Objective Questions Electrical Engineering
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1) Resistivity of a wire depends on
A. material
B. length
C. cross section area
D. all of above
A B C D
Submit RESET EXPLANATION GET ANSWER
Resistivity is a basic property of any materials. It is defined as the resistance offered by a cube of unit volume of the materials. Hence resistivity of a wire depends on its material.
2) When n numbers resistances of each value r are connected in parallel, then the resultant resistance is x. When these n resistances are connected in series, total resistance is
A. nx.
B. n2x.
C. x/n.
D. rnx.
A B C D
CHECK ANSWER RESET Reset GET ANSWER
n numbers resistances of each value r are connected in parallel, then the resultant resistance is x, that means r/n
= x ⇒ r = nx.When these n resistances are connected in series, total resistance is nr = n.(nx) = n2x [Since r = nx
alraedy proved]
A. r/2
B. 4r
C. 2r
D. r/4
A B C D
CHECK ANSWER Reset EXPLANATION Reset
Resistance r = ρl/a. When the wire of length l is stretched to 2l, then the cross - sectional area of the wire becomes a/2. Now new value of resistance, r' = ρ2l/(a/2) = 4ρl/a = 4r.
4) Kirchhoff's second law is based on law of conservation of
A. charge
B. energy
C. momentum
D. mass
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Kirchhoff's voltage law (KVL) is also called Kirchhoff's second law. The principle of conservation of energy implies that the directed sum of the electrical potential differences (voltage) around any closed network is zero.
5) One coulomb of electrical charge is contributed by how many electrons ?
A. 0.625 X 1019.
B. 1.6 X 1019.
C. 1019.
D. 1.6 X 1012.
CHECK ANSWER RESET EXPLANATION GET ANSWER
Electrical charge of one electron is 1.6 X 10 - 19 coulomb, hence one coulomb implies 1/(1.6 X 10 - 19) or 0.625 X
1019 numbers of electrons.
6) Tow bulbs marked 200 watts - 250 V, and 100 watts - 250 V are joined in series to 250 V supply. The power consumed by the circuit is
A. 33 watt
B. 200 watt
C. 300 watt
D. 67 watt.
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
The resistance of first and second bulb are (250)2/200 and (250)2/100 Ω respectively. The total resistance when
the bulbs are connected in series will be (250)2/200 + (250)2/100 Ω. The total power consumption when they
joined in series to 250 V supply. The power consumed in the circuit will be (250)2/{(250)2(1/200 + 1/100)} =
20000/300 = 67 watt.
7) Ampere second is the unit of
A. conductance
B. power
C. energy
D. charge
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Electrical current is transfer electrical charge per second. Therefore Ampere = coulomb/second hence coulomb = ampere X second or ampere second.
A. volt/ampere
B. volt ampere
C. watt
D. joule/second
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Unit of electrical power is watt and watt means joule/second. Again electrical power = voltage X current, hence volt ampere may be another expression for unit of power. But impedance = voltage/current, hence volt/ampere may be expression for unit of impedance not power.
9) One kilowatt hour is same as
A. 36 X 105 watt
B. 36 X 105 ergs
C. 36 X 105 joules
D. 36 X 105 BTU
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Kilowatt hour is the unit of energy and 1 kilowatt hour = 1000 X 1 watt X 3600 second = 36 X 105 watt second
= 36 X 105 joule.
10) An electric current of 6 A is same as
A. 6 joule/second
B. 6 Coulomb/second.
C. 6 watt/second
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Current is rate of charge transferred per second. A current of 6 ampere implies 6 coulomb charge transferred through a cross section of conductor per second. Therefore 6 Ampere = 6 coulomb/second.
11) A circuit contains two un equal resistor in parallel
A. voltage drops across both are same
B. currents in both are same
C. heat losses in both are same
D. voltage drops are according to their resistive value
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Whatever may be the value of resistance the voltage drops, across all the resistors connected in parallel, are always same
12) Conductance of any conductor is expressed as
A. ampere/watt
B. mho
C. volt2/watt
D. watt/ampere2
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Conductance is reciprocal of resistance that means conductance = (resistance) - 1 . Hence unit of conductance will
13) A copper wire of length l and diameter d has potential difference V applied at its two ends. The drift velocity is vd.
If the diameter of the wire is made d/2, then the drift velocity becomes
A. vd.
B. 4vd.
C. vd/4.
D. vd/2.
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
The drift velocity is a basic property of conductor material and hence it does not depend upon the length or diameter of the conductor.
14) Two resistances R1 and R2 give combined resistances 4.5Ω and 1Ω when they are connected in series and
parallel respectively. What would be the values of these resistances ?
A. 3Ω and 6Ω
B. 1.5Ω and 3Ω
C. 3Ω and 9Ω
D. 6Ω and 9Ω
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Here, R1 + R2 = 4.5...(1)
and R1.R2/(R1 + R2) = 1
⇒ R1.R2/4.5 = 1
⇒ R1.R2 = 4.5 ...(2)
Combining (1) & (2) we get R1 = 1.5 Ω or 3 Ω and R2 = 3 Ω or 1.5 Ω
15) Which of the following may be value of resistivity of copper
A. 1.7 X 10 - 6
B. 1.7 X 10 - 5
D. 1.7 X 10 - 3.
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
The value of resistivity of copper is 1.7 X 10 - 6
16) Mass of a proton is how many times greater than mass of an electron
A. 184000
B. 18400
C. 1840
D. 184
A B C D
CHECK ANSWER RESET EXPLANATION GET ANSWER
Mass of a proton is 1840 times greater than mass of an electron.
Multiple Choice Questions on Electric Current
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1. One coulomb of electrical charge is contributed by how many electrons ? A. 0.625 X 1019.
B. 1.6 X 1019.
C. 1019.
D. 1.6 X 1012.
Electrical charge of one electron is - 1.6 X 10 - 19 coulomb, hence one coulomb implies
1 / | - 1.6 X 10 - 19 | or 0.625 X 1019 number of electrons.
2. 5 X 1016 electrons pass across the section of a conductor in 1 minutes and 20 seconds. The current flowing
. 0.1 mA.
A. 1 mA.
B. 10 mA. C. 100 mA.
The charge of an electron is - 1.6 X 10-19 coulomb. Therefore total negative charge passes across the
section of a conductor in 1 minute and 20 seconds is
5 X 1016 X 1.6 X 10-19 = 8 X 10 - 3. Therefore, charge passes across the section in one second, is 8 X 10 -
3 coulomb/80 second = 10 - 4 coulomb/second (or Amp) = 0.1 mA.
3. An electric current of 6 A is same as . 6 joule/second.
A. 6 Coulomb/second. B. 6 watt/second. C. none of the above.
Current is the rate of charge transferred per second. A current of 6 ampere implies 6 coulomb charge transferred through a cross section of conductor per second. Therefore 6 ampere = 6 coulomb/second.
4. Ampere-second is the unit of . conductance.
A. power. B. energy. C. charge.
Electrical current is nothing but rate of flow of charge per second. Therefore, ampere = coulomb/second hence coulomb = ampere × second or ampere-second.
5. The current in a circuit follows the relation i = 100sinωt. If frequency is 25 Hz how long will it take for the current to rise to 50 A ?
. 1 ms.
A. 3.33 ms. B. 10 ms.
C. 20 ms.
Let, at t second the instantaneous value of current i = 50A.
