Alternating Current Class 12 MCQs Questions with
Alternating Current Class 12 MCQ Chapter 7 Question 1.
If the rms current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 s after its value becomes zero is
(A) \(5 \sqrt{2}\)
(B) \(5 \sqrt{\frac{3}{2}}\)
(c) \(\frac {5}{6}\)
(B) \(\frac{5}{\sqrt{2}}\)
Answer:
(B) \(5 \sqrt{\frac{3}{2}}\)
Explanation:
Here, Irms = 5 A, n = 50 Hz and t = \(\frac {1}{300}\) s
= Peak value = \(\sqrt{2} I_{\text {rms }}\) = \(\sqrt{2}\) x 5 = 5\(\sqrt{2}\) A
Now, I = I0 sin ωt = 5\(\sqrt{2}\) sin 2πvt
= 5V2 sin2π x 50 x \(\frac {1}{300}\) = 5\(\sqrt{\frac{3}{2}}\) A
Class 12 Physics Chapter 7 MCQ Question 2.
An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to supply power to a passive load consisting of a resistance Rg and a reactance XL. For maximum power to be delivered from the generator to the load, the value of XL is equal to
(A) zero
(B) Xg
(C) Q – Xg
(D) Rg
Answer:
(C) Q – Xg
Explanation:
As internal resistance of generator is already equal to external resistance Rg. So to deliver maximum power, i.e., to make reactance equal to zero, the reactance in external circuit will be – Xg. In order to deliver maximum power, the generator to the load, the total reactance must be equal to zero, i.e., XL + Xg = 0, XL = – Xg
Chapter 7 Physics Class 12 MCQ Question 3.
When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V. this means
(A) input voltage cannot be AC voltage, but a DC voltage.
(B) maximum input voltage is 220 V.
(C) The meter reads not v but (v2) and is calibrated to read \(\sqrt{\left(v^{2}\right)}\)
(D) The pointer of the meter is stuck by some mechanical defect.
Answer:
(C) The meter reads not v but (v2) and is calibrated to read \(\sqrt{\left(v^{2}\right)}\)
Explanation:
The voltmeter in AC circuit reads value <υ2> and meter is calibrated to rms value <υ2> which is multiplied by \(\sqrt{2}\) to get Vrms. In other words, voltmeter connected to the AC main read root mean square value of
AC voltage, i.e., \(\sqrt{\left\langle v^{2}\right\rangle}\)
Alternating Current MCQ Chapter 7 Question 4.
When frequency of applied alternating voltage very high then
(A) A capacitor will tend to become SHORT
(B) An inductor will tend to become SHORT
(C) Both (A) and (B)
(D) No one will become short
Answer:
(A) A capacitor will tend to become SHORT
Explanation:
XC = l/2πfC
So, as / increases, XC becomes smaller and smaller. For very high value of f, XC will be too small which may be considered as SHORT.
Ch 7 Physics Class 12 MCQ Question 5.
Relation between r.m.s. voltage and instantaneous voltage of an AC
(A) V0 = VRMS / \(\sqrt{2}\)
(B) VRMS = V0 / V2
(C) VRMS = 0.707 V0
(D) Both (C) and (D)
Answer:
(D) Both (C) and (D)
Explanation:
VRMS = V0/\(\sqrt{2}\) = O.707 V0
Physics Class 12 Chapter 7 MCQ Question 6.
The heat produced in a given resistance in a given time by the sinusoidal current I sin cof will be the same as heat produced by a steady current of magnitude
(A) 0.707 I0
(B) 1.4121 I0
(C) I0
(D) \(\sqrt{I}_{0}\)
Answer:
(A) 0.707 I0
Explanation:
Heat produced by AC is Heat produced by DC is I2R
\(\mathrm{I}_{\mathrm{RMS}}^{2}\) = I2R
∴ I = Irms= I0/\(\sqrt{2}\) = 0.707 I0
MCQ Questions For Class 12 Physics Chapter 7 Question 7.
