EENG 31111 Linear Circuits

8/4/2019 EENG 31111 Linear Circuits

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RESISTANCE

Electrical resistance is a ratio of the degree to which an object opposes anelectric current through it, measured in ohms. Its reciprocal quantity iselectrical conductance measured in siemens.

Assuming a uniform current density, an object's electrical resistance is afunction of both its physical geometry and the resistivity of the material it is

made from:

Resistance,

where

l = length of conductor, measured in metersA = cross sectional area, measured in square meters = resistivity of the material, measured in ohmmeter

A

lR


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RESISTANCE

The SI unit of electrical resistance is the ohm, symbol . The resistance of anobject determines the amount of current through the object for a given potentialdiffrence across the object.

Where,R is the resistance of the object, measured in ohmsV is the potential difference across the object, measured in voltsI is the current through the object, measured in amperes

This ratio of Voltage divided by Electric current is also called the ChordalResistance.

For a wide variety of materials and conditions, the electrical resistance does notdepend on the amount of current through or the amount of voltage across theobject, meaning that the resistance R is constant.

I

VR
http://en.wikipedia.org/wiki/%CE%A9http://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/Electric_currenthttp://en.wikipedia.org/wiki/Constanthttp://en.wikipedia.org/wiki/Constanthttp://en.wikipedia.org/wiki/Electric_currenthttp://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/%CE%A9


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SERIES CIRCUITS

Series circuits are sometimes called current-coupled.

The current that flows in a series circuit has to flow through everycomponent in the circuit. Therefore, all of the components in a seriesconnection carry the same current.

To find the total resistance of all the components, add the individualresistances of each component:

For components in series with resistances R1, R2, etc. To find the currentI, use Ohms law.

To find the voltage across a component with resistance Ri, use Ohm's lawagain.

where Iis the current, as calculated above. The components divide thevoltage according to their resistances, so, in the case of two resistors.


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Equivalent resistance of resistors in series : R = R1 + R2 + R3 + ...

A series circuit is shown in the diagram above. The current flows through eachresistor in turn. If the values of the three resistors are:

With a 10 V battery, by V = I R the total current in the circuit is:I = V / R = 10 / 20 = 0.5 A. The current through each resistor would be 0.5 A.


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PARALLEL CIRCUITS

A parallel circuit is a circuit in which the resistors are arranged with theirheads connected together, and their tails connected together.

The current in a parallel circuit breaks up, with some flowing along eachparallel branch and re-combining when the branches meet again.

The voltage across each resistor in parallel is the same.

The total resistance of a set of resistors in parallel is found by adding upthe reciprocals of the resistance values, and then taking the reciprocal ofthe total:

Equivalent resistance of resistors in parallel: 1 / R = 1 / R1 + 1 / R2 + 1 /R3 +...


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A parallel circuit is shown in the diagram above. In this case thecurrent supplied by the battery splits up, and the amount goingthrough each resistor depends on the resistance. If the values ofthe three resistors are:

With a 10 V battery, by V = I R the total current in the circuit is: I= V / R = 10 / 2 = 5 A.

The individual currents can also be found using I = V / R.

The voltage across each resistor is 10 V,

so,I1 = 10 / 8 = 1.25 AI2 = 10 / 8 = 1.25 AI3=10 / 4 = 2.5 A

Note that the currents add together to 5A, the total current.


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CAPACITORS IN SERIES

The overall effect of connecting capacitors in series is to movethe plates of the capacitors further apart. This is shown in figure

below. Notice that the junction between C1 and C2 has both anegative and a positive charge. This causes the junction to beessentially neutral. The total capacitance of the circuit isdeveloped between the left plate of C1 and the right plate of C2.Because these plates are farther apart, the total value of thecapacitance in the circuit is decreased. Solving for the total

capacitance (CT) of capacitors connected in series is similar tosolving for the total resistance (RT) of resistors connected inparallel.


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Note the similarity between the formulas for RT and CT:

If the circuit contains more than two capacitors, use the above formula. If the circuitcontains only two capacitors, use the below formula:

Note: All values for CT, C1, C2, C3,... C n should be in farads. It should be evidentfrom the above formulas that the total capacitance of capacitors in series is less thanthe capacitance of any of the individual capacitors.

Example: Determine the total capacitance of a series circuit containing threecapacitors whose values are 0.01 mF, 0.25 mF, and 50,000 pF, respectively.


