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The correct option is (C) 4

The BEST explanation: The Bode plot is a logarithmic plot which helps in FITTING a large scale of values into a small scale by the application of logarithm. Plotting the slope 40dB/decade on the Bode plot, we get n=4. Hence order of the BAND Pass FILTER is 4.

The CORRECT option is (c) – 1 A, \(\SQRT{2}\) A and \(\sqrt{2}\) A

The EXPLANATION: We know that a PMMC instrument READS only DC value and since it is a centre zero type, so it will GIVE – 1.

So, rms = \(\sqrt{1^2 + (\frac{\sqrt{2}}{\sqrt{2}})^2} = \sqrt{2}\) A

Moving iron also reads rms value, so its reading will also be \(\sqrt{2}\) A.

Right option is (c) 0

For explanation I would SAY: H(s) = \(\frac{V(s)}{J(s)}\)

= \(\frac{1}{s+10}\)

V(s) = \(\frac{1}{s+10}\) . J(s)

J(s) = L (SIN 10t) = \(\frac{10}{s^2+100}\)

V (s) = \(\frac{1}{s^2+100} . \frac{10}{s+10}\)

VSS = lims→0⁡ sV(s)

= 0.

Right answer is (b) 50 KHZ

For EXPLANATION I WOULD say: The bandwidth is defined as the highest cut-off frequency to the LOWEST cut-off frequency. Here the lowest cut-off frequency is Zero.

For a Low-pass filter therefore, Cut-off Frequency = Bandwidth of the filter

∴ Bandwidth = 50 kHz.

Correct option is (b) Chebyshev

To explain: Roll off is a term COMMONLY refers to the steepness of the TRANSMISSION function with respect to the frequency. For a Chebyshev FILTER, the Roll-off value greater than 60. This characteristic feature is useful when a RAPID roll-off is REQUIRED because it provides a Roll-off rate more than 60. On the other hand, both Butterworth and Bessel have the Roll-off rate less than or equal to 60 dB/decade/pole.

Right answer is (a) Low-pass

The best explanation: BANDWIDTH can be calculated by CONSIDERING,

Largest positive value – Smallest Positive Value

Here, in case of the Low-pass filter only, the largest positive value will of course be the critical frequency, BEYOND which FREQUENCIES have to be blocked. Hence, the bandwidth in a Low-pass filter equals the critical frequency.

The correct option is (b) 8

Easiest explanation: The Bode plot is a logarithmic plot which helps in FITTING a LARGE scale of values into a small scale by the application of LOGARITHM. Plotting the slope 80dB/decade on the Bode plot, we get n=8. Hence the order of the BAND Pass FILTER is 8.

Correct answer is (d) \(\frac{1}{4}\)

EXPLANATION: H(s) = \(\frac{KS}{s^2+as+b}\)

Then, quality factor is GIVEN as \(\frac{\sqrt{b}}{a}\)

Here, b = 1, a = 2, K = 1

∴ Q = \(\frac{1}{2}\) = 0.5.

The correct option is (b) High-pass

For explanation I would say: A high-pass filter is one which PASSES all FREQUENCIES above fc by ATTENUATING SIGNIFICANTLY, all frequencies below fc.

Right ANSWER is (b) Temperature ALARM circuit

Easiest explanation: The RESISTANCE of a thermistor DECREASES with increases in temperature. It is used to MONITOR the hot spot temperature of electric machines.

Correct option is (d) 14.53 HZ

Easy explanation: IEQ = L1 + L2 + 2M

LEQ = 5 + 5 + 2 × 1

= 10 + 2 = 12 H

∴ FO = \(\FRAC{1}{2π\sqrt{LC}}\)

= \(\frac{1}{2π\sqrt{12 × 0.1}}\)

= 0.1453 Hz.

Correct answer is (b) 6 Ω

Best EXPLANATION: The resistance of parallel combination is given by,

Req = \(\frac{40}{3}\) – 10 = 3.33 Ω

Or, \(\frac{1}{3.33} = \frac{1}{12} + \frac{1}{15} + \frac{1}{15}\)

Or, R = 6 Ω.