An A.C. source is connected to a resistive circuit. Which of the following statements is true?
(A) Current leads the voltage in phase
(B) Current lags the voltage in phase
(C) Current and voltage are in same phase
(D) Either (A) or (B) depending on the value of resistance.
Answer:
(A) Current leads the voltage in phase
Explanation:
In a pure resistive circuit, current and voltage are always in phase.
MCQ Of Alternating Current Class 12 Question 8.
In which of the following circuit power dissipation is maximum?
(A) Pure capacitive circuit
(B) Pure inductive circuit
(C) Pure resistive circuit
(D) LR or CR circuit
Answer:
(C) Pure resistive circuit
Explanation:
Since in pure resistive circuit the current and voltage are in phase, the power dissipation is maximum.
MCQ On Alternating Current Chapter 7 Question 9.
To reduce the resonant frequency in an L – C – R series circuit with a generator
(A) the generator frequency should be reduced.
(B) another capacitor should be added in parallel to the first.
(C) the iron core of the inductor should be removed.
(D) dielectric in the capacitor should be removed.
Answer:
(B) another capacitor should be added in parallel to the first.
Explanation:
The resonant frequency of L-C-R series circuit is v0 = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
So to reduce resonant frequency, we have either to increase L or to increase C.
To increase capacitance, another capacitor must be connected in parallel with the first.
Physics Chapter 7 Class 12 MCQ Question 10.
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
(A) R = 20 Q, L = 1.5 H, C = 35 pF
(B) R = 25 Q, L = 2.5 H, C = 45 pF
(C) R = 15 Q, L = 3.5 H, C = 30 pF
(D) R = 25 Q, L = 1.5 H, C = 45 pF
Answer:
(C) R = 15 Q, L = 3.5 H, C = 30 pF
Explanation:
Quality factor (Q) of an L-C-R circuit is given by,
Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)
Tuning of an L-C-R circuit depends on quality factor of the circuit. Tuning will be better when quality factor of the circuit is high. For Q to be high, R should be low, L should be high and C should be low. Therefore, option (C) is most suitable.
Class 12 Physics Ch 7 MCQ Question 11.
With increase in frequency of an A.C. supply, the impedance of a series L-C-R circuit
(A) remains constant.
(B) increases.
(C) decreases.
(D) decreases at first, becomes minimum and then increases.
Answer:
(D) decreases at first, becomes minimum and then increases.
Explanation:
The frequency vs. impedance graph of a series LCR circuit is as follows:
With increase in frequency, the impedance decreases at first, becomes minimum and then increases.
AC Class 12 MCQ Chapter 7 Question 12.
The sharpness of tuning of a series LCR circuit at resonance is measured by Q factor of the circuit which is given by
(A) Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)
(B) Q = \(\frac{1}{R} \sqrt{\frac{C}{L}}\)
(C) Q = \(\frac{1}{L} \sqrt{\frac{R}{C}}\)
(D) Q = \(\frac{1}{C} \sqrt{\frac{R}{L}}\)
Answer:
(A) Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)
Explanation:
Q factor of a series LCR circuit is given by Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)
MCQ On Alternating Current Class 12 Question 13.
At resonance, the impedance in series LCR circuit is
(A) maximum.
(B) zero.
(C) infinity.
(D) minimum.
Answer:
(D) minimum.
Explanation:
Impedance of a series LCR circuit is Z = \(\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}\)
At resonance, XC = XL
So, Z is minimum.
Alternating Current MCQ Class 12 Question 14.
The power factor of series LCR circuit at resonance is –
(A) 0.707
(B) 1
(C) 0.5
(D) 0
Answer:
(B) 1
Explanation:
At resonance, LCR circuit behaves I as purely resistive circuit. For purely resistive circuit, power factor is 1.
Alternating Current MCQs Chapter 7 Question 15.