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The total capacitance of 0.008mF is slightly smaller than the smallestcapacitor (0.01mF).


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CAPACITORS IN PARALLEL

When capacitors are connected in parallel, one plate of eachcapacitor is connected directly to one terminal of the source,

while the other plate of each capacitor is connected to theother terminal of the source. Figure 3-14 shows all thenegative plates of the capacitors connected together, and allthe positive plates connected together. C T, therefore,appears as a capacitor with a plate area equal to the sum ofall the individual plate areas. As previously mentioned,capacitance is a direct function of plate area. Connecting

capacitors in parallel effectively increases plate area andthereby increases total capacitance.


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For capacitors connected in parallel the total capacitance is thesum of all the individual capacitances. The total capacitance of thecircuit may by calculated using the formula:

where all capacitances are in the same units.

Example: Determine the total capacitance in a parallel capacitivecircuit containing three capacitors whose values are 0.03 mF, 2.0

mF, and 0.25 mF, respectively.

Q.1 What is the total capacitance of a circuit in which fourcapacitors (10 mF, 21 mF, 0.1 mF and 2 mF) are connectedin parallel?


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KIRCHOFFS CURRENT LAW

Kirchhoff's Current Law, also known as Kirchhoff's Junction Lawand Kirchhoff's First Law, defines the way that electrical current isdistributed when it crosses through a junction - a point wherethree or more conductors meet.

Specifically, the law states that: The algebraic sum of current intoany junction is zero.Since current is the flow of electrons througha conductor, it cannot build up at a junction, meaning that currentis conserved: what comes in must come out.

When performing calculations, current flowing into and out of thejunction typically have opposite signs.

This allows Kirchhoff's Current Law to be restated as: The sum ofcurrent into a junction equals the sum of current out of thejunction.


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KIRCHOFFS CURRENT LAW

In the picture to the right, a junction of four conductors (i.e. wires) is shown.The currents i2 and i3 are flowing into the junction, while i1 and i4 flow outof it. In this example, Kirchhoff's Junction Rule yields the following equation:

i2 + i3 = i1 + i4


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KIRCHHOFFS VOLTAGE LAW

Kirchhoff's Voltage Law describes the distribution of voltage within a loop, orclosed conducting path, of an electrical circuit.

Specifically, Kirchhoff's Voltage Law states that: The algebraic sum of thevoltage (potential) differences in any loop must equal zero.

The voltage differences include those associated with electromagnetic fields

(emfs) and resistive elements, such as resistors, power sources (i.e.batteries) or devices (i.e. lamps, televisions, blenders, etc.) plugged into thecircuit.


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EXAMPLE 1

Kirchoff

Use the Kirchoff laws to determine the current flowing in each branch of the networkshown below.

MethodUse Kirchoff current law to label the current directions remembering that conventionalcurrent flows from the positive side of the battery to the negative side. Divide into twoloops and then use the Kirchoff voltage law. Solve the simultaneous equations to findthe three currents.


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Loop 1

V1 = 5V = -I1xR1 + I2xR2 = -5I1 + 10I2-5I1 + 10I2 = 5 Equation 1

Loop 2

V2 = 10V = I2xR2 + (I1 + I2)xR3 =10I2 + 2I1 + 2I2 = 2I1 +12I22I1+12I2 = 10 Equation 2

Solve the simultaneous equations, multiply equation 1 by 2 and equation2 by 5

-10I1 + 20I2 = 10 Equation 3

10I1 + 60I2 = 50 Equation 4


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Equation 3 + Equation 4

80I2 = 60I2 = 0.75A

Substitute this back into Equation 1 (or any of the otherequations)

-5I1+ (10 x 0.75 )= 5-5I1 = 5 7.5I1 = 0.5A

Finally, the current through R3 = I1 + I2 = 1.25A


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EXAMPLE 2

Kirchoff

Use the Kirchoff laws to determine the current flowing in each branch of the networkshown below.

MethodUse Kirchoff current law to label the current directions start at each voltage sourceand remember that conventional current flows from the positive side of the battery tothe negative side. Divide into two loops and then use the Kirchoff voltage law. Solvethe simultaneous equations to find the three currents.


