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23051.

What is the commercial unit of electric energy? Convert it into joules.

Answer»

Kilowatt hour – commercial unit of electrical energy.
1 kWh = 1000 Wh = 1000 J/S × 3600 sec
= 3600000 J = 3.6 × 106J

23052.

Out of 60 W and 40 W lamps which one has higher electrical resistance when in use?

Answer»

For the same applied voltage P & \(\frac{1}{R}\)

i.e., less the same power of electrical device higher it is electrical resistance.

23053.

Write two points of difference between electric energy and electric power.

Answer»

The work done or energy supplied by the source in maintaining the flow of electric current is called electrical energy. It appears in the from of heat given by
H = VIt = \(\frac{v^2t}{R}\) = I2RT

(i) It is equal to be product of power and time E = P × t

(ii) It SI unit is joule (J)
1J = 1W × 1S

Electric power:
The time rate at which electronic energy is consumed or dissipated by an electrical device is called electric power and is given by
P = VI = \(\frac{v^2}{R}\) = I2R
It is equal to the rate of doing work by an energy source
P = \(\frac{W}{t}\)
Its SI unit is watt(W)
1W = 1 JS-1

23054.

What is meant by electric current? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of conventional current? Give justification for your answer.

Answer»

A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flows through any section of the conductor is 1 second (charge on electron 1.6 x 1019 Coulomb)

Electric current: The amount of charge Q flowing through a particular area of cross section in unit time ‘t’ is called electric current, i.e., Electric current, I = Q/t
SI unit of electric current is ampere

One ampere of current is that current which flow when one coulomb of electric charge flowing through a particular area of cross section of the conductor is one second, i.e, 1A = 1 CS-1

The direction of conventional current is A to B, i.e., opposite to the direction of flow of electrons. In a metal, flow of electrons carrying negative charge constitutes the current. Direction of flow of electrons given the direction of electronic current by convention, the direction of flow of positive charge is taken as the direction of conventional current charge
= q = ne
For q = I Coulomb

\(\frac{1C}{1.6\times 10^{-19}C}\) = \(\frac{10^{19}}{1.6}\) = 6.25 x 1018 electrons

23055.

An isolated particle of mass `m` is moving in horizontal plane` xy` along the `x`-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses `m//4` and `3m//4`. An instant later, the smaller fragment is at `y=+15` cm. The larger fragment at this instant is atA. `-5 cm `B. `-20 cm `C. `+5cm `D. `(-45)/4cm`

Answer» Correct Answer - 1
The particle moving along x-axis, So, centre of mass will not move along y direction.
`y_(cm)=0`
`=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2)) =((m//4).15+(3m//4)y)/m`
y=-5 cm
23056.

The outer surface of the cardboard center of a paper towel roll: A. is a possible Gaussian surface B. cannot be a Gaussian surface because it encloses no charge C. cannot be a Gaussian surface since it is an insulator D. cannot be a Gaussian surface because it is not a closed surface E. none of the above

Answer»

D. cannot be a Gaussian surface because it is not a closed surface

23057.

An electron moves from point i to point f, in the direction of a uniform electric field. During this displacement:A. the work done by the field is positive and the potential energy of the electron-field system increases B. the work done by the field is negative and the potential energy of the electron-field system increases C. the work done by the field is positive and the potential energy of the electron-field system decreases D. the work done by the field is negative and the potential energy of the electron-field system decreases E. the work done by the field is positive and the potential energy of the electron-field system does not change

Answer»

B. the work done by the field is negative and the potential energy of the electron-field system increases

23058.

Correct graph of `y-1=x^(2)` is -A. B. C. D.

Answer» Correct Answer - 1
At y=1, x=0
23059.

`int_(2)^(5)(1)/((2+3x))dx` is -A. `(15)/(2)`B. `(7)/(5)`C. `(1)/(3)l"n"(17)/(8)`D. `l"n"(17)/(8)`

Answer» Correct Answer - 3
`[(log(2+3x))/(3)]_(2)^(5)`
`(1)/(3)(log17-log8)=(1)/(3)l n(17)/(8)`
23060.

Give the characteristics of image formed by a plane mirror.

Answer»
  • The image formed by a plane mirror is virtual, erect, and laterally inverted.
  • The size of the image is equal to the size of the object.
  • The image distance far behind the mirror is equal to the object distance in front of it.
  • If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as, \(n = (\frac{360}{θ}-1)\)
23061.

If the collector resistor decreases to zero in a basebiased circuit, the load line will become a. Horizontal b. Vertical c. Useless d. Fla

Answer»

The correct answer is: b. Vertical

23062.

If the base supply voltage is disconnected, the collector-emitter voltage will equal a. 0 V b. 6 V c. 10.5 V d. Collector supply voltage

Answer»

(d) Collector supply voltage

23063.

If a transistor operates at the middle of the load line, an increase in the base resistance will move the Q point a. Down b. Up c. Nowhere d. Off the load line

Answer»

The correct answer is: (a) Down

23064.

Three different Q points are shown on a load line. The upper Q point represents the a. Minimum current gain b. Intermediate current gain c. Maximum current gain d. Cutoff point

Answer»

(c) Maximum current gain

23065.

If the collector resistor is open, the ac input voltage will a. Decrease b. Increase c. Remain the same d. Equal approximately zero

Answer»

(a) Decrease

23066.

If the emitter resistor is open, the ac input voltage will a. Decrease b. Increase c. Remain the same d. Equal zero

Answer»

(b) Increase

23067.

If any capacitor is open, the ac output voltage will a. Decrease b. Increase c. Remain the same d. Equal zero

Answer»

(a) Decrease

23068.

An open resistor has a. Infinite current through it b. Zero voltage across it c. Infinite voltage across it d. Zero current through it

Answer»

(d) Zero current through it

23069.

