Hall Effect (Electronics devices and circuits)

in this post we will discuss about Hall Effect , this topic is very important for many competitive exams. if you have any doubt regarding this post then comment below, for video lectures on electronics devices and circuit subscribe our you tube channel, all links given below.

 

 

 

 

Hall effect

 

here are the important points for which hall effect is used

  •  To find the type of semiconductor specimen whether semiconductor is P-type or N-type
  •  To find the charge concentration
  •  To find the mobility of charge carrier

 

 

What is Hall Effect

 

Whenever a current carrying semiconductor is subjected to a magnetic field, there will be an induced electric field in the perpendicular direction of the plane consist of I & B, which intern exert a force over the carrier is known ass hall effect.

In above shown figure the direction of current is considers towards X-direction and direction of magnetic field is in Z-direction so induced electric field will have Y-direction because of Lorentz force  

 

 

working

 

we can easily find the types of semiconductor specimen whether it is P-type or N-type by hall effect. if the consider specimen is N-type then electron will be accumulated to the lower surface of the bar which makes hall voltage [latex]{ V }_{ H }[/latex] positive, similarly if the consider specimen is P-type then holes will be accumulated to the lower surface of the semiconductor bar which makes hall voltage [latex]{ V }_{ H }[/latex] negative.

 

NOTE – in any case (either P-type or N-type) the direction of exerted force over the charge carrier is negative Y-direction.

 

 

 Hall Voltage

 

[latex]{ V }_{ H }\quad =\quad \frac { BI }{ \rho W } [/latex]    —————(1)

 

Hall coefficient 

 

It is defined as the reciprocal of charge density (ρ)

 

[latex]{ R }_{ H }\quad =\quad \frac { 1 }{ \rho } [/latex]

 

by equation (1)

[latex]{ R }_{ H }\quad =\quad \frac { W.{ V }_{ H } }{ BI } [/latex]

 

 

Application of Hall Effect

 

  •  To determine type of extrinsic semiconductor (using polarity of [latex]{ V }_{ H }[/latex])
  •  To determine carrier concentration (n)
  •  Used in magnetic field meter
  •  As a hall effect multiplier
  •  To calculate mobility

 

 

NOTE – Hall effect is very less effective in case of metal because in metal n is very large

[latex]{ V }_{ H }[/latex] will be very small in order of micro volts.

 

 

 

 

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Effect of temperatures on semiconductor parameter

 

in this post we will discuss about effect of temperatures over different types of semiconductor parameters, this topic is very important for many competitive exams. if you have any doubt regarding this post then comment below, for video lectures on electronics devices and circuit subscribe our you tube channel, all links given below.

 

 

Effect of temperatures over different types of semiconductor parameters 

 

In this topic we will discuss effect of temperature on different parameters of semiconductor, such as

  1.   Effect of temperature on intrinsic concentration (ni)
  2.   Mobility (μ)
  3.   Conductivity (σ)

 

effect of temperature on intrinsic concentration ([latex]{ n }_{ i }[/latex]) 

 

[latex]{ n }_{ i }=\sqrt { { A }_{ 0 } } \quad { T }^{ \frac { 3 }{ 2 }  }\quad { e }^{ -\frac { { E }_{ GO } }{ 2KT }  }[/latex]

 

Where,  [latex]\sqrt { { A }_{ 0 } } [/latex]  is  constant that depends on material

T is temperature

[latex]{ E }_{ GO }[/latex] is energy gap

 

NOTE –  Energy gap [latex]{ E }_{ GO }[/latex] at 0k = 1.21 ev for Si

= 0.785 ev for Ge

K = Boltzmann constant = [latex]1.38\times { 10 }^{ -23 }\quad j/kelvin[/latex]

 

Having seen above relation we have come to know that intrinsic concentration is heavily depends upon the temperature. that means if we increase temperature then intrinsic concentration also increased and  if we decrease temperature then intrinsic concentration also decreased.