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Loop 1

V1 = 4V = I1xR1 + (I1 + I2)xR3 = 2I1 + 4I1 +4I2 = 6I1 + 4I26I1 + 4I2 = 4 --- Equation 1

Loop 2

V2 = 2V = I2xR2 + (I1 + I2)xR3 = I2 + 4I1 + 4I2 = 4I1 +5I2

4I1 +5I2 = 2 --- Equation 2

Solve the simultaneous equations, multiply equation 1 by 2 and equation 2 by 3

12I1 + 8I2 = 8 Equation 312I1 + 15I2 = 6 Equation 4


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Equation 4 Equation 3

7I2 = -2 therefore I2 = - 0.286A

Substitute this back into Equation 1 (or any of the other equations)

6I1 = 4 - 4 (-0.286)I1 = 0.857AI1 + I2 = 0.571A

Negative current?

Note that the current I2 is negative because it actually flows in the opposite direction

to the direction labeled on the circuit at the beginning!


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EXAMPLE 3

Consider the figure shown below with the following Parameters:

V1 = 15V

V2 = 7V

R1 = 20

R2 = 5

R3 = 10

Find current through R3 using Kirchoff's Voltage Law.
http://en.wikiversity.org/wiki/Image:EE-102-L05-Fig2.jpg


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Solution:

We can see that there are two closed paths (loops) where we can apply KVL in, Loop1 and 2 as shown in the figure.

From Loop 1 we get:

V1 VR3 VR1 = 0

From Loop 2 we get:

V2 VR3 VR2= 0
http://en.wikiversity.org/wiki/Image:EE-102-L05-Fig3.jpghttp://en.wikiversity.org/wiki/Image:EE-102-L05-Fig3.jpghttp://en.wikiversity.org/wiki/Image:EE-102-L05-Fig3.jpghttp://en.wikiversity.org/wiki/Image:EE-102-L05-Fig3.jpg


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The above results can further be simplified as follows:V1 (I1 I2) * R3 I1 * R1 = 0

..... (1)

and

V2 + (I1 I2) * R3 I2 * R2 = 0

..... (2)

By equating above (1) and (2) we can eliminate I2 and hence get the following:

..... (3)

We end up with the above three equations and now substitute the Values given in theabove equations and solve the variables.


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It is clear that: from (3)

Substitute the Above Result into (2).

The Positive sign for I2 only tells us that Current I2 flows in the same direction to our initialassumed direction. Thus now we can calculate Current through R3 as follows:

The Negative sign for IR3 only tells us that Current IR3 flows in the same direction to I2direction.


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SUPERPOSITION THEOREM

In a network with multiple voltage sources, the current in any branch is the sum of thecurrents which would flow in that branch due to each voltage source acting alone with

all other voltage sources replaced by their internal impedances.

The goal of following text is to check superposition theorem.

Example 1

Step 1. Construct following circuit using Circuit Magic then run Node VoltageAnalysis. You can also calculate currents using other techniques.


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R2 = 10Ohms; R1 = 10Ohms; R3 = 10Ohms;E1 = 3V; E3 = 4V;

Solution

V1xG11 = I11G11 = 1/R1+1/R2+1/R3 = 0,3I11 = -E1/R1-E3/R3 = -0.70,3V1 = -0.7

V1 = -2.3333V2 = 0

I1 = (V1-V2+E1)/R1 = 0.0666667I2 = (V1-V2)/R2 = -0.233333I3 = (V1-V2+E3)/R3 = 0.166667

These values are used to check currents determined fromsuperposition theorem


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Step 2. Remove a voltage source from the third branch then run NodeVoltage Analysis.

R2 = 10Ohms; R1 = 10Ohms; R3 = 10Ohms;E1 = 3V;

Solution

V1xG11 = I11G11 = 1/R1+1/R2+1/R3 = 0.3I11 = -E1/R1 = -0.30,3V1 = -0.3V1 = -1V2 = 0

I1(1) = (V1-V2+E1)/R1 = 0.2I2(1) = (V1-V2)/R2 = -0.1I3(1) = (V1-V2)/R3 = -0.1

These values are used to determine current from superposition theorem.