∴ 50 = 100sinΩt
or, sinωt = 1/2
or, ωt = &pi ⁄ 6
Now, ω = 2π.f = 2πX25 = 50π, [Since frequency f = 25 Hz.]
Therefore, t = π ⁄ (50X6Xπ) = 1 ⁄ 300 second = 3.33 ms.
6. The equation of a current is given by i = Imsin2ωt. The frequency of the current in Hz is
. ω ⁄ 2π. A. ω ⁄ 2. B. 2ω ⁄ π. C. ω ⁄ π.
The general equation of a current wave is i = Imsinωt = Imsin2πf.t ...(1)
Where, f is the frequency of the current wave. Here, the given equation is
i = Imsin2ωt...(2)
Comparing, (1) & (2) we get, 2ωt = 2πft or, ω = π.f or, f = ω ⁄ π
7. The equation of alternating current is i = 42.4sin628t. Then the average value of current is . 42.42 A.
A. 27 A. B. 38 A. C. 22 A.
The relation between average value and peak value of a sinusoidal waveform is given by
Here, Im = 42.4 A Therefore, I0 = 0.636X42.4 = 26.97 A
8. If 1 A current is flowing through a series circuit having 100 resistors of each having resistance of 1 Ω. What will be the current in the circuit where, these 100 resistors are connected in parallel ?
. 10 A.
B. 1000 A. C. 10000 A.
Let , the applied voltage across the circuit be V. Therefore, current of
1A = V/(100 X 1 Ω)=100 V
Now when these 100 resistors of each 1 Ω will be connected in parallel, the equivalent resistance of
combined circuit will be 1/100=0.01 ohm
Now, the current will be V/R=100/0.01=100×100= 104 A.
9. In the figure shown, what will be the current passing through 2 Ω resistor ?
. 0.25 A.
A. 0.75 A. B. 0.5 A. C. 1 A.
According to current division law, required current,
10.A copper conductor of one square millimetre can safely carry a current of . 100 A.
A. 50 A. B. 25 A. C. 10 A.
It is rating of copper conductor. For 1 sq mm safe current is 10 A. In case of 1.5 sq mm the safe current is 15 A.
11.For carrying an electric current of 75 A an aluminium conductor should have a minimum cross-section of . 25 mm2.
A. 10 mm2.
B. 15 mm2.
C. 20 mm2.
The current carrying capacity of aluminium conductor is near about 3 A/mm2.
12.A copper wire of length l and diameter d has potential difference V applied across its two ends. The drift velocity is vd. If the diameter of the wire is made d/2, then the drift velocity becomes
. vd.
A. 4vd.
B. vd/4.
C. vd/2.
The drift velocity is a basic property of conductor material and hence it does not depend upon the length or diameter of the conductor.
13.Mass of a proton is how many times greater than mass of an electron? . 184000.
A. 18400.
B. 1840.
C. 184.
Mass of a proton is 1840 times greater than mass of an electron.
14.An electric current of 6 A is same as . 6 joule/second.
A. 6 coulomb/second. B. 6 watt/second.
C. none of the above.
Current is rate of charge transferred per unit time. A current of 6 ampere implies 6 coulomb charges transferred through a cross section of conductor per second. Therefore, 6 amp = 6 coulomb/second.
15.On which factors does the severity of electric shock depends ? . Only on pathway through the body.
A. Only on the type of supply ac/dc. B. Only on magnitude of voltage. C. All of above.
The severity and effects of an electrical shock depends on a number of factors, such as the pathway through the body, the amount of current, the length of time of the exposure, whether the skin is wet or dry, magnitude of voltage, and type of supply ac or dc.
16.Five coulomb of electrical charge is contributed by how many electrons? . 3.125 X 1019.
A. 1.6 X 1019.
B. 1019.
C. 1.6 X 1012.
Electrical charge of one electron is -1.6 X 10 - 19, hence five coulomb implies
5 / | - 1.6 X 10 - 19 | or 3.125 X 1019 numbers of electrons.
17.The transient current in a series AC circuit is given by I(s)=(s+1) ⁄ (s2+s+1). Find the initial current
. 2 Amp.
A. 1 Amp. B. 0 Amp. C. 8734 Amp.
. arc welding.
A. resistance welding. B. gas welding.
C. electric arc welding.
Alternating current can provide any desired combination of voltage and current by means of transformer. So, resistance welding can be suitably controlled using alternating current.
19.The rms value of a half wave rectified symmetrical square wave current of 2A is given by . 0.707 A.
A. 1 A. B. 1.414 A. C. 1.732 A.
Vrms = Vm/√ 2 = 2/√ 2 = 1.414 A.
20.How many coulombs of charge flows through a circuit carrying 5A in 5 min ?
. 1500.
A. 150.
B. 15.
C. 1.
The current means the rate of charge transfer per second. That means current I = Q/t Here, I = 5A, and t = 5 minutes = 5 x 60 = 300 sec. Therefore, total charge flows during 5 minutes is 5 x 300 = 1500 coulomb.
Multiple Choice Questions on Electrical Resistance |
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1. Two resistances R1 and R2 give combined resistances 4.5Ω and 1Ω when they are connected in series and
parallel respectively. What would be the values of these resistances? A. 3Ω and 6Ω.
B. 1.5Ω and 3Ω. C. 3Ω and 9Ω. D. 6Ω and 9Ω. Here, R1 + R2 = 4.5...(1) and R1.R2/(R1 + R2) = 1 ⇒ R1.R2/4.5 = 1 ⇒ R1.R2 = 4.5 ...(2)
Combining (1) & (2) we get R1 = 1.5 Ω or 3 Ω and R2 = 3 Ω or 1.5 Ω
2. Which of the following may be value of resistivity of copper . 1.7 X 10 - 6.
A. 1.7 X 10 - 5.
B. 1.7 X 10 - 4.
C. 1.7 X 10 - 3.
The value of resistivity of copper is 1.7 X 10 - 6
3. Two equal resistors R connected in series across a voltage source V dissipate power P. What would be the power dissipated in the same resistors when they are connected in parallel across the same voltage source ?
. 4P.
A. P.
B. 2P.
C. 16P.
Say R is the resistance of the identical two resistors. When they are connected in series across a voltage source V, the equivalent resistance of the combination is 2R and then total power dissipated by the
resistors will be P = V2/2R. When they are connected in parallel across the same voltage source V, the
equivalent resistance of the combination is R/2 then total power dissipated by the resistors will be V2/R/2
= 4V2/2R = 4P.
4. Two identical resistors are first connected in parallel then in series. The ratio of resultant resistance of the first combination to the second will be
A. 0.25.
B. 2.
C. 0.5.
Let us consider the value resistance of the equal resistors is R. So equivalent resistance of parallel combination of the resistors is R/2, and equivalent resistance of series combination of the resistors is 2R. So ratio of these two combination will be (R/2)/2R = 1/4 = 0.25
5. The ratio of the resistance of a 200W, 230V lamp to that of a 100W, 115V lamp will be . 0.5.
A. 2.
B. 4.
C. 0.25.
Resistance of the first lamp R1 = 2302/200 Ω
Resistance of the first lamp R2 = 1152/100 Ω
Therefore, R1/R2 = (2302/200)/(1152/100) = 2.
6. The resistance of 200W 200V lamp is . 100 Ω.
A. 200 Ω. B. 400 Ω. C. 800 Ω.
Resistance R = V2/W. Here, V = 200 V and W = 200 watts.
Therefore, resistance of 200W 200V lamp is 2002/200 = 200 Ω.