 

 

Energy Gap

 

[latex]Energy\quad gap\quad =\quad { E }_{ GO }-\beta T[/latex]

 

Where,

β is constant and very small

NOTE- Energy gap decreases with respect to temperature

 

 

Mobility 

 

[latex]\mu =\frac { V }{ E } [/latex]

 

where,

V = drift velocity

E = Electric Field

 

NOTE – mobility is inversely proportional to temperature i.e. when temperature increase then mobility decreases and vice versa also valid, because drift velocity in certain direction decreases when temperature increases.

 

 

Conductivity 

 

(1) for intrinsic semiconductor

 

[latex]{ \sigma  }_{ i }={ n }_{ i }q({ \mu  }_{ e }+{ \mu  }_{ h })[/latex]

 

As we already discuss intrinsic concentration is directly proportional to temperature and mobility is inversely proportional to temperature that means if we increases temperature then intrinsic concentration will increase and mobility will decrease, but the rate of decrement  is very small comparison to increment, so when temperature increase then conductivity of intrinsic semiconductor is also increases.

so we can say conductivity of intrinsic semiconductor is directly proportional to temperature.

 

 

For extrinsic semiconductor

N-type    [latex]{ \sigma  }_{ n }={ n }q{ \mu  }_{ e }[/latex]

P-type   [latex]{ \sigma  }_{ p }={ n }q{ \mu  }_{ h }[/latex]

 

NOTE – there is a little impact of the temperature over the majority charge carrier but in extrinsic semiconductor the mobility of charge carrier decreases heavily, hence we see that conductivity of extrinsic semiconductor decreases with respect to temperature.

 

 

 

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Electronics devices and circuits class notes (Semiconductor)

Semiconductors

 

 

N-Type semiconductors

 

n (majority carrier concentration in no of electron in CB/volume) 

p (minority carrier concentration in no of holes in VB/volume)

[latex]{ N }_{ D }[/latex] (donor concentration per unit volume)

 

P-Type semiconductors

 

p (majority carrier concentration in no of holes in VB/volume) 

n (minority carrier concentration in no of electrons in CB/volume)

[latex]{ N }_{ A }[/latex] (donor concentration per unit volume)

 

NOTE –  CB (conduction band)

VB (valance band)

 

 

Mass Action Law

 

Mass action law states that Under thermal equilibrium the product of the free electron concentration and the free hole concentration is equal to a constant equal to the square of intrinsic carrier concentration.

It is also known as low of conservation of charges

so mass action law is [latex]n.p={ n }_{ i }^{ 2 }[/latex]  —————(1)

 

 

Charge Neutrality Equation  

 

Charge neutrality occurs when all the charge in a volume adds to zero, it is neutral, neither positive or negative.

no of +ve charge = no of -ve charge

[latex]n+{ N }_{ A }=p+{ N }_{ D }[/latex]   —————(2)

 

Using mass action law and charge neutrality equation we can find the minority charge carrier concentration

 

For n-type semiconductor (n>>p)

 

so using (2) equation   [latex]n\quad \simeq \quad { N }_{ D }[/latex]

[latex]{ N }_{ A }=0[/latex]

Now by using mass action law

[latex]p\simeq \frac { { n }_{ i }^{ 2 } }{ { N }_{ D } } [/latex]

 

For p-type semiconductor (p>>n)

 

so using (2) equation [latex]p\quad \simeq \quad { N }_{ A }[/latex]

[latex]{ N }_{ D}=0[/latex]

Now by using mass action law

[latex]n\simeq \frac { { n }_{ i }^{ 2 } }{ { N }_{ A} } [/latex]

 

Current component in semiconductor

 

Basically there are two types of current component in semiconductors

  1.  Drift current density
  2.  Diffusion current density

 

Drift current density

 

  • Drift current density is due to free charges or potential gradient

[latex]E=-\frac { { dv } }{ dx } [/latex]

  • It is also known as conduction current density

 

Diffusion current density

 

  • Diffusion current density is due to presence of concentration gradient
  • It is not present in metal or conductors
  • It is only present in semiconductors

 

 

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GATE (2002-2007) questions on Linear Time Invariant system

 

In this post we are providing multiple choice questions on topic linear time invariant, that asked in previous year GATE paper(2002-2007). if you want video solution for this questions then comment below if we get more comments then we will provide video solutions for all this questions on our you tube channel ENGINEER TREE. if you have any GATE related queries then ask on our Facebook group, all links of our other social platforms are given below, you can follow us by them.

