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Circuit Theory (Phy301)
Marks: 25
Q 1:
Find the equivalent resistance R_{T} of given circuit. Write each step of the calculation to get maximum marks. Draw the circuit diagram of each step otherwise you will lose your marks.
Sol.(MISSING IMAGES are in the solution file)
Step-I:
Since 20Ω is parallel with 30Ω, their combined effect is
20Ω ║ 30Ω = (30 x 20)/(30+20) = 600/50 = 12Ω
The circuit adopts the form as
Step-II:
Now 12Ω is in series with 24Ω, so their combined effect = 12Ω + 24Ω = 36Ω
Step-III:
Here 12Ω and 36Ω are in parallel,
so 12Ω || 36Ω = (12 x 36)/(12+36)
= 432/48 = 9Ω
Step-IV:
Since 9Ω and 18Ω are in series, so their combined effect = 9Ω+18Ω = 27Ω
Step-V:
Here 20Ω and 27Ω are in parallel, so
20Ω || 27Ω = (20 x 27)/(20+27) = 540/47 = 11.5Ω
Step-VI:
Since 12Ω and 11.5Ω are in parallel, so
12Ω || 11.5Ω = (12 x 11.5)/(12+11.5) = 138/23.5 = 5.9Ω
Here 8Ω and 5.9Ω becomes in series, so their combined effect = 8Ω+5.9Ω = 13.9Ω
Hence equivalent resistance R_{T} of given circuit = 13.9Ω
Q 2:
Determine V_{R1}, R_{2}, and R_{3} for the given circuit. Mention the units of calculated value.
Sol.
Here converting current value12.3mA into Ampere A, we have
Since all resistors are in series, so same value of current 0.0123A flows through each resistance
So
Current flowing through resistance R_{1} (82Ω) is 0.0123A, therefore voltage drop is
To find R_{2}, we first need to find voltage drop across R_{2} that is. Therefore we have that is
Now
Finally
Q 3:
Answer the following questions.
Sol.
I. Here, let
Using, we have
Now if V_{2} represents the changed volts in the source then using
So we’ll set the source at 30 V.
II.
Here
We know that
III.
Here
Since the resistors are in parallel, so the voltage along all the resistors shall remain same but current will be equally distributed. The following illustrates the idea
R_{1 } R_{2 } R_{3 } R_{4}
Therefore
Now using, we have
Similarly
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Assignment 2(Fall 2012)
(Solution)
Circuit Theory (Phy301)
Marks: 20
(MISSING IMAGES are in the solution file)
Q 1:
Find the current through all meshes of given circuit. Label and identify each mesh. Draw the circuit diagram of each step otherwise you will lose your marks.
Solution:
Since a current source share two meshes, so we can assign meshes and super mesh as
Here
KVL equation for mesh
Writing KVL equation for super mesh:
KVL equation for
Multiplying equ. 1 by 5 and equ 2 by 7 & adding both , we have
Putting in equation (2)
So the calculated mesh currents are
(MISSING IMAGES are in the solution file)
Q 2:
Determine V0 for the given circuit using nodal analysis. Label each mesh otherwise you will lose your marks.
Sol:
There exists a voltage source between node V1 and V2, so labeling for node and super node we have circuit
Constraint Equation of super node is:
V1– V2 = -12V OR
V2 – V1 = 12V………….(1)
KCL equation for Super Node is as:
(V1 – 6) / 5K + V1 / 5K + V2 / 20K= 1mA
Taking L.C.M 20 and solving we have
(4V1 – 24 + 4V1 + V2) / 20K = 1mA
8V1 + V2 = 24 + 20
8V1 + V2 = 44……………(2)
Multiply equation 1 by 8, to make it equal as equation 2, we get:
8V2 – 8V1 = 96V, and adding in equation (2) we have
8V2 – 8V1 = 96V
V2 + 8V1 = 44 (+)
---------------------------------------------------
9V2 = 140V
9V2 = 140V
V2 = 140 / 9 V.
V2 = 15.55V.
This is the voltage drop across 20 K node (sum of series 10k &10k)
So the voltage drop across 10K can be calculated by voltage divider rule. Hence
V10K = (15.55x10/20
V10K = 7.775V
Assignment 3 (Fall 2012)
(Solution)
Circuit Theory (PHY301)
Marks: 20
Due Date: Jan 16, 2013
Q.1:
Using the Source transformation method, find Vo in the network given below. Label and draw each step where it required. Write each step of the calculation to get maximum marks and also mention the units of each derived value.
Solution: (MISSING IMAGES are in the solution file)
We want to calculate the voltage Vo using source transformation method. We proceed as
is in series with , so their combined effect is equal to .
is in series with 12 volts battery. So it can be converted into a current source of value = 12v / 3k (by ohm’s law), which is equal to 4mA.
Now voltage source has transformed in the current source and the modified circuit 4mA current source and resistor 3k will be in parallel. So modified circuit will be as,
Now we will combine current sources as 4m – 1m = 3mA, so our modified circuit will be as,
3mA source is parallel with 3k resistor. So it can be converted into a voltage source. By using ohm’s law we have the value of voltage source = 9 volts which is in series with a resistor of 3k , so our modified circuit will be,
Positive terminal of the 9 volts battery is connected with the negative terminal of
6 volts battery so they will be summed up as shown in the circuit below,
Now 3k resistor is in parallel with 6k resistor so
3k || 6k = (3k x 6k ) / 3k +6k
= 2k
So voltage drop across 2k? will be the same voltage as of parallel 3k and 6k ohm.