7. Two 1 kΩ 1 W resistors are connected in series. Their combine resistance and wattage will be . 2 kΩ, 0.5 W.
A. 1 kΩ, 1 W. B. 0.5 kΩ, 2 W. C. 2 kΩ, 1 W.
Wattage W = V2/R...(I)
Here, for each resistor, W = 1 W and R = 1 kΩ and putting these values in equation (I), we get V2 = 1,
When two 1 kΩ resistance are connected in series, combined resistance will be 2 kΩ and putting this
value and and V2 = 1 in equation (I) we again get, combined wattage W = 1/2 watt.
8. Three 3 Ω resistors are connected to form a triangle. What is the resistance between any two of the corner? . 9 Ω.
A. 6 Ω. B. 3 Ω. C. 2 Ω.
Whenever we look at the said triangle from any two of its corners, we will find that it is just a parallel combination of one 3 Ω and one 6 Ω (3 + 3 = 6) resistor. Thus, the resistance aross these two corner points of the triangle will be 3X6/(3+6) = 18/9 = 2 Ω.
9. A wire of 0.14 mm diameter and specific resistance 9.6 μΩ - cm is 440 cm long. The resistance of the wire will be
. 9.6 Ω.
A. 11.3 Ω. B. 13.7 Ω. C. 27.4 Ω.
Cross - sectional area of the conductor is (π/4)X0.0142 = 0.000154 cm2.
The resistance will be 9.6X440/0.000154 = 27428571 μΩ = 27.4 Ω.
10.A 10 Ω resistor is stretched to increase its length double. Its resistance will now be . 40 Ω.
A. 20 Ω. B. 10 Ω. C. 5 Ω.
The resistance of a conductor is directly proportional to its length and inversely proportional to its cross - sectional area. As the wire is stretched to its double length, it's cross - sectional area will become half, hence, the resistance of the stretched wire will become 4 times.
11.Specific resistance is measured in
. mho.
A. ohm.
B. ohm - cm.
C. ohm/cm.
The resistance R = ρl/a where R is the resistance of any substance in ohm, ρ is the specific resistance of
material of that substance, l and a are length in cm and cross - sectional area in cm2 of that substance
respectively. Therefore, ρ = R.a/l and its unit may be ohm X cm2/cm or ohm – cm.
12.A wire of resistance R has it length and cross - section both doubled. Its resistance will become . 0.5R.
A. R.
B. 2R.
C. 4R.
The resistance of a conductor is directly proportional to its length and inversely proportional to its cross - sectional area. As the length and cross - sectional area both have become double, there will no change in resistance of the wire.
13.A cube of material of side 1 cm has a resistance of 0.002 Ω between its opposite faces. If the same volume of the material has a length of 4 cm and a uniform cross - section, the resistance of this length will be
. 0.128 Ω.
A. 0.064 Ω. B. 0.032 Ω. C. 0.016 Ω.
Here, the cube of material of side 1 cm has a resistance of 0.002 Ω between its opposite faces that means the resistivity of the material is 0.002 Ω. Now the length of the material has become 4 cm, hence for
same volume 1 cm3 the cross - sectional area of the material will be 1/4 or 0.25 cm2. The new resistance
will be 0.002X4/0.25 = 0.032 Ω.
14.Resistance of which material does not change with change in temperature ? . Platinum.
B. Brass. C. Tungsten.
Resistance is constant with changes in temperature in the metal alloy of constantan and manganin. This material has the temperature coefficient zero. They can be used for precision wire wound resistors, which do not change resistance when the temperature increased.
15.The hot resistance of a tungsten lamp is about 10 times the cold resistance. Accordingly, cold resistance of a 100W, 200V lamp will be
. 400 Ω.
A. 40 Ω. B. 4 Ω. C. 800 Ω.
Where, V is supply voltage and W is wattage rating of the lamp. Here, V = 200 V and W = 100 W,
∴ cold resistance of the lamp is 400 ⁄ 10 = 40 Ω
16.For same voltage, the ratio resistance of 100 W lamp to resistance of 25 W lamp is
. 16.
A. 4.
B. 1/4.
C. 1.
Where, V is supply voltage and W is wattage rating of the lamp.
From, above equation it is clear that, for fixed voltage source, resistance of an electric lamp is inversely proportional to its wattage rating. That means, ratio of resistance will be just reverse of their ratio of wattage rating.
17.Which of the following has least resistivity? . Copper.
A. Lead. B. Mercury. C. Aluminium. MATERIALS RESISTIVITY AT 20 °C Copper 1.68 X 10- 8 Ω - m Lead 22 X 10- 8 Ω - m Mercury 98 X 10 - 8Ω - m Aluminium 2.65 X 10 - 8 Ω - m
18.Which of the following lamp has least resistance ? . 200 W, 220 V.
A. 100 W, 220 V. B. 60 W, 220 V. C. 25 W, 220 V.
Where, V is supply voltage and W is wattage rating of the lamp.
From, above equation it is clear that, for fixed voltage source, resistance of an electric lamp is inversely
proportional to its wattage rating.
Here, 200W, 220V lamp has highest wattage rating, therefore its resistance will be least.
19.200 resistors of 200 Ω each are connected in parallel. Their equivalent resistance will be . 1 Ω.
A. 200 Omega;. B. 400 Omega;. C. 4 kΩ.
If N numbers of resistors of RΩ each are connected in parallel. Their equivalent resistance will be R ⁄ N.
20.The resistance of 100W, 200V lamp is . 200 Ω.
A. 400 Ω. B. 800 Ω. C. 1600 Ω.
Where, V is supply voltage and W is wattage rating of the lamp.
Here, wattage of the lamp W = 100W, and supply voltage V = 200V
Multiple Choice Questions on Electrical Resistance |
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Old pagination
1. A 1 kΩ, 1 W resistor can safely pass a current of
A. 30 mA.
B. 60 mA.
C. 40 mA.
D. 100 mA.
The wattage rating of the resistor W = I2.RHere W = 1 W, R = 1 kΩ = 1000 Ω∴ I = √(1/1000) = 0.0316
A = 31.6 mA > 30 mA.
2. Two resistors are connected in parallel across a battery of 2 V and a current flow through the combine resistors is 2 A. It one of the resistors is disconnected, the current will become 1.5 A, then what will be the resistance of that disconnected resistors?
. 2 Ω.
A. 4 Ω.
B. 1 Ω.
C. 0.5 Ω.
Total current is 2A and after disconnection of one, resistors, the current drawn from the battery, is 1.5A. That means the disconnected resistors was sharing 0.5A of currents. So resistance of the disconnected resistor will be 2/0.5 = 4 ohm.
3. Parallel combination of three 3 ohm resistors, connected in series with parallel combination of two 2 ohm resistors, what will be the equivalent resistance of overall combination ?
A. 3 Ω.
B. 5 Ω.
C. 1 Ω.
Three 3 ohm resistor are connected in parallel equivalent resistance will be 3/3=1ohm Two 2 ohm resistor are connected in parallel equivalent resistance will be 2/2=1ohm So, total resistance when these two combinations are series connected, the total resistance will be 1 + 1 = 2 ohm.
4. When a numbers of different valued resistance are connected in series, the voltage drop across each of the resistor is
. proportional to resistance.
A. proportional to current.
B. proportional to square of current.
C. equal.
Let V is the source voltage and R1, R2, R3,...Rn resistances are connected in series, across the source of
voltage V. Therefore, the current through the resistances will be ... Therefore, V1 = IR1, V2 = IR2, V3 =
IR3...Vn = IRn. That means Vn ∝ Rn
So, voltage drop across each resistance will be proportional to their resistive values.
