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BEE

UNIT 1

DC & AC Circuits

Q1) Explain circuit law?

A1) Ohm's Law

Ohm's law is the relationship between voltage, current, and resistance in a circuit. Ohm's law can be written which is commonly used.

The current flowing through a resistance is equal to the voltage across the resistance divided by the resistance (I=V/R).

Voltage is equal to the current flowing through a resistor times its resistance (V=IR).

Resistance is equal to the voltage across a resistor divided by the current flowing through it (R=V/I).

Kirchhoff Voltage Law:

A loop is a path around a circuit that starts and ends in the same place.

Kirchhoff's voltage law (KVL) states that around a loop:
sum of voltage rises=sum of voltage drops--------------------(1)

Applying KVL around the three loops we get

V1 = Vx KVL around loop 1

Vx + V4 = V2 + V3 KVL around loop2

V1 + V4 = V2 + V3 KVL around loop3

Kirchhoff Current Law:

A node is a circuit connection that extends to all its neighboring elements but does not contain them.

Kirchhoff’s current law states that at a node:

Sum of the in currents = sum of out currents

Applying KCL at four nodes

0 = Ix + I1 + I2 KCL at node a

I2 = I3 KCL at node b

I3 + I4 = 0 KCL at node c

Ix + I1 = I4 KCL at node d

Adding equations 6-9 gives the following equation:

Ix + I1 + I2 + I3 + I4 = Ix + I1 + I2 + I3 + I4.

Q2) Explain the fundamental of electrical circuit ?

A2)Charge:

An electron carries a negative charge of 1.602 x 10-19 C. where the unit of charge is the Coulomb (C).

Charges of the same sign repel each other, while charges of opposite sign attract.

For example, consider two point charges, Q1 and Q2, separated a distance r as shown in Figure.

The force exerted by one charge on the other varies directly as the product of the charges and inversely as the square of the distance between them according to Coulomb’s law:

F = k Q1 Q / r 2

The constant of proportionality, k, is approximately 9x109 when the other parameters are given in SI units and the surrounding medium is free space.

Voltage:

Since charges exert forces on other charges, energy must be expended in moving a charge in the vicinity of other charges. Thus, charges produce a type of force field.

The unit of energy is the joule (J), defined as the energy expended in the exertion of ta force of one Newton in moving an object through a distance of one meter (1J = Nm

For example, consider moving a charge q from point a to point b along a chosen path in the presence of another charge Q as illustrated in Fig.

The movement of q from a to b as in Fig. requires work wba on our part then we say that the voltage of point b with respect to point a is the work per unit charge required to move the charge from point a to point b:

V ba = w ba / q -------------------------------(1)

Alternatively, we say that there is a potential difference between points a and b with point b at an assumed higher potential. If this is negative number, it means a is at higher potential.

Determining the voltage between two points that is established by point charge Q is a simple matter.

Current:

Current is the rate of movement of electric charge. If we observe the charge crossing an area of the cylinder in a certain time interval, ∆t, then the current directed to the right is defined as the net positive charge transferred to the right per unit of time:

i = ∆Q / ∆t

The unit of current is the Ampere (A), defined as the movement of one coulomb of net positive charge past a point in one second (1 A=1 C/s) This rate of movement may not be constant but may vary with time. Hence the general definition of current is:

i(t) = d Q(t) /dt

Consider the plot of net positive charge moving past a point shown in Fig. Over the time interval 1 s ≤ t ≤ 3 s the net positive charge passing to the right is increasing at a rate of 1C/s, and 1 A of current results. Over 3 s ≤ t ≤ 5 s the net positive charge passing to the right is decreasing at a rate of -1.5C/s and hence the current at any point in this time interval is -1.5A.

Conversely, the net positive charge passing a point is:

Q(t) = d

so that the net positive charge passing the point between two times is:

Q(t2) – Q(t1) = d

This Equation can be interpreted as the total area under the current -versus-time curve from the beginning of time to the present time.

Q3) State Ohms law ?

A3) Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions remain unchanged. Mathematically it can be represented as,

Potential difference ∝ Current

V ∝ I

( When the value of V increases the value of I increases simultaneously)

V = IR

Where,

V is Voltage in volts (V)

R is Resistance in ohm (Ω)

I is Current in Ampere (A)

The main applications of Ohm’s law are:

To determine the voltage, resistance or current of an electric circuit.

Ohm’s law is used to maintain the desired voltage drop across the electronic components.

Ohm’s law is also used in dc ammeter and other dc shunts to divert the current.

Q4) Explain Kirchoffs law ?

