Match the pair :SET ISET II(a) Chopper(i) Variable speed DC drive(b) Dimmerstat(ii) Variable speed AC drive(c) Inverter(iii) PMDC motor speed control(d) Phase controlled rectifier(iv) Variable AC voltage applications

The correct matching is a-iii, b-iv, c-ii, d-i.A chopper converts fixed DC to variable DC, used for PMDC motor speed control (iii).A dimmerstat (variable autotransformer) gives a variable AC voltage (iv).An inverter produces variable-frequency AC for variable speed AC drives (ii).A phase-controlled rectifier converts AC to variable DC for variable speed DC drives (i).The other codes give the dimmerstat or the rectifier the wrong application, for example treating a phase-controlled rectifier as an AC drive.

Choppers Questions and Answers

Q: In a step down chopper using pulse width modulation, \(T_{ON} = 9 \times 10^{-3}\) s and \(T_{OFF} = 1 \times 10^{-3}\) s. The chopping frequency is :

The chopping frequency is 100 Hz.The chopping period is the on time plus the off time, and the frequency is its reciprocal:\[ \begin{aligned} T &= T_{ON} + T_{OFF} = 9 + 1 = 10\ \text{ms} \\ f &= \frac{1}{T} = \frac{1}{10 \times 10^{-3}} = 100\ \text{Hz} \end{aligned} \]The duty cycle is 9/10 = 0.9, so a step down chopper gives an average output of 0.9 times the input. The option 1000 Hz is \(1/T_{OFF}\), and 200 Hz and 500 Hz do not follow from the given times.

Q: Match the following :Set ISet II(a) Phase controlled converter(i) DC to DC conversion(b) Chopper(ii) DC to AC conversion(c) Inverter(iii) AC to AC conversion(d) Dimmerstat(iv) AC to DC conversion

The correct code is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).A phase-controlled converter (controlled rectifier) turns AC into variable DC (iv). A chopper switches a DC supply on and off rapidly to give a different average DC voltage (i). An inverter produces AC from DC (ii). A dimmerstat, a variable autotransformer, changes the magnitude of an AC voltage, AC to AC (iii).The other codes give the chopper an AC role or treat the dimmerstat as a DC device, which does not fit how these circuits work.

Q: A dc to dc transistor chopper supplied from a fixed voltage dc source feeds afixed-resistive-inductive load and a free-wheeling diode. The chopper operates at 1 kHz and 50% duty cycle. Without changing the value of the average dc current through the load, if it is desired to reduce the ripple content of load current, the control action needed will :

The control action is to increase the chopper frequency keeping the duty cycle constant.The average load voltage of a chopper is \(D \times V_{s}\), so the average load current is fixed by the duty cycle D. The peak-to-peak ripple in an RL load current is inversely related to the switching frequency relative to the load time constant: at a higher frequency the current has less time to rise and fall in each period, so the ripple shrinks.Changing the duty cycle, alone or together with frequency, changes the average voltage and hence the average current, which the question forbids. Decreasing the frequency would increase the ripple, the opposite of what is wanted.

In the given figure, the value of \(R = 0\), the source voltage \(V_s = 300\) V (constant) and the load inductance \(L = 300\ \mu\)H. If the switch S1 is closed till time \(t = 100\ \mu\)s, what is the value of load current I at time \(t = 100\ \mu\)s?

Circuit: a DC source Vs, a switch S1 closing at t = 0 and a diode D1 in series with a load R and L, with a freewheeling diode Dm across the load and load current i

A.0 A
B.50 A
C.100 A
D.200 A

Exam Year(s)

DSSSB AE - 2023,

Answer: C

With \(R = 0\) the load is a pure inductance, and when the switch S1 closes at \(t = 0\) the constant source voltage \(V_s\) is applied across \(L\) (the diode \(D_1\) conducts, and \(D_m\) is reverse biased). The current rises linearly.

\[ V_s = L\frac{di}{dt} \Rightarrow i(t) = \frac{V_s}{L}t \]

\[ \begin{aligned} i(100\ \mu\text{s}) \\ &= \frac{300}{300 \times 10^{-6}} \times 100 \times 10^{-6} \\ &= 100\ \text{A} \end{aligned} \]

The answer is 100 A, option C. The value of 50 A would be at \(t = 50\ \mu\)s and 200 A at \(t = 200\ \mu\)s.

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In a step down chopper using pulse width modulation, \(T_{ON} = 9 \times 10^{-3}\) s and \(T_{OFF} = 1 \times 10^{-3}\) s. The chopping frequency is :

A.200 Hz
B.1000 Hz
C.500 Hz
D.100 Hz

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The chopping frequency is 100 Hz.

The chopping period is the on time plus the off time, and the frequency is its reciprocal:

\[ \begin{aligned} T &= T_{ON} + T_{OFF} = 9 + 1 = 10\ \text{ms} \\ f &= \frac{1}{T} = \frac{1}{10 \times 10^{-3}} = 100\ \text{Hz} \end{aligned} \]

The duty cycle is 9/10 = 0.9, so a step down chopper gives an average output of 0.9 times the input. The option 1000 Hz is \(1/T_{OFF}\), and 200 Hz and 500 Hz do not follow from the given times.

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Match the following :

Set I	Set II
(a) Phase controlled converter	(i) DC to DC conversion
(b) Chopper	(ii) DC to AC conversion
(c) Inverter	(iii) AC to AC conversion
(d) Dimmerstat	(iv) AC to DC conversion

A.(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
B.(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
C.(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
D.(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The correct code is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).