When a capacitor C is charged to a certain potential and connected to an inductor L, then frequency of energy oscillation is given by –
(A) Q = \(\frac{1}{2 \pi \sqrt{L C}}\)
(B) Q = \(\frac{1}{\sqrt{\mathrm{LC}}}\)
(C) Q = \(\frac{2 \pi}{\sqrt{\mathrm{LC}}}\)
(D) Q = \(\frac{1}{2 \pi} \sqrt{\frac{L}{C}}\)
Answer:
(A) Q = \(\frac{1}{2 \pi \sqrt{L C}}\)
Explanation:
When a capacitor C is charged to a certain potential and connected to an inductor L, energy stored in C oscillates between L and C.
XL = XC
∴ f = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
Alternating Current Class 12 MCQ Questions Question 16.
The output of a step-down transformer is measured to be 24 V when connected to a 12 W Light bulb. The value of the peak current is
(A) 1/\(\sqrt{2}\) A
(B) \(\sqrt{2}\) A
(C) 2 A
(D) 2\(\sqrt{2}\) A
Answer:
(A) 1/\(\sqrt{2}\) A
Explanation:
Power associated with secondary Ps = 12 W
Secondary voltage, Vs = 24 V
Current in the secondary, Is = \(\frac{P_{s}}{V_{s}}=\frac{12}{24}\) = 0.5 A
Peak value of the current in the secondary,
l0 = Is\(\sqrt{2}\) = (0.5)(1.414)= 0.707 or \(\frac{1}{\sqrt{2}}\) A
Class 12 Alternating Current MCQ Question 17.
The underlying principle of transformer is –
(A) resonance.
(B) mutual induction.
(C) self induction.
(D) none of the above.
Answer:
(B) mutual induction.
Explanation:
The transformer is based on the principle of mutual induction which state that due to continuous change in current in the primary coil, an emf is induced across the secondary soil.
Physics Chapter 7 MCQ Class 12 Question 18.
The core of a transformer is laminated as
(A) it improves the ratio of voltage in the primary and secondary may be increased.
(B) it checks rusting of the core may be stopped.
(C) it reduces energy losses due to eddy currents.
(D) it increases flux linkage.
Answer:
(C) it reduces energy losses due to eddy currents.
Explanation:
Laminated core means a layered core instead of a single solid core. Eddy currents are current loops generated by changing magnetic fields. They flow in a plane perpendicular to the magnetic field. Laminated magnetic core reduces eddy currents. For this reason, electrically isolated laminations are utilized to manufacture transformers.
Chapter 7 MCQ Class 12 Physics Question 19.
If rotational velocity of an armature is doubled, emf generated in a generator will be
(A) half.
(B) two times.
(C) four times.
(D) unchanged.
Answer:
(B) two times.
Explanation:
emf generated = NBA
Class 12 Physics Chapter 7 MCQ Questions Question 20.
Quantity that remains unchanged in a transformer is –
(A) voltage.
(B) current.
(C) frequency.
(D) none of these.
Answer:
(C) frequency.
Explanation:
Transformer does not change the frequency of the applied AC.
Alternating Current Class 12 MCQs Chapter 7 Question 21.
…………… increases in step-down transformer.
(A) Voltage
(B) Current
(C) Power
(D) Current density
Answer:
(B) Current
Explanation:
Since Vp/Vs = Is/Ip, so as as voltage reduces, the current increases in a step-down transformer.
Question 22.
The efficiency of transformer is very high because
(A) There is no moving part
(B) It uses AC only
(C) It uses the copper wire for the coils
(D) None of the above
Answer:
(A) There is no moving part
Explanation:
Transformer is a static device which transforms power from one circuit to other through electromagnetic induction. In electrical transformer as there are no moving parts, no friction. Losses in the transformer are very less compared to any other rotating machine, hence efficiency of transformers will be very high which is about 95% to 98%.
Assertion And Reason Based MCQs (1 Mark each)
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true
Question 1.
Assertion (A): An alternating current does not show any magnetic effect.
Reason (R): Alternating current changes direction with time.
