductance of three phase symmetrical spacing three conductor

Inductance of a conductor

L = ( 2 × 10 ^{– 7} ) Ln { D / r’ } H / m

( 8 ) Capacitance of conductors

C = Q / V Farad / meter

Where

C = Capacitance in Farad

Q = Charge in coulomb

V = Voltage

( 9 ) Capacitance between two conductors a and b

C_ab = πε / Ln ( D / r )

= 1 / [ 36 Ln ( D / r ) ] Micro Farad / kilo-meter

( 10 ) Capacitance between conductor to neutral

= 2 × Capacitance between two conductors

= 1 / [ 18 Ln ( D / r ) ] Micro Farad / kilo-meter

= 2πε/ Ln ( D / r )

( 11 ) Unsymmetrical conductor spacing in the three phase transposed system

If there are three conductors a, b, c placed unsymmetrical

Capacitance of conductor = 2πε / Ln ( D_eq / r ) Farad / meter

Where D_eq = ( D12 × D23 × D32 )^1/3

( 12 ) Symmetrical conductor spacing in the three phase transposed system

Capacitance per phase = 2πε/ Ln ( D / r )
Where D12 = D23 = D31 = D

( 13 ) Capacitance of single conductor due to effect of earth

C = 2πε₀ / Ln ( 2h / r ) Farad / meter

Where

h = Distance of image ground conductor and earth surface

( 14 ) Capacitance of single phase transmission line due to effect of earth

Capacitance of conductor

= 2πε / Ln ( D / r × 2h / √ 4h² + D² )

Capacitance of single phase between conductors

= 1 / 36 Ln { D × 2h / r × √ ( 4h² + D²) }

Where

h = Distance of ground image conductor and earth surface

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Important Formula of Power Station Engineering - 1

noreply@blogger.com (R D PANCHAL) — Mon, 16 Nov 2020 13:50:00 +0530

( 1 ) Maximum demand

= Greatest demand of load during given period

( 2 ) Connected load

= Sum of all continuous rating of load

If consumer has four equipment and its rating are 500 W, 1 kW, 60 W and 40 W, the connected load = 500 W + 1000 W + 60 W + 40 W = 1600 W

( 3 ) Average load

The average load is given for specify time say daily, monthly or yearly.

Average load ( daily )

= No. of units generated in a day / 24 hours

Average load ( Monthly )

= No. of units generated in a day / Number of hours ( month )

Average load ( Yearly )

= No. of units generated in a day / Number of hours in a year

( 4 ) Load factor

= Average demand / Maximum demand

= Average demand × time / Maximum demand × time

= No. of unit generated in given time / Maximum demand × time

( 5 ) Demand factor

= Maximum demand / Connected load

( 6 ) Diversity factor

= Sum of individual maximum demand / Maximum demand

( 7 ) Plant capacity factor

= Average demand / Plant capacity

= Actual energy produced / Maximum energy have been produced

( 8 ) Plant capacity factor ( Yearly )

= Yearly kWh output / [ Plant capacity × ( 365 × 24 ) ]

( 9 ) Plant use factor

= Station output in kWh /

[ Plant capacity × Plant operation in hours ]

( 10 ) Reserve capacity

= Plant capacity – Maximum demand

If the maximum demand is same, load factor is equal to plant capacity.

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noreply@blogger.com (R D PANCHAL) — Sat, 14 Nov 2020 13:46:00 +0530

( 1 ) Critical disruptive voltage

V_C = m₀g_oδ r Log_e ( d / r )

Where V_C = Phase to neutral voltage

r = Radius of conductor

d = Diameter of conductor

g₀ = Breakdown ( Dielectric ) strength of air at 76 cm of mercury at 25 degree centigrade

= 30 kV / cm ( Max )

= 21.2 kV / cm ( RMS )

m₀= Irregular factor

= 1 for polished conductor

= 0.98 to 0.92 for dirty conductor

= 0.87 to 0.8 for stranded conductor

( 2 ) Air density factor

Air density factor δ = 3.92b / 273 + t

Where b = atmospheric pressure of b cm

t = Temperature in centigrade

Line voltage = √ 3 × V_C

( 3 ) Visual Critical voltage

V_V = m₀g_o δr { 1 + 0.3 / √ δr } Log_e ( d / r )

Where

m₀= Irregular factor

g₀ = Breakdown ( Dielectric ) strength of air

r = Radius of conductor

d = Diameter of conductor

δ = Air density factor

( 4 ) Power loss due to corona

P = 242.2 ( f + 25 / δ ) √ r / d ( V – V_C )² × 10 ^{– 5} kW / km / Phase

Where

f = Supply frequency

V = Phase to neutral voltage ( RMS )

V_C=Critical disruptive voltage ( RMS ) per phase

δ = Air density factor

Power loss due to corona α ( V – V_C )²

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Important Formula : Annual Depreciation Charges

noreply@blogger.com (R D PANCHAL) — Sat, 14 Nov 2020 10:04:00 +0530

The annual depreciation charges are calculated by following methods.