Correct ANSWER is (B) VTH = 7.14 V, RTH = 2.14 Ω

The best explanation: VTH = \(\FRAC{5 X 10}{35}\)X 5 = 7.14 V

Also, RTH = \(\frac{5 X 15}{35}\) = 2.14 Ω.

The CORRECT ANSWER is (c) 18.03 A

Explanation: Currents in RESISTANCE and inductance are out of PHASE by 90°.

Hence, I = \(I_1^2 + I_2^2\)

Or, I = [10^2 + 15^2]^0.5

Or, I = \(\sqrt{100+225} = \sqrt{325}\)

= 18.03 A.

The correct OPTION is (b) High-pass

The explanation: A high-pass filter is ONE which passes all FREQUENCIES above fc by ATTENUATING significantly, all frequencies below fc.

The correct CHOICE is (C) BAND-pass Filter

The explanation: The given circuit is a first order Band Pass Filter. Also, the Roll-off of the filter depends upon the order of the filter. For a first order it is 20dB/decade, for SECOND order it is 40dB/decade, and so on.

Right ANSWER is (b) High-pass Filter

Easy explanation: We know that the position of Resistance (R) and Capacitance (C) determines whether it is Low-pass Filter or High-Pass Filter. If R is connected directly to source and the CAPACITOR connected in PARALLEL to it, then it is a Low-pass Filter and if the position of R and C are inter change then high pass filter is formed.

Here since R and C are in SERIES and also R is not connected directly to the power source, hence the filter is a High-pass Filter.

The correct choice is (a) Square of the CURRENT through the coil

The explanation: We know that TD = \(\frac{1}{2} I^2\frac{dl}{dθ}\)

HENCE, the deflection TORQUE is proportional to the square of the current through the coil.

Correct option is (b) Chebyshev

For explanation I WOULD say: ROLL off is a term commonly REFERS to the steepness of the transmission function WRT to the frequency.

For a Chebyshev filter, the Roll-off value greater than 20. This characteristic feature is useful when a rapid roll-off is REQUIRED because it provides a Roll-off rate of more than 20.

On the other hand, both Butterworth and Bessel have the Roll-off rate less than or equal to 20 dB/decade/pole.

Right OPTION is (d) 80 dB/decade

To explain: The given filter is a first order Band PASS Filter. Also, the Roll-off of the filter is depends upon the order of the filter. For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on.

Therefore, the Roll-off of the filter = 20 dB/decade.

Roll of first order LOW pass BUTTERWORTH filter is 20dB/decade.

Now here two stages of second order Low-pass Butterworth filter are cascaded

∴ Roll-off = 20*4 = 80 dB/decade.

The correct OPTION is (b) 1.23 kHz

For EXPLANATION: The bandwidth is DEFINED as the highest cut-off frequency to the lowest cut-off frequency. Here the lowest cut-off frequency is Zero.

For a Low-pass FILTER therefore, Cut-off Frequency = Bandwidth of the filter

∴ Bandwidth = 1.23 kHz.

Correct answer is (a) 20 dB/decade

Best explanation: The given filter is a first ORDER BAND Pass Filter.

Also, the Roll-off of the filter is depends UPON the order of the filter.

For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on.

Therefore, the Roll-off of the filter = 20 dB/decade.

Right ANSWER is (b) High-pass

Easiest explanation: From the GIVEN circuit, we can infer that Roll off of the Filter circuit is 80dB/decade. This Roll-off value is obtained as second ORDER high pass filter followed by another 2nd order HPF RESULTS in an HPF.

Therefore the circuit represents a High-pass Filter.

Right CHOICE is (c) A0 = \(\FRAC{R_2}{2R_1}\)

Explanation: The total output C = CINPUT + COUTPUT that is the GAIN capacitor.

∴ The total RESISTANCE is equal to the Resistance INPUT and Resistance output.