 

 

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Signal and System (Linear Time Invariant System MCQ’s ) GATE (2002-2007)

 

 

 

 

GATE 2002

(1) Convolution of x(t+5) with impulse function δ(t-7) is equal to\

  1.  x(t-12)
  2.  x(t+12)
  3.  x(t-2)
  4.  x(t+2)

 

 

GATE 2003

(2) Let x(t) be the input to a linear time invariant system. the required output is 4x(t-2). the transfer function of the system should be

  1.  [latex]{ 4e }^{ j4\pi f }[/latex]
  2.  [latex]{ 2e }^{ -j8\pi f }[/latex]
  3.  [latex]{ 4e }^{ -j4\pi f }[/latex]
  4.  [latex]{ 2e }^{ j8\pi f }[/latex]

 

 

GATE 2004

(3) The impulse response h[n] of a linear time invariant system is given by h[n] = u[n+3] + u[n-2] – 2u[n-7] where u[n] is the unit step sequence. the above system is

  1.  stable nut not causal
  2.  stable and causal
  3.  causal but unstable
  4.  unstable and not causal

 

 

GATE 2004

(4) A causal LTI system is described by the difference equation

2y(n) = αy(n-2) – 2x(n) + βx(n-1)  The system is stable only if

  1.  lαl = 2, lβl < 2
  2.  lαl > 2, lβl > 2
  3.  lαl < 2, any value of β
  4.  lβl < 2, any value of α

 

 

GATE 2005

(5) Which of the following can be impulse response of a causal system?

 

 

 

GATE 2005

(6) The sequence

 Will be

 

 

 

GATE 2006

(7) In the system shown below x(t) = (sint)u(t) In steady state, the response y(t) will be

  1.  [latex]\frac { 1 }{ \sqrt { 2 }  } sin(t-\frac { \pi  }{ 4 } )[/latex]
  2.  [latex]\frac { 1 }{ \sqrt { 2 }  } sin(t+\frac { \pi  }{ 4 } )[/latex]
  3.  [latex]\frac { 1 }{ \sqrt { 2 }  } { e }^{ -t }sin(t)[/latex]
  4.  [latex]sin(t)-cos(t)[/latex]

 

 

GATE 2006

(8) The unit step response of a system starting from rest is given by [latex]c(t)=1-{ e }^{ -2t }[/latex]  for t ≥ 0. The transfer function of the system is

  1.  [latex]\frac { 1 }{ 1+2s } [/latex]
  2.  [latex]\frac { 2 }{ 2+s } [/latex]
  3.  [latex]\frac { 1 }{ 2+s } [/latex]
  4.  [latex]\frac { 2s }{ 1+2s } [/latex]

 

 

 

GATE 2007

(9) A Hilbert transform is a

  1.  non-linear system
  2.  non-causal system
  3.  time-varying system
  4.  low-pass system

 

 

GATE 2007

(10) The frequency response of a linear, time-invariant system is given by

[latex]\frac { 5 }{ 1+j10\pi f } [/latex]

 

The step response of the system is

  1.  [latex]5(1-{ e }^{ -5t })u(t)[/latex]
  2.  [latex]5(1-{ e }^{ -\frac { t }{ 5 }  })u(t)[/latex]
  3.  [latex]\frac { 1 }{ 2 } (1-{ e }^{ -5t })u(t)[/latex]
  4.  [latex]\frac { 1 }{ 5 } (1-{ e }^{ -\frac { t }{ 5 }  })u(t)[/latex]

 

 

 

Answer key will be provided within 48 hours of this post

 

 

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GATE previous question on LTI System and its properties for GATE/IES/PSU

 

In this post we are providing multiple choice questions on topic linear time invariant, that asked in previous year GATE paper. if you want video solution for this questions then comment below if we get more comments then we will provide video solutions for all this questions on our you tube channel ENGINEER TREE. if you have any GATE related queries then ask on our Facebook group, all links of our other social platforms are given below, you can follow us by them.

