electrical science circuit analysis

7/30/2019 Electrical Science circuit analysis

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ELECTRICAL AND ELECTRONIC SCIENCE JAMES FLOUNDERS

Q1. Using Thevenins theorem find the current load between points A and B

All resistor values are in

A

10V

B

According to Tooley and Dingle (2012) Thevenins theorem states that:

Any two-terminal network can be replaced by an equivalent circuit

consisting of a voltage source and a series resistance (or impedance

if the source of voltage is an a.c source) equal to the internal resistance

(or internal impedance) see looking into the two terminals.

Using Thevenins theorem to find the current load across A and B we need to take three steps

1. Find the open circuit voltage of A and B (Vth)2. Replace the voltage source with a short circuit, or the current source with an open circuit.3. Find the resistance or impedance seen by the load.

Step 1. To make things easier I first need to re-draw the circuit to show AB open, this also completes

step 2. I can now then mark some new points on the diagram for I1, I2, I3, VX and X and use these

values within Node analysis to work out Vth.

2 6 6

4

9 5.5


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I1 Vx X I3 SOUD1219

I2

Vth

10v

Using node analysis at point X, I1 = I2 + I3The voltage dropped between the 4resistor and the 2 resistor = I1 = 10 Vx

6

I1 = I2 + I3

So 10 Vx = Vx + Vx

6 6 15

as Vx is still unknown we need to rearrange the equation to get Vx on its own, we choose a number

thats a factor of all our current values and times out the equation.

Using 30 as the figure to times out by, this equates to

50 5Vx = 5Vx + 2Vx => 50 = 5Vx + 5Vx + 2Vx = 12Vx

=> Vx = 50 = 4.16v so Vx = 4.16v12

and I3 = Vx = 0.27 so I3 = 0.27

15

so Vth = I3 (0.27) X 9 = 2.5v

To complete the final step we once again re-draw the circuit diagram, just for clarification.

We now need to calculate the impedance of the load, we start with working out the resistors in

parallel

so 6//6 = 6x6 = 36 = 36+6 12

2

4

6 6

9

6

92

4

6


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Now we add the parallel value up with the first series resistor value 3 +6 = 9

now we can work out Rth

9//9 = 9x9 = 81 = 4.5 Rth

9+9 18Now we have Values for Vth and Rth we can find the current between points A and B

IL A

------

Rth

2.5vVth

-------

B

IL = Vth = 2.5 = 0.25a

5.5+4.5 10

So the load between A and B is 0.25 amps.

4.55.5


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Q2. Using Nortons theorem find the current in the load between points A and B

A

60v

1000v

B

All resistor values are in ohms ()

Nortons Theorem states:

The current that flows in any branch of a network is the same as that which

would flow in the branch if it were connected across a source of electrical

energy, the short-circuit current of which is equal to the current that would

flow in a short-circuit across the branch, and the internal resistance of

which is equal to the resistance which appears across the open-circuited

branch terminals. John Bird (2001)

Firstly we need to find the short circuit current, putting a short across points A and B leaves us with

1000v

Now we can work out the resistance (Rsc)

100 + 300 = 400 + 600//60

= 400 + 600 x 60 = 454.54

600 + 60

Now using Ohms law we can work out the short circuit current

1000 = 2.2a

454.54

60

100600

130

170

180

300

100 300

600 60


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We now need to replace the voltage source with a short circuit and a current source to create an

open circuit, then we can find the Internal resistance Rin

A

B

Rin = (130+170)// 60 +(600//[300+100])

= 300//(60+600//400)

= 300//(60 + 600x400)

600+400

= 300//(60+240000) = 300

= 300//300 = 150

We can now find the current between points A and B

IL A

----------

---------

B

IL = 150 x 2.2 = 1 amp

(180+150)

So the current in the load between points A and B is 1 amp

300

100

600

60

130

170

2.2a 150

180


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Q3. Using the principle ofsuperposition, determine the voltage across the 10 resistor in

the circuit below, all resistor values are in

100v

3.4v

According to Tooley and Dingle (2012), superposition theorem states that:

In any network containing more than one voltage source, the current

in, or potential difference developed across, any branch can be found

by considering the effects of each source separately and adding their

effects. During this process, any temporarily omitted source must be

replaced by its internal resistance (or a short circuit if its a perfect

voltage source).