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R2 = 10Ohms; R1 = 10Ohms; R3 = 10Ohms;E3 = 4V;

Solution

V1xG11 = I11G11 = 1/R1+1/R2+1/R3 = 0.3I11 = -E3/R3 = -0.40.3V1 = -0.4V1 = -1.3333V2 = 0

I1(2) = (V1-V2)/R1 = -0.133333I2(2) = (V1-V2)/R2 = -0.133333I3(2) = (V1-V2+E3)/R3 = 0.266667

Superposition theorem checking

I1 = I1(1)+I1(2) = 0.2-0.133333 = 0.0666666I2 = I2(1)+I2(2) = -0.1-0.133333 = -0.233333I3 = I3(1)+I3(2) = -0.1+0.266667 = 0.166667


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Example 2: Solve this example by using superposition theorem.

First turn the voltage source of 20V off (short-circuit with 0V), and get,

Second turn the voltage source of 32V off and get,

The overall currents can then be found to be,


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THEVENIN THEOREM

Any one-port (two-terminal) network of resistance elements and energysources is equivalent to an ideal voltage source in series with a resistor,where

is the open-circuit voltage of the network, and

is the equivalent resistance when all energy sources are turned off(short-circuit for voltage sources, open-circuit for current sources).

If we are only interested in finding the voltage, V across and current, I throughone particular resistor in a complex circuit containing a large number of resistors,voltage and current sources, we can ``pull'' the resistor out and treat the rest ofthe circuit as a Thevenin voltage source , and use Thevenin's theorem tofind V and I.

TV

TR

TT R,V


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Proof:

Assume with the load the network's terminal voltage and current are V and I respectively.

Replace the load by an ideal current source with I while keeping the terminal voltage Vthe same (b), the voltage or current anywhere in the system should not be affected.

Find the terminal voltage V in terms of the internal energy sources inside the network andthe external current source by superposition principle:

When the external current source is open circuit, the terminal voltage is dueonly to the sources internal to the network (c).

When all internal sources are turned off (short-circuit for voltage sources, open-circuit for current sources), the terminal voltage where is the equivalent

resistance of the network with all energy sources off (d).


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The overall terminal voltage is,

As far as the port voltage V and current I are concerned, aone-port network is equivalent to an ideal voltage source,

equal to the open-circuit voltage across the port,

in series with an internal resistance, which can

be obtained as the ratio of the open-circuit voltage and theshort-circuit current.


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EXAMPLE

A demonstration of the thevenin technique to find I1 in the diagram below :

The circuit to the right of points A and B is converted to a Thevenin sourceand resistance. With the 30v battery and left hand 10ohm resistor omitted,the Thevenin voltage becomes:

Vth = 40 x 10/30 = 400/30

Rth = 10||20 = 200/30 (10 ohm parallel with 20 ohm voltage source s/c )

I1 then becomes 30 - Vth / ( 10 + Rth )

I1 = 30 - 400/30 / 10 + 200/30= 900/30 - 400/30 / (300/30 + 200/30)

= 500/30 / 500/30= 1amp


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NORTHON THEOREM

The Norton theorem converts an ordinary circuit to an equivalent parallel circuit whichis a current source in parallel with a resistor. The technique is similar to the thevenintheorem and two points in a circuit must be defined, this is where the analysis willtake place.

As with Thevenin, the equivalent circuit is a current generator In and nortonequivalent resistance, Rn. These must be worked out to use the Norton theorem.The analysis points using Norton are short circuited, whereas using the TheveninMethod they are open circuit.

Value of Vth and Rth

The value of Vth is found by either measuring (if you don't know what's in the circuit)or be using circuit analysis. To find Rth ( with load removed) short circuit voltagesupplies, open current sources and calculate the equivalent resistance.


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EXAMPLE

Norton's theorem is demonstrated to find the current I1 in the diagram below:

The points A and B is where the Norton conversion takes place, the right 50 ohmresistor is removed, A and B are short circuited, see below:

In First the total current is calculated 100 / (50 + 100 || 50) = 1.2 amp. Using thecurrent division rule,In becomes 1.2 x 100 / (50+100) = 0.8 amp


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Rn is 50 + 100 || 50 = 83.3333333 ohm

The Norton equivalent circuit can now be completed with the right hand 50ohm resistor included:

The current I can now be found using the current division rule:

I = 0.8 x (83.3333 / ( 50 + 83.3333)) = 0.5 amp


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CHARGING OF A CAPACITOR

The figure below shows a Capacitor, (C) in series with a Resistor, (R) connectedacross a d.c. battery supply (Vs) via a switch. When the switch is closed, thecapacitor will gradually charge up through the resistor until the voltage across itreaches the supply voltage of the battery. The manner in which the capacitorcharges up is also shown below.