A4) The algebraic sum of currents meeting at a junction or node in a electric circuit is zero or the summation of all incoming current is always equal to summation of all outgoing current in an electrical network.

Explanation

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Assuming the incoming current to be positive and outgoing current negative we have

I e incoming current = ∑ outgoing current thus, the above Law can also be stated as the sum of current flowing towards any junction in an electric circuit is equal to the sum of currents flowing away from that junction

Kirchhoff’s Voltage Law (KVL)

statement : the algebraic summation of all Voltage in any closed circuit or mesh of loop zero.

ie ∑ Voltage in closed loop = 0 the summation of the Voltage rise (voltage sources) is equal to summation of the voltage drops around a closed loop in 0 circuit for explanation from here

determination of sigh and direction of currents (Don’t write in exams just for understanding)

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current entering a resistor is +ve and leaving should be –ve

now

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Potential Rise Potential Drop

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We are reading from +V to –V we are reading from –V to +V

potential drops potential rise

-V +V

Given Circuit

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First identify no of loops and assign direction of current flowing in loop

Note : no of loops in circuit = No, of unknown currents = no, of equations in the circuit

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Note : keep loop direction and current direction same ie either clockwise or anticlockwise for all loops I1 I2

Now according to direction of direction assign signs (+ve to –ve) to the resistors

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Note : voltage sources (V) polarities does not change is constant.

Note: for common resistor between 2 loops appearing in the circuit like R3 give signs according to separate loops as shown

 When considering only loop no 1 (+ R3 - )

- B

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Now consider diagram A and write equations

Two loops two unknown currents two equation

Apply KVL for loop ① [B. Diagram ]

(+ to drop -) = - sign and (- to rise +) = + sign

for drop = -sign

for rise = + sign

-() R2 is considered because in R3,2 currents are flowing and and we have taken () because we are considering loop no 1 and current flowing is in loop no 1

)

Similarly for loop no. 2 currents flowing is resistor R3 it should be )R3

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Consider loop no. 1 apply KVL

- …….①

Consider loop no. 2 apply KVL

-…….②

After solving equation ① and ② we will get branch current and

4. Using nodal analysis find voltage across 5resistor.

For V1

-1

For V2

-2

Solving 1 and 2:

For 5 voltage =

= -50.9 + 57.27

= 6.37V

Q5) Explain mesh current?

A5)

mesh current analysis circuit

Equation No 1 : 10 = 50I1 + 40I2

Equation No 2 : 20 = 40I1 + 60I2

Mesh Current Analysis

An easier method of solving the above circuit is by using Mesh Current Analysis or Loop Analysis which is also sometimes called Maxwell´s Circulating Currents method.

Instead of labelling the branch currents we need to label each “closed loop” with a circulating current.

Any required branch current may be found from the appropriate loop or mesh currents as before using Kirchhoff´s method.

For example: : i1 = I1 , i2 = -I2 and I3 = I1 – I2

Therefore,

mesh current analysis

These equations can be solved quite quickly by using a single mesh impedance matrix Z. Each element ON the principal diagonal will be “positive” and is the total impedance of each mesh. Where as, each element OFF the principal diagonal will either be “zero” or “negative” and represents the circuit element connecting all the appropriate meshes.

[ V] = [I] x [R] or [R] x [I] = [V]

I = V/R = R-1 x V

Inverse of R = [ 60 40

40 50 ]

|R| = (60 x 50 ) – (40 x 40 ) = 1400

R – 1 = 1/1400 [ 60 40

40 50]

having found the inverse of R, as V/R is the same as V x R-1, we can now use it to find the two circulating currents.

[I] = [ R -1] x [V]

[ I1]= 1/1400 [ 60 40 [10]

[I2 ] = 40 50 ] x [ -20]

I1 = (60 x 10 ) + (40 x -20 ) /1400 = -200/1400 = -0.143 A

I2 = (40 x 10 ) + (50 x -20) /1400 = -600/1400 = -0.429A

Where:

[ V ] gives the total battery voltage for loop 1 and then loop 2

[ I ] states the names of the loop currents which we are trying to find

[ R ] is the resistance matrix

[ R-1 ] is the inverse of the [ R ] matrix

and this gives I1 as -0.143 Amps and I2 as -0.429 Amps

As : I3 = I1 – I2

The combined current of I3 is therefore given as : -0.143 – (-0.429) = 0.286 Amp

5. Using superposition theorem find the value of I

1 =

2 =

1 + 2

Q6) Explain Thevenin’s and Norton theorem ?

Thevenin’s equivalent of A

Norton’s equivalent of A

sc = Vth/Rth

Norton’s equivalent is obtained by source conversion of thevenin’s equivalent circuit.