If the bypass capacitor is open, the ac input voltage will a. Decrease b. Increase c. Remain the same d. Equal zero

Answer»

(b) Increase

23070.

If the output coupling capacitor is open, the ac input voltage will a. Decrease b. Increase c. Remain the same d. Equal zero

Answer»

(c) Remain the same

23071.

If the load resistance is open, the ac output voltage will a. Decrease b. Increase c. Remain the same d. Equal zero

Answer»

(b) Increase

23072.

If the input coupling capacitor is open, the ac input voltage will a. Decrease b. Increase c. Remain the same d. Equal zero

Answer»

(d) Equal zero

23073.

A shorted resistor has a. Infinite current through it b. Zero voltage across it c. Infinite voltage across it d. Zero current through it

Answer»

(b) Zero voltage across it

23074.

A VCVS amplifier has no output voltage. A possible trouble is a. Shorted load resistor b. Open feedback resistor c. Excessive input voltage d. Open load resistor

Answer»

(a) Shorted load resistor

23075.

Negative feedback reduces the a. Feedback fraction b. Distortion c. Input offset voltage d. Loop gain

Answer»

(b) Distortion

23076.

The positive clamper adds a. A positive dc voltage to the input b. A negative dc voltage to the input c. An ac signal to the output d. A trip point to the input

Answer»

(a) A positive dc voltage to the input

23077.

A first-order stage cannot have a a. Butterworth response b. Chebyshev response c. Maximally-flat passband d. Rolloff rate of 20 dB per decade

Answer»

(b) Chebyshev response

23078.

If GBW is limited, the Q of the stage will a. Remain the same b. Double c. Decrease d. Increase

Answer»

(d) Increase

23079.

To correct for limited GBW, a designer may use a. A constant time delay b. Predistortion c. Linear phase shift d. A rippled passband

Answer»

(b) Predistortion

23080.

To detect when the input is greater than a particular value, use a a. Comparator b. Clamper c. Limiter d. Relaxation oscillator

Answer»

(a) Comparator

23081.

A window comparator a. Has only one usable threshold b. Uses hysteresis to speed up response c. Clamps the input positively d. Detects an input voltage between two limits

Answer»

(d) Detects an input voltage between two limits

23082.

VDB needs a. Only three resistors b. Only one supply c. Precision resistors d. More resistors to work better

Answer»

(b) Only one supply

23083.

The voltage that starts an oscillator is caused by a. Ripple from the power supply b. Noise voltage in resistors c. The input signal from a generator d. Positive feedback

Answer»

(b) Noise voltage in resistors

23084.

A lead circuit has a phase angle that is a. Between 0 and +90 degrees b. Greater than 90 degrees c. Between 0 and -90 degrees d. The same as the input voltage

Answer»

(a) Between 0 and +90 degrees

23085.

Initially, the loop gain of a Wien-bridge oscillator is a. 0 b. 1 c. Low d. High

Answer»

The answer is: (d) High

23086.

A Wien-bridge oscillator uses a. Positive feedback b. Negative feedback c. Both types of feedback d. An LC tank circuit

Answer»

(c) Both types of feedback

23087.

The Wien-bridge oscillator is useful a. At low frequencies b. At high frequencies c. With LC tank circuits d. At small input signals

Answer»

(a) At low frequencies

23088.

The common-mode rejection ratio is a. Very low b. Often expressed in decibels c. Equal to the voltage gain d. Equal to the common-mode voltage gain

Answer»

(b) Often expressed in decibels

23089.

The input offset current is usually a. Less than the input bias current b. Equal to zero c. Less than the input offset voltage d. Unimportant when a base resistor is used

Answer»

(a) Less than the input bias current

23090.

With Sallen-Key high-pass filters, the pole frequency must be a. Added to the K values b. Subtracted from the K values c. Multiplied by the K values d. Divided by the K values

Answer»

(d) Divided by the K values

23091.

When we trim a resistor, we are a. Making a fine adjustment b. Reducing its value c. Increasing its value d. Making a coarse adjustment

Answer»

(a) Making a fine adjustment

23092.

If an op amp has only a positive supply voltage, its output cannot a. Be negative b. Be zero c. Equal the supply voltage d. Be ac coupled

Answer»

(a) Be negative

23093.

A D/A converter with four inputs has a. Two outputs b. Four outputs c. Eight outputs d. Sixteen outputs

Answer»

(d) Sixteen outputs

23094.

When a JFET is used in an AGC circuit, it acts like a a. Switch b. Voltage-controlled current source c. Voltage-controlled resistance d. Capacitance

Answer»

(c) Voltage-controlled resistance

23095.

Link coupling refers to a. Capacitive coupling b. Transformer coupling c. Resistive coupling d. Power coupling

Answer»

(b) Transformer coupling

23096.

An op amp with a rail-to-rail output a. Has a current-boosted output b. Can swing all the way to either supply voltage c. Has a high output impedance d. Cannot be less than 0 V.

Answer»

(b) Can swing all the way to either supply voltage

23097.

The Q of a Sallen-Key second-order stage depends on the a. Voltage gain b. Center frequency c. Bandwidth d. GBW of the op amp

Answer»

(a) Voltage gain

23098.

The Hartley oscillator uses a. Negative feedback b. Two inductors c. A tungsten lamp d. A tickler coil

Answer»

(b) Two inductors

23099.

To vary the frequency of an LC oscillator, you can vary a. One resistor b. Two resistors c. Three resistors d. One capacitor

Answer»

(d) One capacitor

23100.

The material with the piezoelectric effect is a. Quartz b. Rochelle salts c. Tourmaline d. All the above

Answer»

(d) All the above