Again, the total resistance gain = \(\frac{R_1 R_2}{R_1 + R_2}\)

Hence, the gain = A0 = \(\frac{R_2}{2R_1}\).

The correct answer is (d) Z1Z2 = Ro^2

For explanation I WOULD say: The lattice PHASE equalizer consists of only reactive components. So a lattice phase equalizer is a CONSTANT equalizer if the following EQUATION is SATISFIED. Z1Z2 = Ro^2.

The CORRECT OPTION is (d) Only pure reactance

The explanation is: The bridged-T phase equalizer CONSISTS of only pure reactances. So the bridged T-circuit consists of only inductive or CAPACITIVE ELEMENTS not resistive elements.

Right answer is (b) False

Easy EXPLANATION: The attenuation is SHARP in the stop band for an m-derived FILTER. So the given statement is not TRUE.

Correct choice is (d) Vmax^2/4Ro

Easy explanation: The expression of input power of a SHUNT equalizer is PI=(Vmax/2Ro)^2 Ro=Vmax^2/4Ro.

Right ANSWER is (c) (VmaxjX1)/(2Ro(Ro+jX1))

The explanation: On SUBSTITUTING Is in the load current equation we get the load current in terms of VMAX in the shunt EQUALIZER as Il = (VmaxjX1)/(2Ro(Ro+jX1)).

The correct answer is (d) Is jX1/(2Ro+jX1)

The explanation: The load current in terms of SOURCE current in the shunt EQUALIZER is IL = Is (jX1/2)/(Ro+jX1/2)). On SOLVING, Il= Is jX1/(2Ro+jX1).

The correct answer is (d) 1+ (ω^2 L1^2)/Ro^2

Easiest EXPLANATION: The expression of N in a full series EQUALIZER CONSIDERING Z1 as inductor and Z2 as capacitor is N = 1 + X1^2/Ro^2 = 1+ (ω^2 L1^2)/Ro^2.

Right choice is (c) 1+ Ro^2/(ωL1)^2

For explanation I would say: The EXPRESSION of N in a FULL series equalizer considering Z1 as capacitor and Z2 as INDUCTOR is N = 1+ Ro^2/X2^2 = 1+Ro^2/(ωL1)^2.

The correct choice is (d) [Vmax^2/(4(Ro^2+X1^2))]Ro

To elaborate: The power at the load of a series equalizer is P=(Vmax/√((2RO)^2+(2X1)^2))^2 Ro = [Vmax^2/(4(Ro^2+X1^2))]Ro.

The CORRECT option is (c) Vmax/√((2Ro)^2+(2X1)^2)

The best explanation: When the equalizer is connected, the EXPRESSION of CURRENT flowing in a series equalizer is I1 = Vmax/√((2Ro)^2+(2X1)^2) where Vmax is voltage applied to the network and Ro is RESISTANCE of the load as well as source and 2X1 is the reactance of the equalizer.

Correct choice is (d) Vmax^2/4Ro

Easiest explanation: The expression of INPUT POWER of a series equalizer is Pi=(Vmax/2Ro)^2 Ro=Vmax^2/4Ro.

Correct answer is (B) antilog(D/10)

The EXPLANATION is: The value of N in TERMS of ATTENUATION D is antilog(D/10). N = antilog(D/10) where D is attenuation in DECIBELS.

Correct OPTION is (b) C2R0^2

The explanation is: An inverse network is OBTAINED by CONVERTING each capacitance C into an INDUCTANCE of VALUE CR0^2 where R0 is resistance. The value of the inductor L2^‘ after converting the capacitor into the inductance is L2^‘ = C2R0^2.

Right option is (a) L2/R0^2

Explanation: An inverse NETWORK may be obtained by CONVERTING Each inductance L should be CONVERTED into a capacitance of value L/R0^2 to OBTAIN the inverse network. The value of the capacitance C2^‘ after converting the inductor into the capacitance is

C2^‘ = L2/R0^2.