 

 

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Signal and System (Linear Time Invariant System MCQ’s )

 

 

 

GATE 1990

(1) The impulse response and the excitation of a linear time invariant causal system are shown in figure (A) and (B) respectively. The output of the system at t=2 sec is equal to

 

  1.  0
  2.  1/2
  3.  3/2
  4.  2

 

 

GATE 1995 

(2) Let h(t) be the impulse response of a linear time invariant system, then the response of the system for any input u(t) is

  1.  [latex]\int _{ 0 }^{ t }{ h(\tau )u(t-\tau )d(\tau )}[/latex]
  2.  [latex]\frac { d }{ dt } \int _{ 0 }^{ t }{ h(\tau )u(t-\tau )d(\tau )}[/latex]
  3.  [latex]\int _{ 0 }^{ \tau  }{ h(\tau )u(t-\tau )d(\tau )}[/latex]
  4.  [latex]\int _{ 0 }^{ t }{ { h }^{ 2 }(\tau )u(t-\tau )d(\tau )}[/latex]

 

 

GATE 1996

(3) The auto correlation function of an energy signal has

  1.  no symmetry
  2.  conjugate symmetry
  3.  odd symmetry
  4.  even symmetry

 

 

GATE 1998 

(4) The unit impulse response of a linear time invariant system is the unit step function u(t), For t > 0, the response of the system to an excitation [latex]{ e }^{ -at }u(t)[/latex], a>0 will be

  1.  [latex]{ a.e }^{ -at }[/latex]
  2.  [latex]{ \frac { 1 }{ a } .(1-e }^{ -at })[/latex]
  3.  [latex]{ a.(1-e }^{ -at })[/latex]
  4.  [latex]{ (1-e }^{ -at })[/latex]

 

 

GATE 2000

(5) A linear time invariant system has an impulse response [latex]{ e }^{ 2t }[/latex], t>0. if the initial condition are zero and the input is [latex]{ e }^{ 3t }[/latex], the output for t>0 is

  1.  [latex]{ e }^{ 3t }-{ e }^{ 2t }[/latex]
  2.  [latex]{ e }^{ 5t }[/latex]
  3.  [latex]{ e }^{ 3t }+{ e }^{ 2t }[/latex]
  4.  None of these

 

 

GATE 2000

(6) Let u(t) be the step function. which of the waveform in the figure corresponds to the convolution of u(t)-u(t-1) with u(t)-u(t-2)?

 

 

 

GATE 2001

(7) The transfer function of a system is given by  [latex]H(s)=\frac { 1 }{ { s }^{ 2 }(s-2) } [/latex]

 

The impulse response of the system is

  1.  [latex]{ (t }^{ 2 }*{ e }^{ -2t })u(t)[/latex]
  2.  [latex]{ (t }*{ e }^{ 2t })u(t)[/latex]\
  3.  [latex]{ (t }{ e }^{ -2t })u(t)[/latex]
  4.  [latex]{ (t }{ e }^{ 2t })u(t)[/latex]

 

 

GATE 2001

(8) Let δ(t) denote the delta function. the value of integral [latex]\int _{ -\infty  }^{ \infty  }{ \delta (t)cos(\frac { 3t }{ 2 } )dt } [/latex] is

  1.  1
  2.  -1
  3.  0
  4.  π/2

 

 

GATE 2001

(9) The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by [latex]{ h }_{ 1 }(t)=1[/latex], [latex]{ h }_{ 2 }(t)=u(t)[/latex], [latex]{ h }_{ 3 }(t)=\frac { u(t) }{ (t+1) } [/latex] and [latex]{ h }_{ 4 }(t)={ e }^{ -3(t) }u(t)[/latex] where u(t) is the unit step function. which of these system is time invariant, causal and stable?

  1.  S1
  2.  S2
  3.  S3
  4.  S4

 

 

GATE 2002

(10) If the signal f(t) has energy E, the energy of the signal f(2t) is equal to

  1.  E
  2.  E/2
  3.  2E
  4.  4E

 

Answer key will be provided within 48 hours of this post

 

 

 

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All books mentioned above is very good books for signal and system, we would recommend schoums outline for gate preparation, it is wonderful book to clear basic concepts

 

 

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Fourier series in signal and system for GATE/IES/PSU

 

 

 

In this video i am going to explain about Fourier series basics, i am going to make full series of signal and system lectures in which i cover complete syllabus of signal and system mentioned GATE ECE 2018. more more video lectures subscribe our you tube channel ENGINEER TREE. if you have any queries or doubt than you can post it in to our Facebook group, to join our Facebook group link given below in description.