To make this simple we alter the layout of the diagram

x Vx

I2 I3

I1 Vout

100v V1

3.4v V2

We now short out V2 so we can work out the current using node analysis

I1 = I2 + I3

I1 = 100Vx = Vx + V

25 1 10

as Vx is still unknown we need to rearrange the equation to get Vx on its own, we choose a number

thats a factor of all our current values and times out the equation.

Using 50 as the figure to times out by, this equates to

5

101

20

20

5

110


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200 2Vx = 50Vx + 5Vx

=>200 = 57Vx

=>Vx = 200 = 3.5v

57 So Vx = 3.5vnow we can work out the current I3 using ohms law

3.5v = 0.35amps

10

We now repeat the process but this time short out V1

so I2 = I1 + I3

=> 3.4 Vx = Vx + Vx

1 25 10

We rearrange and times by 50 again leaving us with

170 50Vx = 2vx + 5Vx

=> 170 = 57Vx

=> Vx = 170 = 2.98v

57 so Vx = 2.98v

Now we can work out the current I3 using Ohms law

I3 = 2.98 = 0.298 amps

10

Using superposition the resultant current through the 10 resistor is

0.35 amps + 0.298 amps = 0.648amps

*The question states that we are to determine the voltage across the 10 resistor, I am not certain

on the method, my first thoughts are to add the two values of Vx 3.5 + 2.98 = 6.48v.If we are supposed to be looking for the volt drop across the 10 resistor then using ohms law

10 x 0.648a = 6.48v.


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Q4. An inductor having an inductance of 0.1 henrys is connected in parallel with an 8 resistor. The

frequency of the supply is 30/ hertz calculate:

I. The inductive reactance

To make things more clear I will draw out the circuit

f 30/ 0.1H

Inductive reactance = XL = 2FL

=> 2 x 30 x 0.1 = 6 So XL = 6 or (J6) we use a J to indicate an angle of phase shift.

II. Find the impedance (Z) Z= R+JX

(8)x(J6) x (8)-(J6)

(8)+(J6) (8)-(J6)

8x J6 = J48 J48x8= J384 J48 xJ6 = 288

64(-J48+J48-J36)

=> J384+288 = J384 + 288 = 2.88 + J3.84 so Z = 2.88+J3.84

(64+J36) 100

III. Find the magnitude of Z (IzI)

(2.88) + (3.84) =4.8

IV. Find the inductive susceptance (B)

B = 1 => 1 x by J => -J

XL J3.84 J 3.84 siemens

V. Find the conductance (G)

1 = 1 = 0.347 siemens

R 2.88

VI. Find the admittance (Y)

This is the inverse of the impedance 1 = 1

Z 2.88 + j384

4.8 tan-1(3.84) = 4.8 53.13

(2.88)

Y = 1

4.8 53.13 = 0.208 -53.13

VII. Find the magnitude of the admittance

1 = 0.283

4.8

8

~


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Q5. A parallel R-LC circuit consists of a coil of inductance 0.05H and a resistance of 5 in parallel with

a capacitor of 0.1uF. find the circuits resonant frequency, Q-factor and dynamic impedance.

I. resonant frequency

0.05 0.1

F= 1 1

2 R/L = 2 (0.05x0.1x10-6) = 2250khz

II. Find the Q-factor

Q = 2foL = 141.4

R

III. Find the dynamic impedance

ZD = L = 0.05 = 100k

CR 0.1x10-6 x5

5


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Q6. A series tunes circuit has C= 100nf, L=100uH and R=0.5. the circuit is required to have a

bandwidth of 4khz. Calculate the extra series resistance required to achieve this.

Fo = 1

2

1 ( 100x 10-6 x 100x10-9) = 50329.2

2

Bandwidth = Fw = Fo/Q

= 50329.2 = 12.6

4000

Using Q = 2FoL

R

= 250329.2 = 2.5

12.6

So the extra series resistance needed is 2

electrical science circuit analysis

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