Answer:
(D) A is false and R is true
Explanation:
Current or moving charged particle creates magnetic field irrespective of direct current or alternating current. So assertion is false. Alternating current changes direction with time. So, the reason is true, but cannot explain the assertion.
Question 2.
Assertion (A): Capacitor blocks dc and allows ac to pass.
Reason (R): Capacitive reactance is inversely proportional to frequency.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
Capacitive reactance = \(\frac{1}{2 \pi f C}\)
So, as f (frequency) increases, reactance decreases.
For dc, frequency = 0, hence capacitor offers infinite reactance. So, it blocks dc.
For ac, frequency ≠ 0, hence capacitor offers low reactance and allows ac to pass.
Hence assertion and reason both are true. Assertion is properly explained by reason.
Question 3.
Assertion (A): VRMSvalue of an alternating voltage V = 4\(\sqrt{2}\) sin 314t is 4 volt.
Reason (R): Peak value of the alternating voltage is 442 volt.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Given alternating voltage
V = 40\(\sqrt{2}\) sin 314t.
Where peak value = V0/\(\sqrt{2}\) = 4 volt.
VRMS = V0/\(\sqrt{2}\) = 4 volt.
Hence both assertion and reason both are true. But the reason does not properly explain the assertion.
Question 4.
Assertion (A): Both ac and dc can be measured by hot wire instrument.
Reason (R): Hot wire instrument is based on the principal of magnetic effect of current.
Answer:
(C) A is true but R is false
Explanation:
In both ac and dc, heat generated is proportional to the square of current. Polarity change of ac is immaterial in the case of heat generation. Hence they can be measured by hot wire instrument. Hence, the assertion is true. Hot wire instruments are based on the principle of heating effect of current. Hence the reason is false.
Question 5.
Assertion (A): The dimension of L/R is time.
Reason (R): Time constant (L/R) should be increased to reduce the rate of increase of current through a solenoid.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
For a solenoid, the magnitude of induced emf
e = L\(\frac{d i}{d t}\)
i = \(\frac{e}{R}=\left(\frac{L}{R}\right)\left(\frac{d i}{d t}\right)\)
\(\frac{d i}{d t}=\frac{i}{\frac{L}{R}}\)
In left hand side of the above equation, denominator is time. So, in right hand side, the denominator should be time. So, dimension of L/R is time. So, the assertion is true. If L/R increases, di/dt decreases. So, reason is also true. But reason cannot properly explain the assertion.
Question 6.
Assertion (A): At resonance, the current becomes minimum in a series LCR circuit.
Reason (R): At resonance, voltage and current are out of phase in a series LCR circuit.
Answer:
(D) A is false and R is true
Explanation:
At resonance, XL = XC, so the circuit impedance becomes minimum and resistive and hence the current becomes maximum. So, the assertion is false. At resonance, XL = XC so the circuit impedance becomes resistive. In resistive circuit voltage and current are always in same phase. Hence, reason is also false.
Question 7.
Assertion (A): When capacitive reactance is less than the inductive reactance in a series LCR circuit, e.m.f. leads the current.
Reason (R): The angle by which alternating voltage leads the alternating current in series RLC circuit is
given by tan φ = \(\frac{X_{L}-X_{C}}{R}\)
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
The angle by which alternating voltage leads the alternating current in series RLC circuit is given by tan φ = \(\frac{X_{L}-X_{C}}{R}\)
If XC < XLthen tan φ is positive, φ is also positive. So, e.m.f. leads the current. Assertion and reason both are true. Reason properly explains the assertion.
Question 8.
Assertion (A): Quality factor of a series LCR circuit is Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)
Reason (R): As bandwidth decreases, Q increases in a resonant LCR circuit.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Quality factor of a series LCR circuit is Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\) Assertion is true. Quality factor is also defined as
So, reason is also true. But reason does not explain the assertion.
Question 9.
Assertion (A): Principle of operation of AC generator is electromagnetic induction.
Reason (R): Resistance offered by inductor for AC is zero.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Principle of operation of AC generator is electromagnetic induction. The assertion is true.