A phase-controlled converter (controlled rectifier) turns AC into variable DC (iv). A chopper switches a DC supply on and off rapidly to give a different average DC voltage (i). An inverter produces AC from DC (ii). A dimmerstat, a variable autotransformer, changes the magnitude of an AC voltage, AC to AC (iii).

The other codes give the chopper an AC role or treat the dimmerstat as a DC device, which does not fit how these circuits work.

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Match the pair :

SET I	SET II
(a) Chopper	(i) Variable speed DC drive
(b) Dimmerstat	(ii) Variable speed AC drive
(c) Inverter	(iii) PMDC motor speed control
(d) Phase controlled rectifier	(iv) Variable AC voltage applications

A.a-i, b-ii, c-iii, d-iv
B.a-iv, b-iii, c-ii, d-i
C.a-iii, b-iv, c-ii, d-i
D.a-iii, b-iv, c-i, d-ii

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The correct matching is a-iii, b-iv, c-ii, d-i.

A chopper converts fixed DC to variable DC, used for PMDC motor speed control (iii).
A dimmerstat (variable autotransformer) gives a variable AC voltage (iv).
An inverter produces variable-frequency AC for variable speed AC drives (ii).
A phase-controlled rectifier converts AC to variable DC for variable speed DC drives (i).

The other codes give the dimmerstat or the rectifier the wrong application, for example treating a phase-controlled rectifier as an AC drive.

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A dc to dc transistor chopper supplied from a fixed voltage dc source feeds afixed-resistive-inductive load and a free-wheeling diode. The chopper operates at 1 kHz and 50% duty cycle. Without changing the value of the average dc current through the load, if it is desired to reduce the ripple content of load current, the control action needed will :

A.increase the chopper frequency and duty cycle in equal ratio
B.increase the chopper frequency keeping the duty cycle constant
C.decrease only the chopper frequency
D.decrease only the duty cycle

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The control action is to increase the chopper frequency keeping the duty cycle constant.

The average load voltage of a chopper is \(D \times V_{s}\), so the average load current is fixed by the duty cycle D. The peak-to-peak ripple in an RL load current is inversely related to the switching frequency relative to the load time constant: at a higher frequency the current has less time to rise and fall in each period, so the ripple shrinks.

Changing the duty cycle, alone or together with frequency, changes the average voltage and hence the average current, which the question forbids. Decreasing the frequency would increase the ripple, the opposite of what is wanted.

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Match List – I with List – II and select the correct answer using codes given below :

List – I	List – II
a. Duty cycle	1. Variable voltage regulator
b. AGC	2. Astable multivibrator
c. LM317	3. Voltage regulator
d. Crow-bar protection	4. FET

Codes : a b c d

A.1 2 3 4
B.3 4 1 2
C.4 3 2 1
D.2 4 1 3

Exam Year(s)

UGC NET - 2012,

Answer: D

The duty cycle is a property of the output of an astable multivibrator, so a matches 2. Automatic gain control in a receiver uses a FET as the variable element, so b matches 4. The LM317 is a variable voltage regulator, so c matches 1. A crow-bar protection circuit protects the load of a voltage regulator against overvoltage, so d matches 3. The code is 2 4 1 3. The code 3 4 1 2 swaps the first and last pairs, and 1 2 3 4 and 4 3 2 1 are wrong in every pair. The answer is 2 4 1 3.

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Match List – I with List – II and select the correct answer using the codes given below :

List – I	List – II
a. Choppers	1. pn pn devices
b. SCR	2. V number
c. SMF	3. Speed control of d.c. motors
d. LASER	4. Quantum well

Codes : a b c d

A.3 2 4 1
B.3 1 2 4
C.4 2 1 3
D.4 3 2 1

Exam Year(s)

UGC NET - 2012,

Answer: B

Choppers are used for speed control of d.c. motors, so a matches 3. The SCR is a four layer pnpn device, so b matches 1. A single mode fibre is characterised by its V number, which decides whether only one mode propagates, so c matches 2. A laser diode may use a quantum well structure, so d matches 4. The code is 3 1 2 4. The code 3 2 4 1 swaps the second to fourth pairs, and the codes beginning with 4 give the chopper the quantum well. The answer is 3 1 2 4.

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Step down regulator produces an output

A.low voltage and low current
B.low voltage at high current
C.high voltage at low current
D.high voltage at high current

Exam Year(s)

UGC NET - 2013,

Answer: B

A step-down regulator gives a lower output voltage than its input, and since power is nearly conserved, the output current is higher than the input current. So it produces low voltage at high current, option (B).

Low voltage and low current (option A) would mean that power was lost, and high voltage at low current (option C) describes a step-up converter. High voltage at high current (option D) would give more output power than input power, which no regulator can do.

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Match the following lists :

List - I	List - II
a. Slip	i. Relaxation oscillator
b. SCR	ii. Class-E chopper
c. UJT	iii. Two-transistor model
d. dc drives	iv. Induction motors

A.i ii iii iv
B.iii iv i ii
C.ii iii i iv
D.iv iii i ii

Exam Year(s)

UGC NET - 2013,

Answer: D

The correct match is slip with induction motors (iv), SCR with the two-transistor model (iii), UJT with the relaxation oscillator (i) and dc drives with the class-E chopper (ii), so the code is iv iii i ii, option (D).

Slip is a term of the induction motor, an SCR is explained by two coupled transistors, a UJT is the basis of a relaxation oscillator, and the class-E chopper is used in dc drives. Options A, B and C exchange some pairs.

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