5. All the resistances in figure shown below are 1 Ω each. The value of current in Ampere through the battery is
A. 2/15.
B. 4/15.
C. 8/15.
If look at the circuit from right side, we will see that the equivalent resistance of the circuit is 15 / 4 Ω. Current through the battery will be 1 V / (15 / 4) Ω = 4 /15 A.
6. Two wires A and B of same material and length l and 2l have radius r and 2r respectively. The ratio of their specific resistance will be
. 1 : 4.
A. 1 : 2.
B. 1 : 1.
C. 1 : 8.
The resistivity of any substance depends upon its material not upon its dimensions.
7. If the length of a wire of resistance R is uniformly stretched n times its original value, its new resistance is
. n.R.
A. n2
.R.
C. R/n2
.
Let's cross-section of the wire is A = πr2, length of the wire is l therefore volume of the wire is A.l. Now
if the length of the wire is stretched to n times of its original length i.e. now length of the wire becomes l'
= n.l. Now if r' is the new radius of the cross-section of the wire then new cross-sectional area A' = πr'2.
If the volume of the wire is same before and after stretching, A.l = A'.l' ⇒ πr'2.n.l = πr2.l ⇒ r'2 = r2/n ⇒
πr'2 = πr2/n ⇒ A' = A/n. Thus resistance of the wire after stretched is ρ(l'/A')= ρ{n.l/(A/n)} = n2ρ(l/A) =
n2.R.
8. The resistance between the opposite faces of 1 m cube is found to be 1 Ω. If its length is increased to 2 m, with its volume remaining the same, then its resistance between the opposite faces along its length is
. 1 Ω.
A. 2 Ω.
B. 4 Ω.
C. 8 Ω.
Volume = lengthXarea so if length is increased by 2 times then area will be decreased by 1/2, if resistance R = ρ(l/A) = 1 Ω ⇒ R' = ρ(2l/0.5A) = 4ρ(l/A) = 4R = 4 Ω.
9. A wire of length l and of circular cross - section of radius r has a resistance of R ohms. Another wire of same material and of x-section radius 2r will have the same R if the length is
. 2l.
A. l/2.
B. l2
.
C. 4l.
The cross-section of the first wire is πr2. The cross-section of the second wire is π(2r)2 = 4πr2. The
resistance of any wire depends upon the ratio of its length to area. So if the cross-section of the second wire is 4 times of that of first wire, the length of the second wire must also be 4 times of that of first if the resistance of both wires are same.
10. The insulation resistance of a cable of 10 km is 1 MΩ. For a length of 100 km of the same cable, the insulation resistance will be
. 1 MΩ.
A. 0.1 MΩ.
C. 0.01 MΩ.
Conductor resistance is directly proportional to length. But insulation resistance is the resistance to the flow of leakage current to ground. Since the flow of leakage current is directly proportional to the length of the conductor as because with length conductor inner and outer surface are of the insulation layer of the conductor increases. So insulation resistance is inversely proportional to the length of conductor.
11. The hot resistance of the filament of a bulb is higher than the cold resistance because the temperature coefficient of the filament is >
. positive.
A. negative.
B. zero.
C. infinite.
Positive temperature coefficient refers to materials that experience an increase in electrical resistance when their temperature is raised.
12. The temperature coefficient of resistance of an insulator is
. positive and independent of temperature.
A. negative and dependent on temperature.
B. negative and independent on temperature.
C. positive and dependent on temperature.
.
13. Four resistances 80 Ω, 50 Ω, 25 Ω and R are connected in parallel. Current through 25 Ω resistance is 4 A. Total current of the supply is 10 A. The value of R will be
. 36.36 Ω.
A. 66.66 Ω.
B. 40.25 Ω.
The current through 25 Ω resistor is 4 A hence voltage across it is 4X25 = 100 V and this is the voltage across the supply as well as other resistors. hence current through 50 Ω and 80 Ω resistors will be 100/50 = 2A
and 100/8 = 1.25 A. Therefore current through Rx will be 10 - 4 - 2 - 1.25 = 2.75 A and then Rx =
100/2.75 = 36.36 Ω
14. Three parallel resistive branches are connected across a DC supply. What will be the ratio of the branch current
I1:I1:I1 if the branch resistances are in the ratio R1:R2:R3 :: 2:4:6
. 6:4:2.
A. 6:3:2.
B. 2:4:6.
C. 3:2:6.
Current is inversely proportional to resistance. Then 1/2:1/4:1/6=3:3/2:1=6:3:2.
15. Two resistors R1 and R2 given combined resistance of 4.5 Ω when in series and 1 Ω when in parallel, the resistance are
. 1.5 Ω and 3 Ω.
A. 2 Ω and 2.5 Ω.
B. 1 Ω and 3.5 Ω.
When in series R1 + R2 = 4.5...(1) when in parallel (R1*R2)/(R1 + R2) = 1 (R1*R2)/4.5 = 1 R1*R2 = 4.5...(2)
COMBINING (1) AND (2),WE GET,
R1 = 1.5 or 3 and R2 = 3 or 1.5
16. When a resistor R is connected to a current source, it consumes a power of 18 W. When the same R is connected to a voltage source having the same magnitude as the current source, the power absorbed by R is 4.5 W. The magnitude of the current source and the value of R are
. √18 A and 1 Ω.
A. 1 A and 18 Ω.
B. 3 A and 2 Ω.
C. 6 A and 0.5 Ω.
For resistance R, connected to the current source, the consumed power is 18w i.e 18 = I2R (1) and for
second condition 4.5 = V2/R (2) and current and voltage having same magnitude that is V = I (3).
By solving these 3 equations we get R = 2 ohms and I = 3 A
17. When all the resistances in the circuit are of 1 Ω each, the equivalent resistance across the points A and B will be
. 1 Ω.
A. 0.5 Ω.
C. 2 Ω.
All the resistances are same. There is no potential difference between central vertical resistance [like Wheatstone Bridge], so it can be imagined that it is opened. Then the equivalent resistance between A &
B is :
Req = (1+1) || (1+1) || 1 Ω
Req = 2 || 2 || 1 Ω
Req = 0.5 Ω
18. Resistivity of metals is expressed in terms of
. μ Ω. A. μ Ω - cm ⁄ °C. B. μ Ω - cm. C. μ Ω. The resistivity ρ = R.A
L The resistivity also changes with temperature so for expressing resistivity one should mentioned temperature too. Hence, the most appropriate unit of resistivity is μ Ω - cm ⁄ °C.
Resistivity of copper is of the order of
D. 17.2 μ ohm-cm.
E. 1.72 μ ohm-cm.
F. 0.172 μ ohm-cm.
G. 172 μ ohm-cm.
Multiple Choice Questions of Analog Electronics
1. The conduction loss verses device current characteristics of power MOSFET in best approximately by A. a parabola.
B. an exponentially decaying function. C. a rectangular hyperbola.
I = Device current, Ron = on state Resistance of Power Mosfet, hence conduction loss P = I2Ron. Therefore
conduction loss verses device current characteristics can be approximately parabola.
2. A 3phase diode bridge rectifier is fed from a 400V R.M.S, 50 Hz, 3 phase AC source. If the load is purely resistive, then peak instantaneous output voltage is equal to
. 400√2 V. A. 400 V. B. 400√(2/3) V. C. 400/√3 V.
