Maths sem 1 blueprint expected questions



Q.1
a)Problems on basic partial derivatives (03 marks)

b) Relation between circular and hyperbolic function (03 marks)

C)Jacobians (03 marks)

D) Expansion standard series (03 marks)

E) Properties of matrices (04 marks)

F) Problems on standard formula of successive derivatives (04 marks)

Q.2
A)Complex Numbers- Powers & Roots (06 marks)

B)Matrices PAQ/normal form (06 marks)

C)Eulers theorem with deduction (08 marks)


Q.3
A)Linear homogenous and non-homogenous equations (06marks)

B)Maxima and minima/Lagranges method (06 marks)

C)Separation of real & imaginary parts (08 marks)

Q.4
A)Jacobian of implicit functions/partial derivative of implicit functions using Jacobian (06 marks)

B)Logarithm of complex numbers (06 marks)

C)Numerical methods (08 marks)


Q.5
A)Expansion of sine and cosine, etc (06 marks)

B)Expansion of series/indeterminate forms (06 marks)

C)Problems on Leibnitzs theorem (08 marks)

Q.6
A)Linear independent & dependent vectors /Numerical methods (06 marks)

B)Composite/Implicit functions (06 marks)

C)Fitting of curves/Regression (08 marks)
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Question from Blueprint
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All the best :)

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Complex numbers Formula Sheet

Here is an easy way to solve the complexity of complex numbers.
All important formulas pertaining to Complex numbers under one sheet. :)


Enjoy!!


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Maths - Sem1 - Matrices Solutions

Solutions for the questions on Matrices :










Enjoy!! :)

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Maths- Sem 1 - Matrices Questions

Here are few questions on matrices.


Enjoy Solving !! 

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Maths- Sem1 - Important Questions - Answer set (1)

Here are the answers for the questions posted earlier.










Cheers!
Enjoy!

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Maths - Sem 1 - Imp Questions (1)

 Here are few questions which would surely help you strengthen your concepts.




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BEE important questions and answers

Here are few important questions with answers.
Do check it out.

1. A 3 φ 4 pole 50 hz induction motor runs at 1460 r.p.m. find its % of slip.
Solution
N s = 120f/p
= 120*50/4
= 1500r.p.m. Running speed of motor
n= 1460r.p.m. Slip S=( N s–N)/ N s*100
=(1500-1460) x 100 / 1500
= 2.667%

2. Explain the working principle of Transformer.

A Transformer is a device that transfers electrical energy from one circuit to another by electromagnetic induction (transformer action). The electrical energy is always transferred without a change in frequency, but may involve changes in magnitudes of voltage and current. Because a transformer works on the principle of electromagnetic induction, it must be used with an input source voltage that varies in amplitude. There are many types of power that fit this description; for ease of explanation and understanding, transformer action will be explained using an ac voltage as the input source.
The amount of power used by the load of an electrical circuit is equal to the current in the load times the voltage across the load, or P = EI. If, for example, the load in anelectrical circuit requires an input of 2 amperes at 10 volts (20 watts) and the source is capable of delivering only 1 ampere at 20 volts, the circuit could not normally be used with this particular source. However, if a transformer is connected between the source and the load, the voltage can be decreased (stepped down) to 10 volts and the current increased (stepped up) to 2 amperes. Notice in the above case that the power remains the same. That is, 20 volts times 1 ampere equals the same power as 10 volts times 2 amperes.
A Transformer consists of the following parts
• A primary coil or winding.
• A secondary coil or winding.
• A core that supports the coils or windings.
The primary winding is connected to a 50 hertz ac voltage source. The magnetic field (flux) builds up (expands) and collapses (contracts) about the primary winding. The expanding and contracting magnetic field around the primary winding cuts the secondary winding and induces an alternating voltage into the winding. This voltage causes alternating current to flow through the load. The voltage may be stepped up or down depending on the design of the primary and secondary windings.