 

 

 

 

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Fourier series

 

In communication system Fourier series is tool which is use to decompose continuous periodic into its sinusoidal component. or we can say it decompose any periodic function or periodic signal into the sum of sin and cosine waves. by our video given above you can get clear idea how it decompose periodic function into sum of sin and cosine wave.

Types of Fourier Series

  • trigonometric or quadrature Fourier Series
  • polar Fourier Series
  • Exponential Fourier Series

 

Dirichlet condition

  • f(t) should be deterministic in -T/2 ≤ f(t) + T/2
  • Finite number of discontinuities
  • Finite number of minima and maxima
  • f(t) must be absolute integral over period T

 

important points 

 

Case 1.  when f(t) is even signal

  • for even signal only cosine functions are present (a0,an ≠ 0) (bn = 0)
  • for odd signal only sin functions are present (a0,an = 0) (bn ≠ 0)

 

Case 2. when f(t) is half wave symmetry

                f(t) = -f(t ± T/2) 

  • for half wave symmetry only odd harmonics present
  • in Fourier series on sin term with odd harmonics present

 

Case 3. when f(t) is full wave symmetry 

                     f(t ± T/2) = f(t)

  • for full wave symmetry only even harmonics are present

 

 

To buy signal and system practice book by kanodia click here

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All books mentioned above is very good books for signal and system, we would recommend schoums outline for gate preparation, it is wonderful book to clear basic concepts.

 

keep visiting for more practice questions

 

GATE previous year questions on Time Response- 2

 

 

 

In this post we are providing some MCQ on control system that asked in previous year GATE examination. if you have any queries or doubts then you can ask in our Facebook group, and for detail theory lectures of control system subscribe our you-tube channel ENGINEER TREE. all links given below to follow us

 

 

 

answer key will be provided within 48 hour of this post 

 

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Time response previous year gate questions

 

 

 

 

GATE 2000

(1) An amplifier with resistive negative feedback has two left half plane poles in its open loop transfer function. the amplifier

  1.  will always be unstable at high frequency
  2.  will be stable for all frequency
  3.  may be unstable, depending on the feedback factor
  4.  will oscillate at low frequency

 

 

GATE 2001 

(2) If the characteristic equation of a closed loop system is [latex]{ s }^{ 2 }+2s+2=0[/latex], then system is

  1.  over damped
  2.  critically damped
  3.  under damped
  4.  undamped

 

 

GATE 2002

(3) consider a system with transfer function [latex]G(s)=\frac { (s+6) }{ k{ s }^{ 2 }+s+6 } [/latex]. its damping ratio will be 0.5 when the value of k is

  1.  2/6
  2.  3
  3.  1/6
  4.  6

 

GATE 2002

(4) The transfer function of a system is

[latex]G(s)=\frac { 100 }{ (s+1)(s+100) } [/latex]

 

For a unit step input to the system the approximate settling time for 2% criterion is

  1.  100 sec
  2.  4 sec
  3.  1 sec
  4.  0.01 sec

 

 

GATE 2004

(5) A system described by the following differential equation [latex]\frac { { d }^{ 2 }y }{ d{ t }^{ 2 } } +3\frac { dy }{ dt } +2y=x(t)[/latex] is initially at rest. for input x(t)=2u(t), the output y(t) is

  1.  [latex](1-2{ e }^{ -t }+{ e }^{ -2t })u(t)[/latex]
  2.  [latex](1+2{ e }^{ -t }-2{ e }^{ -2t })u(t)[/latex]
  3.  [latex](0.5+{ e }^{ -t }+1.5{ e }^{ -2t })u(t)[/latex]
  4.  [latex](0.5+{ 2e }^{ -t }+2{ e }^{ -2t })u(t)[/latex]

 

 

GATE 2005

(6) In the derivation of expression for peak percent overshoot [latex]{ M }_{ P }={ e }^{ \frac { -\delta \pi  }{ \sqrt { 1-{ \delta  }^{ 2 } }  }  }\times 100%[/latex], which of the following condition is not required?

