FINALTERM EXAMINATION
Fall 2009
PHY301- Circuit Theory (Session - 2)
Time: 120 min
Marks: 70
ExamDate: | 3/1/2010 12:00:00 AM |
Question No: 1 ( Marks: 1 ) - Please choose one
Charge of 2c and 5c will
► attract each other
► repel each other
► no effect
► cancel each other
Question No: 2 ( Marks: 1 ) - Please choose one
Which battery applies a greater potential difference?
► 12v
► 1.5v
► 10v
► 0.5v
Question No: 3 ( Marks: 1 ) - Please choose one
When one resistance in a series string is open
► The current is maximum in the normal resistances.
► The current is zero in all resistances
► The voltage is zero across the open resistance
► The current increases in voltage source
Question No: 4 ( Marks: 1 ) - Please choose one
Addition of impurities in semiconductor material to produce more current is called
► dopping
► bonding
► excitation
► intrinsic
Question No: 5 ( Marks: 1 ) - Please choose one
For a P-N junction under farward bias
► more farward current flows
► no farward current flows
► more reverse current flows
► infinite reverse current flows
Question No: 6 ( Marks: 1 ) - Please choose one
A diode is a
► one way conductor
► two way conductor
► three way conductor
► insulator
Question No: 7 ( Marks: 1 ) - Please choose one
In farward biased Characteristics of junction diode, current value is
► i=ISe^V/V
► Is=ie^V/VT
► i=Ise^VT/V
► i=Ise^V/VT
Question No: 8 ( Marks: 1 ) - Please choose one
The primary and secondary winding of transformer are
► physically touched
► physically isolated
► touched with conductor
► largely separated
Question No: 9 ( Marks: 1 ) - Please choose one
A certain transformer has 500 turns in the primary winding and 2500 turns in the secondary winding.The turn ratio is
► 0.5
► 25
► 25v
► 2500
Question No: 10 ( Marks: 1 ) - Please choose one
A transformer will be out of phase when
► output voltage is 0 degree out of phase with input voltage
► output voltage is 180 degree out of phase with input voltage
► output voltage is 360 degree out of phase with input voltage
► output voltage is same as input voltage
Question No: 11 ( Marks: 1 ) - Please choose one
If N denotes the total number of nodes, then number of nodal equations in nodal analysis will be
► Number of equations=N-1
► Number of equations=N-1/2
► Number of equations=N
► Number of equations=N-2
Question No: 12 ( Marks: 1 ) - Please choose one
Using superposition theorem, for a circuit containing independent sources, any remaining current source is
► replaced by capacitor
► replaced by short circuit
► made zero by replacing them by open circuit
► replaced by close circuit
Question No: 13 ( Marks: 1 ) - Please choose one
The PIV of a half wave rectifier circuit with a shunt capacitor filter is
► 2Vm
► Vm
► Vm/2
► 3Vm
Question No: 14 ( Marks: 1 ) - Please choose one
The primary function of a filter is to
► Minimize ac input variations
► Suppress odd harmonics in the rectifier output
► Stabilize dc level of the output voltage
► Remove ripples from the rectified output
Question No: 15 ( Marks: 1 ) - Please choose one
The base region of a p-n-p transistor is
► Very thin and heavily doped with holes
► Very thin and heavily doped with electrons
► Very thin and lightly doped with holes
► Very thin and lightly doped with electrons
Question No: 16 ( Marks: 1 ) - Please choose one
Voltage multipliers are usually used in low current high voltage applications e.g
► Stabilizer of refrigerator
► Microcontroller
► Cathode ray tube in television
► Remote control
Question No: 17 ( Marks: 1 ) - Please choose one
The current flow across the base-emitter junction of a p-n-p transistor consists of
► Mainly electrons
► Equal numbers of holes and electrons
► Mainly holes
► The leakage current
Question No: 18 ( Marks: 1 ) - Please choose one
In an P- type semicondutor material , the majority carriers are
► holes
► electrons
► ions
► protons
Question No: 19 ( Marks: 1 ) - Please choose one
Which light bulb will not glow in given circuit
► bulb 1 and 2
► all bulbs will not glow
► all bulbs will glow
► bulb 3
Question No: 20 ( Marks: 1 ) - Please choose one
Three bulbs are connected in series of a battery, what would happen if any one bulb is opened.
► half of current will flow
► same current will flow
► no current will flow
► 2/3 current will flow
Question No: 21 ( Marks: 3 )
A small light bulb with a resistance of 25Ω is connected across the same 220 volt power line. How much is the current ‘I’
Solution:
R=25 Ω
V=220 V
From ohm law
V=IR , I=V/R
So,
I=220/25=8.8 Ampere
Question No: 22 ( Marks: 3 )
Describe Source Transformation method for simplifying circuit..