( 1 ) Straight line method

Annual depreciation charge = P – S / n

Where P = Initial cost of equipment / plant

S = Salvage / Scrap value of equipment / plant

n = Life of equipment

( 2 ) Diminishing value method

Value of equipment after one year

S = P ( 1 – x )

Value of equipment after 10^th year

S = P ( 1 – x )¹⁰

Value of equipment of n years

S = P ( 1 – x )ⁿ

Annual depreciation

x = 1 – ( S / P )ⁿ

Where

P = Initial cost of equipment / plant

S = Salvage / Scrap value of equipment / plant

n = Life of equipment

r = Annual rate of interest

x = Annual depreciation

( 3 ) Sinking Fund method

Annual depreciation in the sinking fund method

q = ( P – S ) [ r / ( 1 + r )ⁿ – 1 ]

Where

P = Initial cost of equipment / plant

S = Salvage / Scrap value of equipment / plant

n = Life of equipment

r = Annual rate of interest

x = Annual depreciation

Cost of equipment after n year

= P – S

= q [ ( 1 + r )ⁿ – 1 / r ]

Sinking fund factor = r / { ( 1 + r )ⁿ – 1 }

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noreply@blogger.com (R D PANCHAL) — Sun, 1 Nov 2020 13:11:00 +0530

( 1 ) Current amplification factor

β= I_C/ I_B

( 2 ) Collector current for common emitter configuration

I_C = βI_B + I_CBO

Where

I_CBO = Collector – base current with emitter open

I_CEO = Collector cut off current ( Collector current when

base is open circuited )

I_CEO >> I_CBO

( 3 ) Stability Factor

S = dI_C / dI_CO at constant I_B and β

Where

dI_CO = Collector leakage current

General equation of stability factor

S = ( β + 1 ) / 1 - β ( dI_B / dI_C )

Base Resistor Method

( 1 ) Base resistor method

R_B = V_CC – V_BE / I_B

= V_CC / I_B

V_BE is very small so it is neglected

Stability factor S = ( β + 1 ) / 1 - β ( dI_B / dI_C )

The base current I_B is independent of I_Cin the fixed bias method so dI_B / dI_C = 0

Stability factor S = ( β + 1 )

( 2 ) The Q point is extremely depends on β in a base resistor method therefore the base bias circuit is unstable. It is not used in the linear operation.

Emitter Bias Circuit

( 1 ) Emitter bias circuit

The emitter bias circuit requires one positive voltage source ( + V_CC ) and negative voltage source ( - V_EE ) and these two voltage sources are equal.

V_CC = + 10 V DC

V_EE = - 10 V DC

Collector current

I_C = V_EE – V_BE / ( R_E + R_B / β )

R_E >> R_B / β

Therefore I_C = V_EE – V_BE / ( R_E )

Similarly V_EE >> V_BE

I_C = V_EE / R_E

As I_C is independent of β and V_BE, the Q point is not affected so much.

Biasing with collector feedback resistor

( 2 ) Biasing with collector feedback resistor

The stability factor

S < ( β + 1 )

It provides better stability than fixed bias method.

V_CC = I_CR_C + I_BR_B + V_BE

Voltage divider Bias Method

( 1 ) Voltage divider bias method

Current I₁ passes through resistor R₁

I₁ = V_CC / ( R₁ + R₂ )

Voltage across R₂ is

V₂ = ( V_CC / R₁ + R₂ ) R₂

V₂ = V_BE + V_E

= V_BE + I_ER_E

I_E = V₂ – V_BE / R_E

As I_E ≈ I_C

I_C = V₂ – V_BE / R_E

The value of I_C does not depend on β. The value of IC independent of transistor parameters therefore good stabilization is provided.