CONDITIONS FOR APPLICATION

For network A:

Network A should contain linear elements.

Network A can have independent and dependent current and voltage source.

If network A has dependent source than controlling parameter must lie in network A itself.

Network A should not have any source coupling and magnetic coupling.

For network B:

It can have linear and non linear elements.

It can have dependent and independent voltage and current sources.

It should not have any source and magnetic coupling with network A.

 Method for finding Rth :

Firstly, open circuit terminal A and B.

If network is operating with only independent sources:

Make all sources zero in network A.

Find out the equivalent resistance across terminal A and B.

2. If network A is operating with independent and dependent sources:

Make all independent sources zero in network A.

Connect a generation between A and B.

3. If network is operating with only dependent sources:

Connect generation between A and B

 Method for Vth:

First open circuit terminal A and B.

Find out the voltage between A and B this is Vth

 Method for Isc:

Isc =

Remove network B and S.C. the terminal A and B and current from terminal A to B Isc.

Q7) Explain single phase circuit?

A7) A purely or a non-inductive circuit which has inductance so small that at normal frequency its reactance is negligible compared to its resistance. If the circuit is purely non-inductive no reactance emf is set up and whole of the applied voltage is utilized in overcoming the ohmic resistance of the circuit.

The current flowing through a pure resistance changes, no back emf is set up, therefore, applied voltage overcomes the ohmic drop of i R only:

i.e i R = v

i = v/R = Vmax/R sin wt

current will be maximum when wt = π/2 or sinwt =1

Imax = Vmax/R

and instantaneous current may be expressed as:

i = Imax sin ωt

From the expressions of instantaneous applied voltage and instantaneous current, it is evident that in a pure resistive circuit, the applied voltage and current are in phase with each other, as shown by wave and phasor diagrams in (b) and (c) respectively.

The instantaneous power delivered to the circuit in question is the product of the instantaneous values of applied voltage and current.

i.e p = v i = Vmax sin wt . I max sin wt = Vmax I max sin 2 wt

p = Vmax . Imax / 2 (1 -cos 2 wt) sin 2 wt = 1 – cos 2 wt/2

p = Vmax I max /2 (1 – cos 2 wt)

= Vmax I max /2 - Vmax Imax /2 cos 2 wt

Average Power = Vmax I max / 2 – average of Vmax . I max /2 cos 2 wt

Since average of Vmax I max/2 cos 2wt over a complete cycle is zero

P = Vmax . I max/2 = Vmax / . I max / = VI watts

Where V and I are the rms values of applied voltage and current respectively.

Purely inductive circuit.

An inductive circuit is a coil with or without an iron core having negligible resistance.

When an alternating voltage is applied to purely inductive coil an emf known as self-induced emf is induced in the coil which opposes the applied voltage. Since the coil has no resistance at every instant of the applied voltage has to overcome this self-induced emf only.

Let the applied voltage v = Vmax sin wt

And self- inductance of coil = L henry

Self- induced emf in the coil eL = - L di/dt

Since the applied voltage at every instant is equal and opposite to the self induced emf

v = - eL

Vmax sin wt = -(-L di/dt)

di = Vmax/ L sin wt dt

Integrating both sides we ger

i = Vmax/L = Vmax / wL sin (wt – π/2)

current will be maximum when sin(wt – π/2) = 1 hence maximum value of current

I max = Vmax / wL

And the instantaneous current is expressed as

i = I max sin (wt – π/2)

Purely inductive circuit

ωL in the expression Imax = Vmax/ωL is known as inductive reactance and is denoted by XL i.e., XL = ω L

If L is in henry and co is in radians per second then XL will be in ohms.

Power in Purely Inductive Circuit:

Instantaneous power, p = v × i = Vmax sin ω t Imax sin (ωt – π/2)

Or p = – Vmax Imax sin ω t cos ω t = Vmax Imax/2 sin 2 ωt

The power measured by wattmeter is the average value of p which is zero since average of a sinusoidal quantity of double frequency over a complete cycle is zero. Hence in a purely inductive circuit power absorbed is zero

Capacitive:

The charging current will flow only during the period of “build up” and will cease to flow as soon as the capacitor has attained the steady voltage of the source. This implies that for a direct current, a capacitor is a break in the circuit or an infinitely high resistance.

During the first quarter-cycle, the applied voltage increases to the peak value, and the capacitor is charged to that value. The current is maximum in the beginning of the cycle and becomes zero at the maximum value of the applied voltage, so there is a phase difference of 90° between the applied voltage and current. During the first quarter-cycle the current flows in the normal direction through the circuit; hence the current is positive.