Resistance offered by inductor = 2πfL. For AC, f ≠ 0. So, 2nfL ≠ 0. So, the reason is false.
Question 10.
Assertion (A): An alternator is a machine which converts mechanical energy into electrical energy.
Reason (R): When a coil rotates in a magnetic field an e.m.f. is induced in it.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
Alternator is basically a generator in which a coil rotates in a strong magnetic field and according to laws of electromagnetic induction e.m.f. is generated. So, assertion and reason both are true and reason explains the assertion.
Question 11.
Assertion (A): A transformer does not work on DC.
Reason (R): DC neither change direction nor magnitude.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
Transformer has two coils. If current fluctuates in one coil, e.m.f. is induced in the other coil. For DC supply current does not change, so there is no induced e.m.f. Hence both assertion and reason are true and reason explains the assertion.
Question 12.
Assertion (A): A step-up transformer converts input low AC voltage to output high AC voltage.
Reason (R): It violate the law of conservation of energy.
Answer:
(C) A is true but R is false
Explanation:
Step-up transformer means it converts input low AC voltage to output high AC voltage. So, the assertion is true.
For step up transformer, VOUT/ VIN >1, but simultaneously IOUT /IIN < 1 and PIN = POUT (ideally). Hence, the law of conservation of energy is not violated.
Case-Based MCQs
Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:
Tuning a radio set:
In essence the simplest tuned radio frequency receiver is a simple crystal set. Desired frequency is tuned by a tuned coil / capacitor combination, and then the signal is presented to a simple crystal or diode detector where the amplitude modulated signal, is demodulated. This is then passed straight to the headphones or speaker. In radio set there is an LC oscillator comprising of a variable capacitor (or sometimes a variable coupling coil), with a knob on the front panel to tune the receiver.
Capacitor used in old radio sets is gang capacitor. It consists of two sets of parallel circular plates one of which can rotate manually by means of a knob. The rotation causes overlapping areas of plates to change, thus changing its capacitance. Air gap between plates acts as dielectric. The capacitor has to be tuned in tandem corresponding to the frequency of a station so that the LC combination of the radio set resonates at the frequency of the desired station.
When capacitive reactance (XC) is equal to the inductive reactance (XL), then the resonance occurs and the resonant frequency is given by ω0 = \(\frac{1}{\sqrt{L C}}\) current amplitude becomes maximum at the resonant frequency. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the \(\frac{V_{m}}{R}\) Current amplitude is the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.
Question 1.
Name the phenomenon involved in tuning a radio set to a particular radio station.
(A) Stabilization
(B) Rectification
(C) Resonance
(D) Reflection
Answer:
(C) Resonance
Explanation:
Phenomenon involved in tuning a radio set to a particular radio station is resonance. The capacitor has to be tuned in tandem corresponding to the frequency of a station. So, that the LC combination of the radio set resonates at the frequency of the desired station.
Question 2.
Resonance may occur in:
(A) RL circuit.
(B) RC circuit.
(C) LC circuit.
(D) circuit having resistor only.
Answer:
(C) LC circuit.
Explanation:
A simple radio receiver is a simple crystal set with a coil and capacitor combination. Desired frequency is tuned by tuning the coil – capacitor combination. Tuning means to make capacitive reactance (XC) equal to the inductive reactance (XL), so that the resonance occurs.
Question 3.
Resonance frequency is equal to:
(A) \(\frac{1}{\mathrm{LC}}\)
(B) \(1 / \sqrt{\mathrm{LC}}\)
(C) \(\sqrt{\frac{L}{C}}\)
(D) \(\sqrt{\frac{C}{L}}\)
Answer:
(B) \(1 / \sqrt{\mathrm{LC}}\)
Explanation:
The resonant frequency is given by CQ0 = 1/VLC
Question 4.