As the load is resistive the peak instantaneous output voltage Vm = 400√2 V.
3. An SCR is considered to be a semi controlled device because- . it can be turned OFF but not On with a gate pulse.
A. it conducts only during one half cycle of an alternating current wave. B. it can be turned ON but not OFF with a gate pulse.
C. it can be turned ON only during one half cycle of an AC.
During positive half cycle SCR is in forward Blocking mode. By applying gate pulse the SCR can be turned ON during forward Blocking mode. But SCR can be turned OFF by applying gate pulse. That is why it is called semi controlled rectifier.
4. Dual slope ADC has R = 1 KΩ & C = 0.22 nanofarad has charging & discharge times for some voltage is 9 ns & 3 ns respectively. The reference Voltage is 2.2 V. What is the peak voltage reached by triangular wave during charging?
. 90 mV.
A. 30 mV. B. 300 mV. C. 900 mV.
5. An SCR has half cycle surge current rating of 3000A for 50 Hz supply. One cycle surge current will be . 1500 A.
A. 6000 A. B. 2121.32 A. C. 4242.64 A.
6. The typical value of SCR for modern alternator is- . 1.5.
A. 0.5. B. 1.0. C. 1.2.
Always the typical value of SCR for modern alternator is 0.5.
7. A zener diode voltage regulator has load requirement of 12 V & 2 Amp. The zener diode's minimum current requirement is 0.2 A. The minimum voltage at input is 24 V. What is Maximum efficiency of circuit?
. 34.3%.
A. 45.5%. B. 52.8%. C. 66.3%.
η = (2*12) ⁄ (2.2*24)*100 % = 24 ⁄ 52.8 = 45.5 %
8. In n type semiconductor elements of which group of periodic table is added as dopant . group 5.
A. group 2. B. group 3. C. group 4.
With group 5 elements like phosphorous, arsenic, antimony are added to Ge or Si crystal which produces free electrons.
9. Darlington connection is achieved in 2 transistors by connecting . both emitter.
A. both collector. B. both base.
C. grounding both collector.
in darlington connection 2 transistor's collectors are connected and emitter of 1st transistor to the base
of 2nd is also connected. This provides a three terminal device that in actual operation can be regarded as
2 cascaded emitter followers.
10.FM stands for
. frequent modulation.
A. frequency modulation. B. frequency moderator. C. frequent moderator.
n telecommunications and signal processing, frequency modulation (FM) conveys information over a carrier wave by varying its instantaneous frequency. This contrasts with amplitude modulation, in which the amplitude of the carrier is varied while its frequency remains constant.
11.An opamp comparator circuit employs . no feedback.
A. +ve feedback. B. -ve feedback. C. both b and c.
comparator is a device used for comparison of 2 voltage levels. Output indicates which of the 2 voltages is greater.No feedback is used.
12.A single phase full wave midpoint thyristor uses a 230/200 V Transformer with central tap on the secondary side. The PIV per thyristor is-
. 282.8 V.
A. 200 V. B. 100 V. C. 141.4 V.
PIV=√(2)*Vm = √(2)*200 = 282.84 V.
13.In a rectifier circuit, the diode converts . alternating voltage to direct voltage.
A. both (A)&(C) options are connect. B. alternating voltage to direct current. C. alternating current to direct voltage.
Diode converts alternatring voltage to unidirectional current which is then converted in to voltage by load resistance connected across the output node.
14.A single phase one pulse controlled circuit has a resistance & counter emf load &400 sin(314t) as the source voltage for a load counter emf of 200 V, the range of firing angle control is-
. 30 degree to 150 degree.
A. 30 degree to 180 degree. B. 60 degree to 120 degree. C. 60 degree to 180 degree.
400 sinθ=200 or, θ=30 degree, so control range is &theta to(Ï€-θ) i.e. 30 degree tp 150 degree.
15.A single phase full bridge inverter can operated in load commutation mode in case load consist of-
. RL.
A. RLC underdamped. B. RLC overdamped. C. RLC critically damped.
In this case reading current will flow in circuit & it will became zero so thyristor will be load commutated.
16.A junction transistor with β=49 & Ic0 = IcB0 = 1µA has IB=10µA .The value of Ic is given in µA by- . 540.
A. 440. B. 539. C. 490.
Ic=βIB+(β+1)*Ic0=49*10+50*1=540µA.
17.A step up chopper has input voltage 110 V & output voltage 150 V .The vqalue of duty cycle is- . 0.32.
A. 0.67. B. 0.45.
C. none of these.
150=110(1⁄1-α) or, 150 -150α=110 or, α=40⁄150=0.267(α-duty cycle).
18.A schottky diode is a- . majority carrier device.
A. minority carrier device. B. fast recovery diode.
C. both a majority & a minority carrier diode.
Although a schottky diode behaves a p-n junction diode, there is no physical junction and as a result a schottky diode is a majority carrier diode.
Objective Questions on Battery
A. which can be recharged.
B. which cannot be reconditioned by replacing chemical. C. which cannot be reused.
D. which cannot be recharged.
An electro-chemical cell or battery is such which can not be recharged but the chemical has to be replaced for reconditioning, is called primary battery.
2. The first electro - chemical cell was invented by . Luigiri Galvani.
A. Alessandro Volta. B. Deniel.
C. Lechanche.
The first electrochemical cell was invented by Luigiri Galvani in 1791. It was greatly improved by Alessandro Volta in the year of 1800.
3. The secondary battery is such a battery . which cannot be recharged.
A. which can be recharged.
B. which can be reused after replacing its chemical. C. which is charged by primary cells.
A secondary battery cell is one in which chemical energy is converted into electrical energy but they do only when they are charged by passing current through them by some source.
4. An example of secondary battery cell is . Edison Alkali cell.
A. Daniel cell. B. Lachanche cell. C. Bunsen cell.
Voltiac cell, daniel cell, Lachanche cell, Bunsen cell, fuel cell are some well known example of primary battery cells. Whereas, Lead Acid cell and Edison Alkali cell are two well known example of secondary cells.
5. Internal resistance of a battery cell increases with . increases in concentration of electrolyte.
A. increase in distance between two electrodes. B. increases in area of the plates inside the electrolyte. C. increase in size of the electrodes.
Internal resistance of a battery cell
1. increases with increase in distance between two electrodes,
2. decreases with increase in concentration of electrolyte,
3. decreases with increase in area of the plates inside the electrolyte,
4. decreases with increase in size of the electrodes.
6. Internal resistance of a battery cell decreases with . increase in area of the plates inside the electrolyte.
A. increase in distance between two electrodes. B. decrease in size of the electrodes.
C. increase of age of the battery.
Internal resistance of a battery cell
1. increases with increase in distance between two electrodes,
2. decreases with increase in concentration of electrolyte,
3. decreases with increase in area of the plates inside the electrolyte,
4. decreases with increase in size of the electrodes.
7. Unit of electro - chemical equivalent of the substance of electrolyte . Kg - Coulomb.
A. Kg / Coulomb. B. Coulomb / Kg. C. Kg / °C.
According to Faraday's law of electrolysis,the mass (Δm) of ions liberated at an electrode is directly
proportional to the quantity of charge (Δq) passing through the electrolyte.
Δm ∝ Δq ⇒ Δm = Z.Δq ⇒ Z = Δm ⁄ Δq
This Z is proportionality constant and also called Electro - Chemical Equivalent of the substance of the electrolyte.