3. Calculate the amount of resistance (R) in a circuit, given values of voltage (E) and current (I):

clip_image001
the amount of resistance (R) offered by the lamp
clip_image002


4. calculate the amount of voltage supplied by a battery, given values of

current (I) and resistance (R):
clip_image003
the amount of voltage provided by the battery
clip_image004


5. Calculate the electric power in the given circuit and discuss the effect ofincreasing the battery voltage.

The formula for determining the power in an electric circuit: by multiplying the voltage in "volts" by the current in "amps" we arrive at an answer in "watts." Let's apply this to the given circuit.
clip_image006
In the above circuit, we know we have a battery voltage of 18 volts and a
lamp resistance of 3 Ω.
 Using Ohm's Law to determine current, we get:
clip_image007
Now that we know the current, we can take that value and multiply it by
the voltage to determine power:

clip_image008


6. What is meant by DEFLECTING TORQUE ?

The deflecting torque is produced by making use of one of the magnetic, chemical, electrostatic and electromagnetic induction effects of current or voltage and causes the moving system of the instrument to move from its zero position when the instrument is connected in an electrical circuit to measure the electrical quantity. The method of producing this torque depend upon the type of instrument. In attracting the type of instrument, this torque to equal to
Td = 1/2 I2 dL/dθ
Whereas in Pmmc instruments
Td = Bilur
Where B - magnetic density
i - current flowing l - length of coil
u - number of turn r - radius of coil



7. Find the voltage across each resistors in the following circuit.

clip_image009
clip_image011



8. The effective resistance of two resistors connected in series is 100 clip_image013.

When connected in parallel, then effective value in 24 ohm’s. Determine the value of two resistors
Series R1+R2=100 clip_image012 => R2 =100 - R1
R1R2/R1+R2 = 24
R1R2/100 = 24
R1R2 =2400
R1 (100-R1) = 2400
100 R1-R1^2-2400 = 0
R1^2-100 R1 + 2400 = 0 (R1-60)(R1-40) = 0
There Fore R1 = 60; R1 = 40
When R1 = 60 clip_image013[1] ; R2 = 100 – 60 = 40 clip_image013[2]

clip_image020
When R1 = 40 clip_image013[3] ; R2 = 100 - 40 = 60 clip_image013[4]
9. Find the Req between two points A & B.
1/Req = ½+1/3+1/3 = 1.17 (Req = 1/1.17= 0.8547)
1/Req = 2+.85+4
Req = 7.2 clip_image021


10. Explain about Krichoffs voltage and current laws.

Kirchhoff’s Current Law

The sum of current flowing towards a function is equal to the current
flowing away from it.
clip_image028clip_image030

Consider a function formed by 6 conductors. The current in these conductors are i1, i2, .i6.Some of these currents are flowing towards a 8 other’s away from A
According to Kirchhoff’s Law, i1+i4+i5+i6 = i2+i3
(Flowing towards) (Flowing away from A)
Kirchhoff’s Voltage Law (II Law)
In a closed circuit, the sum of the potential drops is equal to the sum of the potential resistance
clip_image043
ABCDA forms a closed circuit.
From A -> B, We have a potential drop of IR1. From D -> A, We have a potential drop of V. Sum of potential drops = IR1+IR2+IR3



Potential rise from D -> A =V IR1+IR2+IR3 = V

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BEE - circuit analysis

CIRCUIT ANALYSIS TECHNIQUES


Ohm’s Law
Temperature remaining constant, the potential difference (E) across the ends of a conductor is proportional to the current (I) flowing through it.
Mathematically, V=IR


Kirchhoff's Current Law (KCL)
"The algebraic sum of all currents entering and exiting a node must equal zero"
SIin = SIout
Similarly, at any instant the algebraic sum of all the currents at any circuit node is zero.
SI = 0


Kirchhoff's Voltage Law (KVL)
"The algebraic sum of all voltages in a loop must equal zero"
SE = SIZ
Similarly, t any instant the algebraic sum of all the voltages around any closed circuitis zero:
SE - SIZ = 0



Series and Parallel Resistor Combinations
There are two basic ways in which to connect more than two circuit components:
Series and Parallel.
• For analysis, series resistors/impedances can be replaced by an equivalent resistor/ impedance.
• Parallel resistors/impedances can be replaced by an equivalent resistor/ impedance.


Series Resistance
Two elements are in series if the current that flows through one must also flow through the other.
SERIES Req = R1 + R2 + R3
clip_image001
Req is equivalent to the resistor network on the left in the sense that they have the same i-v characteristics.