Resonance occurs only when:
(A) XC = R
(B) XL = R
(C) XL = XC
(D) XC>XL
Answer:
(C) XL = XC
Explanation:
At resonance, capacitive reactance (XC) is equal to the inductive reactance (XL). Circuit is totally resistive and the current amplitude becomes maximum.
Question 5.
Capacitor used in radio set for tuning is a:
(A) parallel plate capacitor.
(B) spherical capacitor.
(C) paper capacitor.
(D) electrolytic capacitor.
Answer:
(A) parallel plate capacitor.
Explanation:
Capacitor used in old radio is a parallel plate capacitor. It consists of two sets of parallel circular plates, one of which can rotate manually by means of a knob. The rotation causes overlapping areas of plates to change, thus changing its capacitance.
II. Read the following text and answer the following questions on the basis of the same:
At power plant, a transformer increases the voltage of generated power by thousands of volts so that it can be sent of long distances through high-voltage transmission power lines. Transmission lines are bundles of wires that carry electric power from power plants to distant substations.
At substations, transformers lower the voltage of incoming power to make it acceptable for high- volume delivery to nearby end-users.
Electricity is sent at extremely high voltage because it Emits so-called line losses. Very good conductors of electricity also offer some resistance and this resistance becomes considerable over long distances causing considerable loss.
At generating station, normally voltage is stepped up to around thousands of volts. Power losses increase with the square of current. Therefore, keeping voltage high current becomes low and the loss is minimized. Another option of minimizing loss is the use of wires of super-conducting material. Super-conducting materials are capable of conducting without resistance, they must be kept extremely cold, nearly absolute zero, and this requirement makes standard super-conducting materials impractical to use.
However, recent advances in super-conducting materials have decreased cooling requirement. In Germany recently 1 km super-conducting cable have been installed connecting the generating station and the destination. It has eliminated the line loss and the cable is capable of sending five times more electricity than conventional cable. Using super-conducting cables Germany has also get rid of the need of costly transformers. Transformers generate waste heat when they are in operation and oil is the coolant of choice.
It transfers the heat through convection to the transformer housing, which has cooling fins or radiators similar to heat exchangers on the outside. Flush point is a very important parameter of transformer oil. Flashpoint of an oil is the temperature at which the oil ignites spontaneously. This must be as high as possible (not less than 160° C from the point of safety). Fire point is the temperature at which the oil flashes and continuously burns. This must be very high for the chosen oil (not less than 200° C).
Question 1.
Which of the following statement is true for long distance transmission of electricity?
(A) Step-down transformer is used at generating station and step-up transformer is used at destination substation.
(B) Step-down transformers are used at generating station and destination substation.
(C) Step-up transformers are used at generating station and destination substation.
(D) None of the above
Answer:
(D) None of the above
Explanation:
At power plant, a step-up transformer increases the voltage of generated power by thousands of volts, so that it can be sent of long distances through high-voltage transmission power lines. At substations, step-down transformers lower the voltage of incoming power to make it acceptable for high-volume delivery to nearby end-users.
Question 2.
Super-conducting transmission line has the following advantages:
(A) Resistance being zero, there is no I2,R loss.
(B) There is no requirement of costly step-up and step-down transformers.
(C) Cable is capable of sending more electricity.
(D) All of the above
Answer:
(D) All of the above
Explanation:
Super-conducting materials are capable of conducting without resistance. So, this eliminates the line loss and the cable is capable of sending more electricity than conventional cable. Using super-conducting cables, one can get rid of the need of costly transformers.
Question 3.
Why does stepping up voltages reduce power loss?
(A) Since resistance of conductor decreases with increase of voltage
(B) Since current decreases with increase of voltage
(C) Both of the above
(D) None of the above
Answer:
(B) Since current decreases with increase of voltage
Explanation:
At generating station, normally voltage is stepped up to around thousands of volts. Power losses increase with the square of current. Therefore, keeping voltage high, current becomes low and the loss is minimized.
Question 4.
Oil transfers heat from transformer winding by the process of:
(A) convection.
(B) conduction.
(C) radiation.
(D) All of these
Answer:
(A) convection.