Equation for collector side

V_CC = I_CR_C + V_CE + I_CR_E

Therefore V_CE= V_CC – I_CR_C – I_CR_E

Stability factor S = ( β + 1 )( R₀ + R_E) / R_O + R_E + β R_E

Where

R_O = R₁ × R₂/ ( R₁ + R₂ )

S = ( β + 1 ) [ 1 + ( R_O / R_E )] / ( β + 1 ) + R_O / R_E

The ratio of R_O / R_E is very small therefore it is neglected.

S = ( β + 1 ) / ( β + 1 )

= 1

The smallest value of stability factor leads to thermal runaway. The ratio of R₀ / R_E cannot be neglected.

The stability factor should be around 10 in this method.

( 1 ) Mid point Biasing

As the Q point lies at centre

I_C = ½ I_{C MAX}

V_CE = V_CC / 2

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Important Formula - Negative Feedback Amplifier

noreply@blogger.com (R D PANCHAL) — Sun, 1 Nov 2020 12:11:00 +0530

Negative voltage feedback

( 1 ) Gain without feedback

A_v = V_o / V_in

Where

A_v = Gain

V_o = Output voltage

V_in = Input voltage

( 2 ) Feedback fraction ( mv )

= Output voltage of feedback / Output voltage of amplifier

< 1

( 3 ) Power gain

A_p = A_v × A_i

Where

A_p = Power gain

A_v = Voltage gain

A_i = Current gain

( 4 ) Gain of negative voltage feedback amplifier

A_vf = A_v / ( 1 + A_vm_v )

Where

A_vf = Gain of amplifier with feedback

A_v = Gain of amplifier without feedback

m_v = Feedback fraction

( 5 ) Non linear distortion

D_vf = D / ( 1 + A_vm_v )

Where

D_vf = Distortion of amplifier with feedback

D = Distortion of amplifier without feedback

A_v = Gain of amplifier without feedback

m_v = Feedback fraction

( 6 ) Input impedance

Z_in’ = Z_in ( 1 + A_vm_v )

Where

Z_in’ = Input impedance without feedback

Z_in= Input impedance with feedback

( 7 ) Output impedance

Z_out’ = Z_out/ ( 1 + A_vm_v )

Where

Z_out’ = Input impedance without feedback

Z_out= Input impedance with feedback

( 8 ) Gain stability

A_vf = A_v / ( 1 + A_vm_v )

A_vf ≈ A_v / A_vm_v

A_vf ≈ 1 / m_v

Parameters	Negative voltage feedback
Gain stability	A_vf ≈ 1 / m_v
Input impedance	Z_in’ = Z_in ( 1 + A_vm_v )
Output impedance	Z_out’ = Z_out/ ( 1 + A_vm_v )
Non linear distortion	D_vf = D / ( 1 + A_vm_v )
Voltage Gain	A_vf = A_v / ( 1 + A_vm_v )

Negative current feedback

( 1 ) Feedback fraction of negative current feedback

Feedback Fraction = Feedback current / Output current

m_i = I_f / I_out

Where

I_f = Feedback current

I_out = Output current

( 2 ) Current gain with negative feedback

A_if = A_i / ( 1 + A_im_i )

Where

A_if = Current gain with feedback

A_i = Current gain without feedback

m_i= Current feedback fraction

( 3 ) Input impedance with negative current feedback

Z_in’ = Z_in / ( 1 + A_im_i )

Where

Z_in’ = Input impedance with negative current feedback

Z_in = Input impedance without negative current feedback

( 4 ) Output impedance with negative current feedback

Z_out’ = Z_out ( 1 + A_im_i )

Where

Z_out’ = Output impedance with negative current feedback

Z_out = Output impedance without negative current feedback

( 5 ) Bandwidth with negative current feedback

BW’ = BW ( ( 1 + A_im_i )

BW’ = Bandwidth with negative current feedback

BW = Bandwidth without negative current feedback

Parameters	Negative current feedback
Feedback Fraction	m_i = I_f / I_out
Current gain	A_if = A_i / ( 1 + A_im_i )
Input impedance	Z_in’ = Z_in / ( 1 + A_im_i )
Output impedance	Z_out’ = Z_out ( 1 + A_im_i )
Bandwidth	BW’ = BW ( ( 1 + A_im_i )

Emitter Follower

( 1 ) Voltage gain of emitter follower

A_v = R_E / r_e’ + R_E

Where

A_v = Voltage gain of emitter follower

r_e' = 25mV / I_E

R_E = Emitter resistance

( 2 ) Input impedance of Emitter follower

Z_in = R1 ‖ R2 ‖ Z_in _{( base )}

Where

Z_{in
( base )} = β ( r_e' + R_E^’)

r_e' = 25mV / I_E

R_E’ = R_E ‖ R_L

Generally, the impedance of base is very large as compared to R1 ‖ R2 so Z_in _{( base )}is neglected.