In the second quarter-cycle, the voltage applied across the capacitor falls, the capacitor loses its charge, and current flows through it against the applied voltage because the capacitor discharges into the circuit. Thus, the current is negative during the second quarter-cycle and attains a maximum value when the applied voltage is zero.

The third and fourth quarter-cycles repeat the events of the first and second, respectively, with the difference that the polarity of the applied voltage is reversed, and there are corresponding current changes.

In other words, an alternating current flows in the circuit because of the charging and discharging of the capacitor. As illustrated in Figs.(b) and (c) the current begins its cycle 90 degrees ahead of the voltage, so the current in a capacitor leads the applied voltage by 90 degrees – the opposite of the inductance current-voltage relationship.

Let an alternating voltage represented by v = Vmax sin ω t be applied across a capacitor of capacitance C farads.

The expression for instantaneous charge is given as:

q = C Vmax sin ωt

Since the capacitor current is equal to the rate of change of charge, the capacitor current may be obtained by differentiating the above equation:

i = dq/dt = [ C Vmax sin wt] = w C Vmax cos wt = V max / 1/wC sin (wt + π/2)

Current is maximum when t=0.

Imax = Vmax / 1/wC

i = Imax sin (wt + π/2)

From the equations of instantaneous applied voltage and instantaneous current flowing through capacitance, it is observed that the current leads the applied voltage by π/2, as shown in Figs. (b) and (c) by wave and phasor diagrams respectively.

Capacitive Reactance:

1/ω C in the expression Imax = Vmax/1/ω C is known as capacitive reactance and is denoted by XC i.e., XC = 1/ω C

If C is in farads and ω is in radians/s, then Xc will be in ohms.

Power in Purely Capacitive Circuit:

p = v I = Vmax sin wt . I max sin (wt + π/2)

= Vmax . I max/2 x average of sin 2 wt

Over complete cycle =0

Hence power absorbed in a purely capacitive circuit is zero. The same is shown graphically in Fig. (b). The energy taken from the supply circuit is stored in the capacitor during the first quarter- cycle and returned during the next.

The energy stored by a capacitor at maximum voltage across its plates is given by the expression:

Wc = ½ C V2 max.

Q8) Derive emf equation?

A8) Single phase EMF generation:

Alternating voltage may be generated

1) By rotating a coil in a magnetic field

2) By rotating a magnetic field within a stationary coil .

The value of voltage generated depends upon

1) No. of turns in the coil

2) field strength

3) speed Equation of alternating voltage and current

N= No. of turns in a coil

Φm= Maximum flux when coil coincides with X-axis

ω= angular speed (rad/sec) = 2πf

At θ=ωt, Φ= flux component ⊥ to the plane =Φm cos ωt

According to the Faraday’s law of electromagnetic induction,

e -N dɸ/dt = - N d/dt ɸm cos wt = w N ɸm sin wt ----------------------------(1)

When e is maximum value of Em when sin Ɵ = sin 90 = 1

Ie Em = w N ɸm -----------------------------------------------------------------------(2)

From equation (1) and (2)

e = Em sin wt volt

Now current (i) at any time in the coil is proportional to the induced emf€ in the coil.

Hence i = Im sin wt amp

Q9) Explain

Real power

Apparent power

Reactive power

A9) Real power/ True power/Active power/Useful power : (P) it is defined as the product of rms value of voltage and current and the active component or it is the average or actual power consumed by the resistive path (R) in the given combinational circuit.

It is measured in watts

P = VI Φ watts / KW, where Φ is the power factor angle.

Apparent power : (S):- it is defined as product of rms value of voltage (v) and current (I), or it is the total power/maximum power

S= V × I

Unit - Volte- Ampere (VA)

In kilo – KVA

Reactive power/Imaginary/useless power [Q]

It is defined as the product of voltage, current and sine B and I

Therefore,

Q= V.I Φ

Unit –V A R

In kilo- KVAR

As we know power factor is cosine of angle between voltage and current

i.e. Φ.F= CosΦ

In other words also we can derive it from impedance triangle

Now consider Impedance triangle in R.L.ckt

From triangle ,

Now Φ – power factor=

Power factor = Φ or

Q10) Derive the expression for R-L circuit.

A10) Series R-L Circuit

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Consider a series R-L circuit connected across voltage source V= Vm sin wt

As some I is the current flowing through the resistor and inductor due do this current voltage drops arcos R and L R VR = IR and L VL = I X L

Total V = VR + VL

V = IR + I X L V = I [R + X L]

Take current as the reference phasor : 1) for resistor current is in phase with voltage 2) for Inductor voltage leads current or current lags voltage by 90 0.

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