Explanation:
Transformers generate waste heat when they are in operation and oil is the coolant of choice. It transfers the heat through convection to the transformer housing.
Question 5.
Flush point of an oil is –
(A) the temperature at which the oil flashes and continuously burns.
(B) the temperature at which the oil ignites spontaneously.
(C) the temperature at which the oil starts boiling.
(D) The temperature at which the oil forms fumes.
Answer:
(B) the temperature at which the oil ignites spontaneously.
Explanation:
Flush point is a very important parameter of transformer oil. Flashpoint of an oil is the temperature at which the oil ignites spontaneously. This must be as high as possible (not less than 160° C from the point of safety).
III. Read the following text and answer the following questions on the basis of the same:
Losses of transformer:
There are 4 types of losses in a transformer: Core loss, Ohmic loss, Stray load loss and dielectric loss.
(1) Core loss :
Core loss has two components – hysteresis loss and eddy current loss. These together are called no-load losses of a transformer and are calculated by open circuit test.
(a) Hysteresis loss: This loss mainly depends on the core material used in the transformer. To reduce this loss, the high-grade core material can be used. CRGO- Cold rolled grain oriented Si steel is commonly used for this purpose.
(b) Eddy current loss: This loss can be reduced by designing the core using slight laminations.
These losses are present even when no load is connected. So, these are also known as no-load loss.
(2) Copper Loss:
Copper losses occur because of the Ohmic resistance in the windings of the transformer. If the currents in primary and secondary windings of the transformer are I1 and I2, and if the resistances of these windings are R1 & R2 then the copper losses that occurred in the windings are I12R1 & I22R22 respectively. So, the entire copper loss will be I12R1 + I22R2 This loss is also called variable or ohmic losses because this loss changes based on the load.
(3) Stray Loss :
These types of losses in a transformer occur because of the occurrence of the leakage flux. As compared with copper and iron losses, the percentage of stray losses are less, so these losses can be neglected.
(4) Dielectric Loss :
This loss mainly occurs within the oil of the transformer. Oil is an insulating material. Once the oil quality in the transformer deteriorates then the transformer’s efficiency is affected. Efficiency of Transformer It is the ratio of output power and input power. Efficiency = Output Power / Input Power. The transformer is a highly efficient device which ranges between 95% – 98.5%.
Question 1.
What is the relationship among core loss, hysteresis loss and eddy current loss?
(A) Eddy current loss = Core loss + Hysteresis loss
(B) Core loss = Hysteresis loss + eddy current loss
(C) Hysteresis loss = Core loss + eddy current loss
(D) Core loss = Hysteresis loss X eddy current loss
Answer:
(B) Core loss = Hysteresis loss + eddy current loss
Question 2.
Which of the following losses in transformer is also known as no-load loss?
(A) Copper loss
(B) Stray loss
(C) Dielectric loss
(D) Core loss
Answer:
(D) Core loss
Explanation:
Core loss is present even when no load is connected. So, these are also known as no-load loss.
Question3.
Which of the following losses in transformer is also known as variable loss?
(A) Copper loss
(B) Stray loss
(C) Dielectric loss
(D) Core loss
Answer:
(A) Copper loss
Explanation:
If the currents in primary and secondary windings of the transformer are I1 and I2 respectively and the resistances of these windings are R1 and R2 then the copper losses that occurred in the windings are I12R1 and I22R2 respectively. So, the entire copper loss will be I12R1 + I22R2. This loss is also called variable or ohmic losses because this loss changes based on the load.
Question 4.
How hysteresis loss can be reduced?
(A) Using core of Si Steel
(B) Using laminated core
(C) Using core of non-ferro magnetic material
(D) Using oil of higher dielectric constant
Answer:
(A) Using core of Si Steel
Question 5.
Specify the range of transformer efficiency.
(A) 10 – 15%
(B) 95 – 98%
(C) 50 – 60%
(d) 40 – 50%
Answer:
(B) 95 – 98%