Source Transformation:
If we have any source embedded within a network, say this source is a current source having a value I & there exists a resistance having a value R, in parallel to it. We can replace it with a voltage source of value V=IR in series with same resistance R.
The reverse is also true that is a voltage source V, in series with a resistance R can be replaced by a current source having a value I= V/R in parallel to the resistance R.
Parameters within circuit, for example an output voltage remain unchanged under these transformations.
Question No: 23 ( Marks: 3 )
Given below are two figures (a) and (b) having Diode, which diode is forward biased or reversed biased, tell reason.
Both A and B are reversed biased
In fig. a the negative of the battery is connected to the anode through a resistor and positive terminal of the battery is connected the cathode therefore no current will flow and we can say that diode is not existing and it will act as an open circuit.
In fig. b the negative terminal of the battery is connected to the anode through a resistor and positive terminal of the battery is connected the cathode
Question No: 24 ( Marks: 3 )
Crude logic gates circuits may be constructed out of nothing but diodes and resistors. Take for example this logic gate circuit:
Identify what type of logic function is represented by this gate circuit
This is an OR gate circuit.
Question No: 25 ( Marks: 3 )
Differentiate between Half wave & Full Wave Rectifier.
Half Wave Rectifier:
A half wave rectifier is a special case of a clipper. In half wave rectification, either the positive or negative half of the AC wave is passed easily, while the other half is blocked, depending on the polarity of the rectifier. Because only one half of the input waveform reaches the output, it is very inefficient if used for power transfer. Half-wave rectification can be achieved with a single diode in a one phase supply.
Full Wave Rectifier:
A full-wave rectifier converts the whole of the input waveform to one of constant polarity (positive or negative) at its output. Full-wave rectification converts both polarities of the input waveform to DC (direct current), and is more efficient. However, in a circuit with a non-center tapped transformer, four diodes are required instead of the one needed for half-wave rectification.
Question No: 26 ( Marks: 5 )
Keeping in mind the Source transformation method, how we will convert 10v voltage source into current source and 1mA into voltage source in the following circuit, Draw diagrams of converted circuit.
4 is in parallel with 3 so it becomes
4*3/4+3=1.71 ohm
0.71 again in series with 2 to give 2.71 ohm
1mA multiply with 12 k by source transformation to give 12 volts
2.71 is in series with 12 to give 12.71.
10 V is in series with 2.71 so, by source transformation
So ,
I= 10/12.71=0.786A
Question No: 27 ( Marks: 5 )
Did the power remain same or not, in either case provide a proof?
The power will not remain same.if primary side is equal to 120V than the secondaryside will be
120/10
=12V
Question No: 28 ( Marks: 5 )
Calculate I1, I2 and I3 values.
Solution:
I1=2mA….Ans
I2=-3mA….Ans
KVL for loop 3
2kI3 + 2k(I3 + I2) = 8
2kI3 + 2kI3+ 2kI2= 8
4kI3 – 8 =8
4kI3= 16
I3=4mA….Ans
Question No: 29 ( Marks: 5 )
For the circuit shown, find IE & RC
Question No: 30 ( Marks: 5 )
A particular diode, for which n=1, is found to conduct 3mA with a junction voltage of 0.7V to 0.8V. Find diode current..
Solution:
Id=Ise Vd/nVt
If Id=3*10^-3 A when Vd=0.7 then
3*10^-3= Ise0.7/25.2*10^-3
= Is=2.5905*10^-15
when Vd=0.7 Id=4.4619mA
when Vd=0.8 Id=0.15868mA
Question No: 31 ( Marks: 10 )
Design the circuit in the fig. so that V0 = 3V when IL =0, and V0 changes by 40mV per 1mA of load current .Find the value of R. (assume four diodes are identical) relative to a diode with 0.7v drop at I mA current. Assume n =1
SOLUTION:
V0 = 3V, when IL =0, therefore each diode should exhibit a drop of 0.75V. If IL =1mA, then Vo changes by 40mV and a change due to each diode is 10mV.
Hence
rd = 10mV/1mA
=10 Ohms
but
rd = nVT/ID
10 = 1 x 25m/ID
ID = 2.5mA
Hence
15 – 3 – IDRD = 0
R = (15 – 3)/ID
= (15 – 3)/2.5m
= 4.8k Ohms.
No comments:
Post a Comment