Parallel Resistance
Two elements are in parallel if they are connected between (share) the same two (distinct) end nodes.
clip_image003
PARALLEL
clip_image004


Inductors

Series and parallel inductances
clip_image006
clip_image008
Where, L = Inductance in henrys





Capacitors

Series and Parallel Capacitances

clip_image010
Where, C = Capacitance in farads




Mesh current method
The Mesh Current Method uses simultaneous equations, Kirchhoff's Voltage Law, and Ohm's Law to determine unknown currents in a network. It differs from the Branch Current method in that it does not use Kirchhoff's Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equations.


Steps to follow for the .Mesh Current method of analysis:
  1. Draw mesh currents in loops of circuit, enough to account for all components.
  2. Label resistor voltage drop polarities based on assumed directions of mesh currents.
  3. Write KVL equations for each loop of the circuit, substituting the product IR for E in each resistor term of the equation. Where two mesh currents intersect through a component, express the current as the algebraic sum of those two mesh currents.
  4. Solve for unknown mesh currents (simultaneous equations).
  5. If any solution is negative, then the assumed current direction is wrong!
  6. Algebraically add mesh currents to find current in components sharing multiple mesh currents.
  7. Solve for voltage drops across all resistors (E=IR).


Node voltage method
The node voltage method of analysis solves for unknown voltages at circuit nodes in terms of a system of KCL equations. This analysis looks strange because it involves replacing voltage sources with equivalent current sources.


Node voltage rules:
  1. Convert voltage sources in series with a resistor to an equivalent current source with the resistor in parallel.
  2. Change resistor values to conductance.
  3. Select a reference node(E0)
  4. Assign unknown voltages (E1)(E2) ... (EN)to remaining nodes.
  5. Write a KCL equation for each node 1,2, ... N. The positive coefficient of the first voltage in the first equation is the sum of conductances connected to the node. Repeat for coefficient of second voltage, second equation, and other equations. These coefficients fall on a diagonal.
  6. All other coefficients for all equations are negative, representing conductances between nodes. The first equation, second coefficient is the conductance from node 1 to node 2, the third coefficient is the conductance from node 1 to node 3. Fill in negative coefficients for other equations.
  7. The right hand side of the equations is the current source connected to the respective nodes.
  8. Solve system of equations for unknown node voltages.


NETWORK THEOREMS:

Thevenin's Theorem:
• Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor.
• Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis.


Norton's Theorem:
Any circuit with voltage sources, resistances (impedances) and open output terminals can be replaced by a single current source in parallel with single resistance (impedance), where the value of current source is equal to the current passing through the short circuit output terminals and the value of the resistance (impedance) is equal to the resistance seen into the output terminals.


Super Position Theorem:
In a linear, lumped element, bilateral electric circuit that is energized by two or more sources the current in any resistor is equal to the algebraic sum of the separate currents in the resistor when each sources acts separately. While one source is applied, the other sources are replaced by their respective internal resistances.

Super Position Theorem is not valid for power responses. It is applicable only for computing voltage and Current responses.

Maximum Power transfer Theorem:
The maximum Power transferred to a load resistor occurs when it has a value equal to the resistance of the network looking back at it from the load terminals (all sources being replaced by their respective internal resistances).

Duality:
Two electrical networks which are governed by the same type of equations are called duality.
For the networks to be duals it is necessary that the variables & elements of one network should also be the duals of variables & elements of other networks.
Method of drawing duality (or) dual network:
a) Place a dot in each independent loop of the given network. These dots placed inside the loops correspond to the independent node in the dual network.
b) A dot is placed outside the given network. This corresponds to the reference node of the dual network.
c) All the dots are connected by dotted lines crossing all the branches. The dotted lines should cross only one branch at a time.
d) The dual elements will form the branches connecting the corresponding nodes in the dual network.
Star-Delta Transformation:
The Star-Delta transformation techniques are useful in solving complex network.
A star network of three resistances RARB and RC connected together at common node N can be transformed into a delta network of three resistances RABRBC andRCA by the above equations:
clip_image012
In general terms:


Rdelta = (sum of Rstar pair products) / (opposite Rstar)

Delta-Star Transformation
A delta network of three impedances RABRBC and RCA can be transformed into a star network of three impedances RARB and RC connected together at common node N by the following equations:
clip_image014
In general terms:
Rstar = (adjacent Rdelta pair product) / (sum of Rdelta)

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