Therefore Z_in ≈ R1 ‖ R2

( 3 ) Output impedance of Emitter follower

Z_out = R_E ‖ ( r_e' + R^’_in / β )

The value of R_Eis very large in the practical circuit

Z_out = ( r_e' + R^’_in / β )

Where r_e' = 25mV / I_E

R^’_in = R1 ‖ R2 ‖ R_s

Where

R_s = Source resistance

β = Current amplification factor

Darlington Transistor

( 1 ) Current gain of Darlington transistor

β = β1 β2

Where

β1 = Current gain of transistor 1

β2 = Current gain of transistor 2

( 2 ) Input impedance of Darlington transistor

Z_in= ( β1 )( β2 )( R_E )

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Important Formula of Multistage Transistor Amplifiers

noreply@blogger.com (R D PANCHAL) — Thu, 29 Oct 2020 19:42:00 +0530

( 1 ) Gain of amplifier

G = G1 × G2 × G3

Where

G1 = Gain of first stage

G2 = Gain of second stage

G3 = Gain of third stage

( 2 ) Relation between “Bel” and “dB”

1 Bel = 10 dB

( 3 ) Gain of Amplifier in dB

G = G1 + G2 + …… + G_n

Where

G1 = Gain of first stage ( dB )

G2 = Gain of second stage ( dB )

G_n = Gain of nth stage ( dB )

( 4 ) Power gain

= 10 Log10 ( P_out / P_in ) dB

( 5 ) Voltage gain

= 20 Log10 ( V_out / V_in ) dB

( 6 ) Current gain

= 20 Log10 ( I_out / I_in ) dB

( 7 ) Bandwidth of amplifier

BW = f2 – f1

Where f2 = Upper cut off frequency

f1 = Lower cut off frequency

f1 and f2 is called as 3 dB frequency or half power frequency.

( 8 ) Voltage gain of RC Coupled Amplifier

Voltage gain = β ( R_ac / R_in )

Where R_ac = Effective load of first stage

= R_c ‖ R_in

( 9 ) The load reflected at the primary side of the transformer coupled amplifier

R_L’ = ( N_P / N_S ) R_L

Where R_L = Load resistance

N_P = Primary turns

N_S = Secondary turns

R_L’ = ( I_S / I_P )² R_L

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Important Formula of Transistor Tuned Amplifiers

noreply@blogger.com (R D PANCHAL) — Mon, 26 Oct 2020 10:26:00 +0530

( 1 ) Parallel resonance frequency

fr = ( 1 / 2π ) √ ( 1 / LC – R² / L² )

( 2 ) Dynamic Impedance of parallel resonance circuit

Z = L / CR

( 3 ) Q factor of parallel resonance circuit

Q = X_L / R

= ωL / R

= Energy Stored / Energy Lost

= Reactive Power / Resistive Power

( 4 ) Voltage gain of tuned amplifier

Voltage gain = βZ_C / Z_in

Where β = Current gain of the transistor

Z_C = Effective collector load

Z_in = Input impedance of amplifier

( 5 ) Relation between bandwidth and Q factor

Q = f_r / BW

Where

f_r = Resonance frequency

= 1 / [ 2π √ ( LC ) ]

( 6 ) Bandwidth of double tuned circuit

BW_dt = kf_r

Where k = Coefficient of coupling

dt = It is used for double tuned circuit

( 7 ) Critical coupling in the double tuned circuit

K_critical= 1 / √ ( Q₁ × Q₂ )

Where

Q₁ = Quality factor for primary resonant circuit

Q₂ = Quality factor for secondary resonant circuit

( 8 ) The power losses in the tuned Class C amplifier is very small in high Q resonant circuit.