From the above relation it is clear that unit of Z may be Kg/Coulomb *unit of mass ⁄ unit of charge+.
8. If Z is the electro - chemical equivalent of a substance of the electrolyte. E is the chemical equivalent of the same substance, then the relation between Z & E will be,
. Z &porp; E. A. Z = E. B. Z < E. C. Z > E.
Faraday's second law of electrolysis states Electro - Chemical Equivalent of a substance is directly proportional to its chemical equivalent.
9. Negative electrode or anode of simple voltaic cell is made of . copper.
A. zinc. B. lead. C. carbon.
In Voltaic battery cell, zinc atoms in contact with dilute sulfuric acid give up electrons and forms
Zn++ ions which pass into the electrolyte. As a result zinc electrodes get high concentration of electrons
and it get negatively charged. Hence this zinc electrode acts as negative electrode or anode.
10.For all substances, [Chemical Equivalent / Electro - Chemical Equivalent] = . 96500 Coulombs.
A. 9650 Coulombs. B. 965 Coulombs. C. 96.5 Coulombs.
Faraday's second law of electrolysis states Electro - Chemical Equivalent ( Z ) of a substance is directly proportional to its chemical equivalent ( E ).∴ Z ∝ E ⇒ E ⁄ Z = F, a constant called Faraday's constant and value of F = 96500 Coulomb.
11.A 10V battery with an internal resistance of 1 Ω is connected across a non-linear load whose v-i characteristic is given by 7i = v2 + 2v. The current delivered by the battery is
. 7 A. A. 6 A. B. 5 A. C. 4 A. . 12.Nickel is used in
. electrodes of thermionic valves. A. bulb filaments.
B. automatic voltage regulators. C. pressure sensitive elements.
The electrodes of a thermionic valve are generally made of nickel.
13.A cell has an Ah efficiency of 80%. It has an average terminal voltage on discharge and charge of 1.2 V and 1.6 V respectively. The Watt- hour efficiency of the cell is ... %.
. 60%.
A. 80 %.
B. 100%.
C. 50%.
14.Effect of temperature on internal resistance of a battery is . directly proportional.
B. no effect. C. none of these.
Internal resistance Ri is inversely proportional to temperature t.
15.E.C.E stands for
. electrovalent chemical equivalent.
A. electron chemical equivalent. B. electro chemical equivalent. C. electrolysis cathode equivalent.
E.C.E stands for ELECTRO CHEMICAL EQUIVALENT. E.C.E is the mass of ions in grams which is liberated /deposited by chemical action by the passage of 1C of electricity i.e 1A for 1 sec.
Objective Questions on Capacitor
1. A capacitor
A. passes ac but blocks dc. B. passes dc but blocks ac. C. passes both ac and dc. D. blocks both ac and dc.
The impedance of capacitor can be expressed as
.
Hence, impedance of capacitor is inversely proportional to the supply frequency(f). In ac frequency has non - zero finite value, so impedance will have finite value but in DC f = 0, so impedance will have infinitely large value. Thus a capacitor blocks DC but passes AC.
2. A 100 μF capacitor supplied from 3 V source with a frequency of 50 Hz. The capacitive reactance is . 63.68 Ω.
A. 15.92 Ω. B. 31.84 Ω.
C. 7.96 Ω.
The capacitive reactance can be expressed as
3. A capacitor passes a current of 12.6 mA when supplied with 20 V ac with a frequency of 1000 Hz. The capacitance will be
. 0.001 μF. A. 0.01 μF. B. 0.1 μF. C. 1 μF.
Where, Vc is the voltage across capacitor, Ic current through the capacitor, f & C are frequency and
capacitance.
Here, Vc = 20 V, Ic = 12.6 mA and f = 1000 Hz.
4. A 10 μF capacitor and 100 W, 220 V lamp is connected in series across a 220 V alternating supply. In which frequency of the supply the lamp will glow brightest?
. 1000 Hz. A. 100 Hz. B. 10 Hz. C. 1 Hz.
The impedance of capacitor is inversely proportional to its supply frequency. The impedance offered by the capacitor to the circuit is less when supply frequency is more. If impedance is less, current flows through the circuit (i.e. lamp) is more which results to glow the lamp brighter.
5. A 20 μF capacitor and 200 W, 220 V lamp is connected in series across a 220 V alternating supply. In which frequency of the supply the lamp will glow dimmest?
A. 1 Hz. B. 10 Hz. C. 100 Hz.
The impedance of capacitor is inversely proportional to its supply frequency. The impedance offered by the capacitor to the circuit is more when supply frequency is less. If impedance is more, current flows through the circuit (i.e. lamp) is less which results to glow the lamp dimmer.
6. The capacitive reactance of a capacitor of 1 / 2π F at 103 Hz is
. 10 6 Ω.
A. 10- 3 Ω.
B. 10 3 Ω.
C. 10- 6 Ω .
The capacitive reactance can be expressed as
7. When ac flows through a pure capacitance then the current . leads the emf by 90°.
A. lags the emf by 90°. B. leads the emf by - 90°. C. is in phase with emf.
When ac flows through an capacitance, the current leads the emf by 900.
8. It a capacitors of capacitance 100 μF is connected across a voltage source of 10 V, then what will be the energy stored in that capacitor
. 5 × 10 - 3 Joule.
A. 10 × 10 - 3 Joule.
B. 10 × 10 6 Joule.
It a capacitor of capacitance value C have voltage difference V between its parallel plates then the energy
stored in the capacitor is expressed as
9. A capacitor carries a charge of 0.3 C at 20 V. Its capacitance is . 1.5 F.
A. 0.015 F. B. 1.5 μF. C. 15 μF.
The capacitance of a capacitor is expressed as
Where C is the capacitance, Q is charge & V is the voltage.
10.A parallel plate capacitor has a capacitance of C farad. It area of the plates is doubled and the distance between them is half, the capacitance of the capacitor is
. 1 C farad.
A. 2 C farad. B. 4 C farad. C. 16 C farad.
A capacitor consists of two parallel places separated by a dielectric material. It the area of the plates is A
m2 and the distance between them is d meter, the capacitance C is given by
Where A is the area of the plates & d is distance between the plates. From the above expression of capacitance it is obvious that, if area is doubled and distance is half the capacitance will become 4 times.
11.Which of the followings is the expression for energy stored in a capacitor
. Cv.
A. C dv/dt. B. C/v. C. (1/2)Cv2.
The instantaneous power in the capacitor is given by p=iv
12.A capacitor is connected to supply with switch and the switch is connected between capacitor and supply. Initially switch is open at time zero, and then switch is closed. Then how capacitor behaves at time t=0+ . short circuit.
A. open circuit.
B. dielectric losses decreases. C. dielectric losses increases.
In case of a capacitor voltage across it does not change instantaneously. If an uncharged capacitor is connected to an energy source, at the time of switching, the capacitor will behave like a short circuit.
13.C eq of two capacitors connected in series is given by
. C1 C2.
A. C1 =C2.
B. C1 C2 /C1 +C2.
C. C1 +C2.
SERIES combination of capacitors is same as PARALLEL of resistance.
14.A capacitor has a capacitance of 6 μF. Calculate the stored energy in it if a dc voltage of 100 V, is applied across it
. 3 × 10 − 2 joules.
A. 2.5 × 10 − 2 joules.
B. 6 × 10 − 2 joules.
C. 4 × 10 − 2 joules.
. 5.