  1.  system is linear and time invariant
  2.  the system transfer function has a pair of complex conjugate poles and no zeros
  3.  there is no transportation in this system
  4.  the system has zero initial condition

 

 

GATE 2005

(7) A ramp input applied to an unity feedback system results in 5% steady state error, the type number and zero frequency gain of the system are respectively

  1.  1 and 20
  2.  0 and 20
  3.  0 and 1/20
  4.  1 and 1/20

 

 

GATE 2006 

(8) The unit step response of a system starting from rest is given by

[latex]c(t)=1-{ e }^{ -2t }\quad for\quad t\le 0[/latex]

 

The transfer function of the system is

  1.  [latex]\frac { 1 }{ 2+2s } [/latex]
  2.  [latex]\frac { 2 }{ 2+s } [/latex]
  3.  [latex]\frac { 1 }{ 2+s } [/latex]
  4.  [latex]\frac { 2s }{ 1+2s } [/latex]

 

 

GATE 2006

(9) Consider two transfer functions

[latex]{ G }_{ 1 }(s)=\frac { 1 }{ { s }^{ 2 }+as+b } \quad and\quad { G }_{ 2 }(s)=\frac { s }{ { s }^{ 2 }+as+b } [/latex]

 

The 3-db bandwidths of their frequency response are, respectively

  1.  [latex]\sqrt { { a }^{ 2 }-4b } ,\sqrt { { a }^{ 2 }+4b } [/latex]
  2.  [latex]\sqrt { { a }^{ 2 }+4b } ,\sqrt { { a }^{ 2 }-4b } [/latex]
  3.  [latex]\sqrt { { a }^{ 2 }-4b } ,\sqrt { { a }^{ 2 }-4b } [/latex]
  4.  [latex]\sqrt { { a }^{ 2 }+4b } ,\sqrt { { a }^{ 2 }+4b } [/latex]

 

 

GATE 2007

(10) A control system with PD controller is shown in the figure, if the velocity error constant Kv =1000 and the damping ratio δ=0.5, the values of Kp and Kd are

 

  1.  Kp=100, Kd=0.09
  2.  Kp=100, Kd=0.9
  3.  Kp=10,  Kd=0.09
  4.  Kp=10,  Kd=0.9

 

 

 

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Linear Circuit problem with solution video (Network Analysis)

 

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Difference between Linear Network and Bilateral

Network 

Linear network

Linear circuit is a circuit in which all the passive elements parameter like resistance, capacitance, inductance are constant with voltage and current source. and the circuit in which voltage and current sources are independent or directly proportional to the other voltage and current sources or their derivatives is also known as linear circuits.

 

Bilateral Network

Bilateral network is the network in which all the passive elements like inductor, resistor, capacitor gives response in both direction of sources. or in other word we can say properties and characteristics of elements are independent of direction of current flow.

 

Theory related to above given question 

 

 

Linear circuit always follows superposition principle that means if we apply linear combination of signal Ax1(t) + Bx2(t) then output F(t) of the circuit will be equal to the linear combination of the outputs due to the signals x1(t) and x2(t) applied separately.

 

                                               F(Ax1(t) +Bx2(t)) = A.F(x1(t) ) + B.F(x2(t))

 

 

Superposition theorem 

 

 

Superposition theorem state that the response (voltage or current) in any branch of a linear bilateral network contain multiple sources is equal to the algebraic sum of the response obtain by each source acting separately, during which time all other sources are made zero.

 

Note – voltage sources is made zero by short circuit and current source is made zero by open circuit.

Note – superposition theorem can not be applied to the power across the elements.

 

Fig-1

 

 Fig-2

 

  Fig-3

                                                             I = I1 + I2

 

 

In above figure you can see superposition principle. in fig-1 we have linear circuit and we have to find current I across R2 so to find this current we use superposition principle. in fig to we find current I1 by voltage source by made current source zero then we find current I2 by made voltage source zero. now by superposition principle total current will be

                                                             I = I1 + I2

 

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