Now applying voltage division rule,
Q.2:
Find IO in the network given below using Norton’s theorem.
Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.
Solution: (MISSING IMAGES are in the solution file)
As we have current dependent voltage source, so first we find Vth
Removing load resistances to calculate Vth
Using mesh analysis to find Vth
KVL for loop 1,
HERE I2 = -4mA
Ix = I1 – I2
Ix = -2.33 mA – [-4mA]
Ix = 1.66 mA
And according to ohm’s Law:
V2k = Ix(2 k)
V2k = (1.66 mA)(2 k)
V2k = 3.33 V
VTH = V2k + 2kIx
Here
Ix = 1.66 mA
VTH = 3.33 V + 2k[1.66 mA]
VTH = 3.3 V + 3.3 V
VTH = 6.6 V
Calculating ISC
2000Ix
by nodal analysis V1 V2
6 k? 4 k?
2 k? 4 mA ISC
6 V
Ix I Ix
Equation for super node:
V1 + 6 V1 V2
+ + = 4 mA
4 k 2 k 4 k
3V1 + 6 + V2
= 4mA
4 k
3V1 + 6 + V2 = (4 mA)(4 k)
3V1 + V2 = 10 ………… (A)
Constraint Equation:
V2 – V1 = 2000Ix
V1
Ix =
2 k
V1
V2 – V1 = 2000
2 k
V2 – V1 = V1
V2 – V1 - V1 = 0
V2 – 2V1 = 0
V2 = 2V1
V1 = 0.5V2
Substituting the value of V1 in equation (A)
3[0.5V2] + V2 = 10
1.5V2 + V2 = 10
2.5V2 = 10
V2 = 4 V
So V2
ISC =
4 k
4 V
ISC =
4 k
ISC = 1mA
VTH
RN =
ISC
6.6 V
RN =
1mA RN = 6.6 k?
Norton’s equivalent circuit is:
According to current divider rule:
6.6 k
I4k = × 1mA
6.6 k + 6 k
I6k = 0.52348 mA
Assignment 4 (Fall 2012)
(Solution)
Circuit Theory (PHY301)
Marks: 25
Due Date: Feb 01, 2013
Q.1:
Calculate the peak load current for the circuit shown below.
Sol: (MISSING IMAGES and FORMULAS are in the solution file)
Here the input voltage is given an rms value. So this value is converted to a peak value V1(pk) = V1(rms)/0.707
= 150Vac/0.707
V1(pk) = 212.16 Vpk
We can now find V2(pk) as
V2(pk) = N2/N1 V1(pk)
= (5/10)(212.16 Vpk)
V2(pk) = 106Vpk
And the load voltage is:
VL(pk) = V2(pk) – VF ( where VF is forwarding voltage)
= 106 – 0.7
VL(pk)= 105.38Vpk
Now the peak load current will be
IL(pk) = VL(PK)/RL
= 105.38 Vpk/5k
IL(pk) = 21mApk
Q.2:
A silicon diode has n=1.Determine the change in voltage if the current changes from 0.5mA to 15mA.
Sol: (MISSING IMAGES and FORMULAS are in the solution file)
The general diode equation in forward biased is given as
i = Is eV/nVT (Where VT is constant and at room temp. its value is 25mv)
For varying currents I1 and I2
I1 = Is eV1/ nVT
I2 = Is eV2/ nVT
I1/I2 = e(v1 –v2)nVT
Which can be written as
V1–V2 = nVTln(I1/I2)………….(A)
Here,
VT= 25 mV n=1 I1=15mA and I2=0.5mA
Putting in (A)
V1–V2 = 1 x25 ln( 15/0.5)
So the change in the voltage will be
V1–V2 = 85mV
Q.3:
Give the brief answers;
1) Name the two possible failures in a transformer, state the most likely one.
Ans:
(i) Open winding is a most common transformer failure; short windings are much less common.
(ii)Operating the transformer above rated values will also cause transformer failure.
2) Name any four semiconductor devices and their use in electronic industry.
Ans:
Semiconductor devices are made by semiconductor material and are manufactured both as single discrete devices and as integrated circuits (ICs), which consist of few (as low as two) to billions—of devices manufactured and interconnected on a single semiconductor substrate, or wafer.
Following are given some important semiconductor devices
• Electronic transistor, junction transistor, transistor used for amplification
• Junction rectifier, semiconductor diode, diode used for rectification.
• Microprocessor chip, silicon chip, chip used in computers
• Laser diode used for optical storage, laser pointers, fiber optics, surgery.
• Light-emitting diode (LED) used as indicated lamps
• Solar cell used for converting solar energy to electrical energy.
3) Usually resistance of a conductor increases by increasing the temperature, how the conductivity of semiconductor is affected at increased temperature?
Ans:
By increasing temperature the conductivity of semiconductor increases due to breaking of covalent bonds and availability of more free electrons.
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