A. 0.05.
B. 10.
C. 20.
16.A p.d. of 300 V is applied across series combination of 3 μF and 9 μF capacitors. The charge on each capacitor is ... μC.
. 675.
A. 3600.
B. 240. C. 7.5.
17.A 50 μF capacitor is charged to retain 10 MJ of energy by a constant charging current of 1 A. Determine the voltages across the capacitor:-
. 30 V.
A. 20 V. B. 50 V. C. 60 V.
V being the voltage developed across the capacitor of capacitance(C).
18.The capacitance of a conductor is varying from 2 microfarad to zero in 1 sec linearly if the voltage applied to it is 6 V the energy stored in 0.5 sec in the condenser is
A. 18 μ joules. B. 10μ joules. C. 20μ joules.
Here C is taken as μ F only since capacitor varies linearly.
19.What will be the capacitance when distance between the 2 plates of a condenser of capacitance 8 micro farad is reduced from 10 mm to 4 mm?
. 20 microfarad.
A. 3.2 microfarad. B. 8 microfarad. C. 1 microfarad.
Where, A is common area of conductor plates.
Where, d is the distance between two conductor plates.
20.Purpose of using capacitor is/are . increase p.f of inductive load circuit.
A. to do phase split in ac 1 phase motor. B. effect dc filter in electronic circuit. C. all of these.
Purposes of using capacitors are
1) increasing p.f. of inductive load circuit
2) to do phase split in ac 1 phase motor
3) effect dc filter in electronic circuit
4) also helps in tuning in radio and TV sets.
Objective Questions on Circuit Theory | Page – 1
A. E1 lags E2 by θ ⁄ 2.
B. E1 leads E2 by θ.
C. E2 leads E1 by θ.
D. E2 leads E1 by θ.
At Ωt = 0, E1 = 0 & E2 = A.sin(-θ) = -A.sinθ
From, the expression of E1 = A.sinΩt & E2 = A.sin(Ωt - θ), it is clear that, E2 crosses zero t = θ ⁄ Ω sec
after that of E1
Therefore, it can be concluded that E1 leads E2 by θ.
2. ) Two sinusoidal quantities are said to be phase quadrature, when their phase difference is
. 0°.
A. 30°.
B. 45°.
C. 90°.
Two sinusoidal quantities are said to be phase quadrature, when their phase difference is 90°.
3. The equation for 25 cycles current sine wave having rms value of 30 amps, will be
. 42.4sin50πt.
A. 30sin50πt.
B. 30sin25πt.
C. 42.4sin25πt.
General equation of sinusoidal current quantity is ImsinΩt = Imsin2.πf.t.
Where, Im is the maximum ampletude of the current wave, f is the frequency or cycle per second.
Here, rms value of current is 30A
∴, maximum amplitude of the current wave form Im = √2 X 30 = 42.4 A and frequency f is here 25 Hz.
Hence, the current equation will be 42.4sin2.π.25.t = 42.4sin50πt.
4. What will be the rms value of rectangular wave with amplitude 10V
. 5√2 V.
B. 11.2 V.
C. 7.7 V.
RMS value means, root mean square value of a wane. A rectangular voltage wave has constant amplitude in both positive and negative direction. Hence the mean value and its amplitude will be identical. Therefore square root of square of the mean value of amplitude is same as amplitude of the wave.
5. The equation of an emf is given by e = Im[(R2 + 4Ω2L2)½]sin2Ωt. The amplitude of the wave will be
. Im[(R 2 + 4Ω2 L2 )½ ]. A. √2Im[(R2 + 4Ω2L2)½]. B. [Im(R2 + 4Ω2L2)]½. C. 2Im[(R 2 + 4Ω2 L2 )½ ].
The general form of emf equation is e = VmsinΩt ...(1)
Where, Vm is the amplitude of the voltage wave
Here, the given equation is e = Im[(R2 + 4Ω2L2)½+sin2Ωt ...(2)
Now, Comparing equation (1) & (2) we can conclude that amplitude of the given emf equation is Im[(R2 + 4Ω2L2)½].
6. The RMS value of sinusoidal voltage wave v = 200sinΩt, is
. 100√2 V.
A. 200 V.
B. 100 V.
C. 200√2 V.
The peak value of given voltage wave is 200 V. Therefore the RMS value will be 200 ⁄ √2 = 100√2 volts.
7. If one cycle of ac waveform occurs every milli - second, the frequency will be
. 100 Hz.
A. 1000 Hz.
B. 50 Hz.
One cycle of ac waveform occurs every milli second means 1000 cycles of that waveform occur in one second. The numbers of cycles of waveform per second is the frequency of that waveform.
8. If emf in a given circuit is given by e = 100sin628t, then maximum value of voltage and frequency will be
. 100 V, 100 Hz.
A. 100 V, 50 Hz.
B. 100√2 V, 100 Hz.
C. 50√2 V, 100 Hz.
The emf equation is e = 100sin628t ...(1)
Again the general form of emf equation is e = Vmsin2π.f.t ...(2)
Where, Vm is the voltage amplitude and f is the frequency.
Comparing, equations (1) & (2) we get,
Vm = 100 V and 2πf = 628 ⇒ 628/(2X3.24) ⇒ f = 628/6.28 = 100 Hz.
9. The value of supply voltage for 400W, 4 ohm load is
. 40 V.
A. 400 V.
B. 20 V.
C. 200 V.
Where, V is supply voltage and W is wattage rating of the lamp.
Here, W = 400W and R = 4Ω
⇒ V = 40V.
10. Say A point has an absolute potential of 40V. and point B has an absolute potential of -10V, then what will be the value
of VBA ?
. -50 V.
A. 50 V.
B. 30 V.
VBA is defined as VB − VA = -10 - 40 = -50V.
11. The rms value of the voltage U(t)= 3+4cos(3t)
. 5 V. A. 4.123 V. B. 7 V. C. {3+2(1.141)} V. Urms = √(9+(16/2))= √17= 4.123 V. .
12. In the figure, the potential difference between points P and Q is
. 6 V.
A. − 6 V.
B. 10 V.
C. 12 V.
.
13. A coil of negligible resistance has an induction of 100 mH. The current passing through the coil changes from 2 A to 4 A at a uniform rate in 0.1 sec the voltage across the coil during this time would be ___ V.
. 2.
A. 8.
B. 36.
V = L X di/dt = 100 X 10-3 X 2/0.1 = 2 votls.
14. 1518) What is representated by the hypotenuse of impedance triangle
. impedance drop.
A. resistance drop.
B. reactance drop.
C. apparent power .
Impedance triangle means the right angle triangle formed by the vectors representing the resistance drop, reactance drop & the impedance drop of the circuit carrying an alternating current.
15. The phase angle difference between current and voltage is 90°, the power will be
. minimum.
A. maximum.
B. zero.
C. V.I.
The expression of active power P = V.I.cosθ.
Where, V is voltage, I is current and θ is the angle between current and voltage.
here, this θ = 90°
∴ Power P = V.I.cos90° = 0 [Since, cos90°= 0].
16. Kirchhoff's laws are valid for
. linear circuit only.
A. passive time invariant circuits.
B. non-linear circuits only.
C. both linear and non-linear circuits.
17. For the circuit shown below the value of R is adjusted so as to make the current in RLequal to zero. Calculate the value of R. . 1 Ω. A. 2 Ω. B. 3 Ω. C. 4 Ω. As per Wheatstone Bridge principle: 10 / 4 = 5 / R ⇒ R = 4 / 10 X 5 ohms ⇒ R = 2 ohms.