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Important Formula of Sinusoidal Oscillators

noreply@blogger.com (R D PANCHAL) — Mon, 26 Oct 2020 09:46:00 +0530

( 1 ) Frequency of tank circuit

f = 1 / 2π √ ( LC )

Where L = Inductance of tank circuit

C = Capacitance of tank circuit

( 2 ) Barkhausen criteria

m_vA_v = 1

Where

m_v = Feedback fraction

A_v = Voltage gain of amplifier without feedback

( 3 ) Voltage gain of positive feedback amplifier

A_vf = A_v / ( 1 – m_vA_v )

Where

A_vf = Voltage gain with positive feedback

m_v = Feedback fraction

A_v= Voltage gain without feedback

( 4 ) Frequency of Colpitt oscillator

f = 1 / 2π √ ( LC_T )

Where C_T = C1 ‖ C2

= C1 × C2 / ( C1 + C2 )

Feedback fraction m_v = C1 / C2

= V_f / V_out

( 5 ) Frequency of Hartley oscillator

f = 1 / 2π √ ( CL_T )

Where L_T= L1 + L2 + 2M

M = Mutual Coupling between L1 and L2

Feedback Fraction mv = L2 / L1

= V_f / V_out

( 6 ) Frequency of Phase shift oscillator

f_o = 1 / 2πRC √ 6

Where
R = R1 = R2 = R3

C = C1 = C2 = C3

( 7 ) Frequency of Wien bridge oscillator

f = 1 / 2π √ R₁C₁R₂C₂

Where

R₁ = R₂ = R and C₁ = C₂ = C

f = 1 / ( 2π √ R²C² )

f = 1 / 2π √ RC

( 8 ) Frequency of Piezo electric crystal

f = k / t

Where

k = Constant

t = Thickness of crystal

( 9 ) When the crystal is not vibrating, it is equal to mounting

capacitance. When the crystal is vibrating, it is equal to R – L – C series circuit.

( 10 ) Q factor of Crystal

Q = 1 / ( R √ L / C )

( 11 ) Series resonance frequency of Crystal

f_s = 1 / ( 2π √ LC )

( 12 ) Parallel resonance frequency of Crystal

f_p = 1 / ( 2π √ LC_T )

Where CT = ( C × C_m ) / C + C_m

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Important Formula of Electron Emission

noreply@blogger.com (R D PANCHAL) — Sat, 24 Oct 2020 17:32:00 +0530

( 1 ) Richardson Dushman equation relates for thermionic emission and emitter temperature

J_S = AT²e ^{–
b / T} Amp / meter²

Where

J_S = Current density in ampere per meter square

T = Absolute temperature of emitter ( ^oK )

A = Constant ( Amp / meter² / K² )

= Depend upon type of emitter

b = Constant for emitter

e = Natural Logarithmic base

The value of b for metal is

b= φe / K

Where

e = Charge of electron

= 1.602 × 10 ^{– 19} Coulomb

φ = Emitter work function

K = Boltzmann’s Constant

= 1.38 × 10 ^{– 23} J / K

( 2 ) If the temperature increases twice, the electron emission increases up to 10⁷ times.

( 3 ) Electron Mobility

The ability of electron moves through semiconductor or metal in the presence of applied electric field is called as electron mobility.

µ_n= V_n / E

Where

µ_n= Mobility of electron

V_n = Drift velocity of electron

E = Applied electric field

( 4 ) Hole Mobility

The ability of holes moves through semiconductor or metal in the presence of applied electric field is called as electron mobility.

µ_H= V_H / E

Where

µ_H= Mobility of electron

V_H = Drift velocity of electron

E = Applied electric field

( 5 ) Diffusion

The process by which electron / holes in the semiconductor moves from higher concentration region to lower concentration region is called as diffusion.

The concentration gradient for n type semiconductor is given by

J_n α ( dn / dx )

Where

J_n = Current density of electrons

If both donors and acceptors are added to semiconductor, the semiconductor is called compensated.

If N_d > N_a , the compensated semiconductor is N type.

N_effective = N_d – N_a

If N_a > N_d, the compensated semiconductor is P type.

N_effective = N_a – N_d

( 6 ) The hole concentration in the valance band is given by

P = N_V e ^{– ( EF – EV
) / KB * T}

Where

T = Absolute temperature of intrinsic semiconductor

Nv = Effective current density in the valance band

KB = Boltzmann’s Constant

EF = Fermi level

EV = Valance band

( 7 ) The electron carrier concentration is hole carrier concentration in the intrinsic semiconductor

P = n = n_i

Where

P = Hole carrier concentration

n = Electron carrier concentration

n_i= Intrinsic carrier concentration

( 8 ) Fermi level for intrinsic semiconductor

E_F = ( E_C + E_V) / 2

Where

E_F = Fermi level

E_C = Conduction band

E_V = Valance band

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