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RC TIME CONSTANT

Consider a capacitor initially charged to Vinit, and connected to a resistor: Here wewill apply the Kirchoff Voltage Law for this loop:

Vr+Vc=0 or Vr=-Vc

Because the same current runs through each element: Ir = Ic.

Using the constitutive law of the capacitor and resistor, Vc= 1/C Qc

Vr = R x Ir --- (1)

and the relation between charge and current, qc' = ic , together with the voltage andcurrent relations next to the figure above,

Qc' = - (1/RC) Qc --- (2)

To see it in terms of current, take its time derivative: Ic' = - (1/RC) Ic --- (3)

Or in terms of voltage, using (1), Vc' = - (1/RC) Vc --- (4)

All of these equations have an easy solution, e.g. --- (5)

We must choose A such that we match the stated initial condition that Vc = Vinit att=0. For that to be so, A must be Vinit, so the specific solution is

RC

t

CAeV

RC

t

initC

eVV


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What happens if the capacitor is initially discharged, then placed in a circuit with aresistor and a battery? Here is such a circuit. Note that there is a switch in the circuitabove the battery which will be closed at time t=0. At t=0, VC=0. To analyze thiscircuit, we will apply the Kirchoff Voltage Law that we will see in more detail in thenext section. The polarities of my meters has been assigned consistently with ourconvention:

Vr + Vc - Vb = 0

Because the same current runs through each element: Ir = Ic = I.Our constitutive laws of the two elements are:

Vc = 1/C Qc ; Vr = R I x r

Again, starting with our current/charge relation for the capacitor,Qc' = Ic = Vr/R = ( Vb - Vc )/R ; Qc' = Vb /R - (1/RC) Qc

Since we are not so interested in charge per se, but more so in the voltage changingacross the capacitor, we can rewrite this diffeq using the constitutive law of the

capacitor again: --- (7)

Again, we have a differential equation in which the derivative of something is equal toa constant times itself, plus another constant. We anticipate an exponential solutionfor this and the following is the solution for (7):


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EXAMPLE

Calculate the time constant of thefollowing circuit. The time constant,T of the circuit is found using thefollowing formula T = R x C givenin seconds.

Therefore, the RC circuits time constant:

T = R x C = 100k x 22uF = 2.2 Sec

a) What value will be the voltage across the capacitor at 0.7 timeconstants?At 0.7 time constants (0.7T) Vc = 0.5Vc.Therefore, Vc = 0.5 x 10V = 5V

b) What value will be the voltage across the capacitor after 1 timeconstant?

At 1 time constant (1T) Vc = 0.63Vc.Therefore, Vc = 0.63 x 10V = 6.3V

c) How long will it take for the capacitor to "fully charge" itself (5 timeconstants)?1 time constant (1T) = 2.2 seconds. Therefore, 5T = 5 x 2.2 = 11 Sec


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RL CIRCUIT

When switch is connected, the R-L combination is suddenly put acrossthe voltage Vvolt. The applied voltage Vmust, at any instant, supplynot only the ohmic drop iR over the resistance R but must alsoovercome the e.m.f. of self inductance i.e.

--- (1)

--- (2)

--- (3)

--- (4)

Multiplying both sides by (-R), we get, --- (5)

Integrating both sides, we get, --- (6)

dt

diL

dt

diLiRvvV

LR

dt

diLiRV

L

di

iRV

di

dt

L

R

iRV

diR

dtL

R

iRV

diR


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--- (7)

To begin with, when , t=0, i=0, hence putting these values in equation (7) above, we

get,--- (8)

Substituting this value of Kin the above given equation, we have,

--- (9)

--- (10)

--- (11)

where, time constant.

KtL

RiRVe

log

KV

elog

V

e

iRV

e tL

R

loglog

tL

RVe

iRV

e loglog

tt

LR

ViRV

e log

R

L


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RL Circuit


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--- (12)

--- (13)

Now, represents the maximum steady value of current that would eventually

be established through the R-L circuit.

--- (14)

The graph for the rate of rise of current with the time constant of the R-L circuit isshown in figure below

t

e

V

iRV

t

eR

Vi 1

t

meIi 1

R

V


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100

63.2

0

Sw. Closed Time

RVIm

tm

eIi 1

R

L

% ofMax.

Current,Im

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EENG 31111 Linear Circuits