18. In the circuit shown in figure if I1 = 1.5A, then I2 will be
. 2 A.
B. 1 A.
C. 0.5 A.
19. In the circuit shown in the figure the voltage across the 2 Ω resistor is
. 1 V.
A. 2 V.
B. 3 V.
C. 4 V.
20. The value of current I flowing in the 1 Ω resistor in the circuit shown in the given figure will be
. 5 A.
A. 6 A. B. 0 A. C. 10 A.
1. The voltage across the 1 kΩ resistor between the nodes A and B of the network shown in the given figure is
A. 1 V.
B. 2 V.
C. 3 V.
D. 4 V.
2. In the network shown, what is the current I in the direction shown
. 0 A.
A. 1/3 A.
B. 5/6 A.
C. 4 A.
3. An electrical circuit with 10 branches and 7 nodes will have
. 10 loop equations.
A. 7 loop equations.
B. 3 loop equations.
C. 4 loop equations.
4. In given figure, the value of resistance R in Ω is
. 10.
A. 20.
B. 30.
C. 40.
The given circuit can be simplified by replacing voltage source by equivalent current source as below.
The current
through 5 Ω resistor is 8 A hence voltage across it is 8X5 = 40 V and this is the voltage across unknown resistor R. As current through R is 2 A and then R = 40/2 = 20 Ω.
5. In given figure, the value of resistance R in Ω is
. 2.5 Ω.
A. 5 Ω.
B. 7.5 Ω.
C. 10 Ω.
The given circuit can be simplified by replacing voltage source by equivalent current source as below.
8X(R + 5) = 100 V
6. A 35 V source is connected to a series circuit of 600 Ω and R as shown. If a voltmeter of internal resistance 1.2 kΩ is
connected across 600 Ω resistor, it reads 5V. The value of R is
. 2.4 kΩ.
A. 1.2 kΩ.
B. 3.6 kΩ.
C. 7.2 kΩ.
As the voltmeter of internal resistance 1.2 KΩ is connected across the 600 Ω resistor and it gives 5 V reading, the circuit current I = 5/600 +5/1200 A = 0.0125 A. As supply voltage is 35 V, the voltage cross resistor R is 35 - 5 = 30 V
7. A certain network consists of large number of ideal linear resistances, one of which is designated as R and two constant
ideal source. The power consumed by R is P1 when only the first source is active and P2 when only second source is
active. In both sources are active simultaneously then the power consumed by R is
. √P1 ± √P2.
A. (√P1 ± √P2)2.
B. P1 ± P2.
C. (P1 ± P2)2.
. 4 A.
A. 3 A.
B. 2 A.
C. 1 A.
9. In the circuit shown below, what is the voltage across 5Ω resistor ?
. − 30 V.
A. 30 V.
B. 1250 V.
C. − 1250 V.
10. For the circuit shown in the given figure the current I is given by
. 2 A.
A. 1 A.
B. 3 A.
11. For the circuit given in the figure the power delivered by the 2 V source is given by . 4 W. A. 2 W. B. − 2 W. C. − 4 W. .
12. In the circuit shown in the figure, the value of Vs is 0, when I = 4A. The value of I when Vs = 16 V, is
. 6 A.
A. 12 A.
B. 10 A.
C. 8 A.
13. Consider the following circuit: In this circuit, when Vs = 3V, I = 4A, when is the value of I when Vs = 12V ?
. 5 A.
A. 10 A.
B. 15 A.
C. 20 A.
14. In the figure given, the value of R is
. 12 Ω.
A. 18 Ω.
B. 24 Ω.
15. In the given figure the value of the source voltage is
. 12 V.
A. 30 V.
B. 44 V.
C. 24 V.
16. ) Three resistance of two ohms each are connected in star in the equivalent delta representation each resistance will have a value of ________ohms.
. 3.4.
A. 6.
B. 0.6.
C. 5.2.
RA = RB = RC = 2 X 2 + 2 X 2 + 2X2 / 2 = 6 ohms .
17. Consider the following circuit: What is the value of current I in the 5 Ω resistor in the circuit given in the figure ?
. 0 A.
A. 2 A.
C. 4 A.
18. The value of V in the circuit shown in the given figure is
. 2 V.
A. 3 V.
B. 4 V.
C. 5 V.
19. In the given figure, the Thevenin equivalent voltage and impedance as seen from the terminals P-Q is given by
. 2 V and 5 Ω.
A. 2 V and 7.5 Ω.
B. 4 V and 5 Ω.
2 V, 5 Ohm
20. For the network shown below when I = 0, V = 20 V and when R = 0, I = 10 A. If now R = 3 Ω what is the
value of the current I?
. 6 A.
A. 4 A. B. 5 A. C. 10 A.
1. For the circuit given in the figure, the Thevenin voltage and resistance as seen at AB are represented by A. 5 V 10 Ω. B. 10 V 10 Ω. C. 5 V 5 Ω. D. 54 V 15 Ω. Same as 2065 V 10 Ω
2. Form factor is defined as ratio of . average value to RMS value.
A. RMS value to average value. B. maximum value to RMS value. C. RMS value to maximum value.
Form factor = rms value / avg value.
3. Peak factor is defined as ratio of . maximum value to RMS value.
A. average value to RMS value. B. RMS value to average value. C. RMS value to maximum value.
Peak factor = maximum value/RMS value.
4. At resonance
. magnitude of capacitive reactance > magnitude of inductive reactance.
A. magnitude of capacitive reactance = magnitude of inductive reactance. B. magnitude of capacitive reactance < magnitude of inductive reactance. C. none of above.
Resonance is occurs at when inductive, capacitive reactance both are same i.e. , circuit is having only resistive components.
5. A series R-L-C circuit has R=50 Ω, L=100 μH and C = 1 μF. The lower half power frequency of the circuit is . 30.55 KHz.
A. 3.05 KHz. B. 51.92 KHz. C. 1.92 KHz.
6. The period of the function cosπ ⁄ 4(t-1) is . 8 Second.
A. 1 ⁄ 8 Second. B. 4 Second. C. 1 ⁄ 4 Second.
7. A T-section lowpass filter has series inductance 80 mH and shunt capacitance 0.022 µF. Determine cutoff frequency . 7 KHz. A. 7.58 KHz. B. 7.8 KHz. C. 8 KHz. L = 80 mH, C = 0.022 µF , fc (cutoff frequency) = 1 ⁄ π√(LC) = 1 ⁄ π√(80X10 − 3)X(0.022X10 − 6) = 7.58 KHz.
8. In a constant-k high pass filter having cutoff frequency of 12 kHz. Find out phase constant at 24 kHz . 75°.
A. 60°. B. 90°. C. 45°.
β ( phase constant ) = 2sin − 1(fc/f) = 2sin − 1(12X103/24X103) = 60°.
9. In Constant K high- pass filter having cutoff frequency of 12 Khz, Findout attenuation at 4 Khz- . 3.5.
A. 3.525. B. 3.425. C. 3.4.
α ( attenuation ) = 2cosh − 1(fc/f) neper = 2cosh − 1(12X103/4X103) = 3.525 neper.
10.Find out the series arm capacitance & shunt arm inductance in a constant K-high pass filter, when impedance R0 = 600 Ω & cut off frequency = 4 Khz (Π - sec)
. 0.033 μF & 11.937 mH. A. 0.4 μF & 12mH. B. 0.05 μF & 13mH. C. 0.02 μF & 10.9mH.
