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The correct option is (a) 110000/110V P.T.

The explanation: An 110000/110V potential transformer can be USED for MEASURING a VOLTAGE of upto 110000V by making use of a voltmeter with 110V voltage reading.

The correct answer is (d) 1000/5A C.T.

Explanation: A 1000/5A current transformer can be USED for measuring a current of UPTO 1000A by making use of an ammeter with 5A current READING.

Correct option is (a) 0.225

To EXPLAIN: Logarithmic decrement, δ = \(ln \frac{x_1}{x_2} = ln \frac{120}{96}\) = 0.225

Now, δ is related to DAMPING ratio K as, K = \(\frac{1}{(1 + (\frac{2π}{δ})^2)^{0.5}}\)

K = \(\frac{1}{(1 + (\frac{2π}{0.225})^2)^{0.5}}\)

∴ K = 0.0357.

Right CHOICE is (d) 9 kΩ/V

For explanation I WOULD say: Sensitivity = \(\frac{1}{100 × 10^{-6}}\) = 10 kΩ/V

For full-wave RECTIFICATION SAC = 0.9 + SDC = 0.9 × 10 = 9kΩ/V.

The correct ANSWER is (C) 0.0357

The best EXPLANATION: Logarithmic decrement, δ = \(LN \frac{x_1}{x_2} = ln \frac{120}{96}\) = 0.225

Now, δ is related to damping ratio K as, K = \(\frac{1}{(1 + (\frac{2π}{δ})^2)^{0.5}}\)

K = \(\frac{1}{(1 + (\frac{2π}{0.225})^2)^{0.5}}\)

∴ K = 0.0357.

The CORRECT ANSWER is (a) 10 kΩ/V and 10 kΩ/V

Explanation: SV = \(\frac{1}{I_{SS}}\) = current REQUIRED for full scale deflection

SV = \( \frac{1}{100 × 10^{-6}}\) = 10 kΩ/V.

The correct OPTION is (a) 0.0216 N-m

Explanation: Given, N = 250, L = 40 × 10^-3, d = 30 × 10^-3m, I = 160 × 10^-3A, B = 450 × 10^-3 T

Torque = 250 × 450 × 10^-3 × 40 × 10^-3 × 30 × 10^-3 × 160 × 10^-3 = 200 × 10^-6N-m = 0.0216 N-m.

The correct answer is (b) 4.8 μH/rad

Best explanation: At FULL scale POSITION, \(\frac{1}{2} I^2 \frac{DL}{dθ}\) = TC

\(\frac{1}{2} 10^2 \frac{dL}{dθ}\) = 240 × 10^-6

∴ \(\frac{dL}{dθ}\) = 4.8 μH/rad.

The correct answer is (B) \(\frac{1.11 V_a}{V_b}\)

For explanation: Form FACTOR of the wave = \(\frac{RMS value}{Mean value}\)

Moving iron instrument will show rms value. Rectifier VOLTMETER is calibrated to READ rms value of the sinusoidal voltage that is, with form factor of 1.11.

∴ Mean value of the applied voltage = \(\frac{V_b}{1.11}\)

∴ Form factor = \(\frac{V_a}{V_b/1.11} = \frac{1.11 V_a}{V_b}\)

Correct answer is (c) 3 A ± 1.7 %

The best I can explain: I = [2 + \(\sqrt{2}\) sin (314t +30°) + 2\(\sqrt{2}\) cos (952t + 45°)]

Thermocouple MEASURE the RMS value of current.

Irms = \([2^2 + (\FRAC{\sqrt{2}}{\sqrt{2}})^2 + (\frac{2\sqrt{2}}{\sqrt{2}})^2]^{1/2} = \sqrt{9}\) = 3 A ± 1.7%.

The correct option is (c) 89.83 kΩ

For explanation I would say: VOAverage = 0.636 × \(\sqrt{2}\) Vrms = 0.8993 Vrms

The deflection with AC is 0.8993 times that with DC for the same value of voltage V

SAC = 0.8993 SDC

SDC of a rectifier type instrument is \(\frac{1}{I_{fs}}\) where Ifs is the CURRENT required to produce full scale deflection, Ifs = 1 mA; Rm = 100 Ω; SDC = 10^3 Ω/V

SAC = 0.8993 × 1000 = 899.3 Ω/V. Resistance of MULTIPLIER RS = SAC V – Rm – 2Rd, where Rd is the resistance of diode, for ideal diode Rd = 0

∴ RS = 899.3 × 100 – 100 = 89.83 kΩ.

The correct CHOICE is (C) – 8, 10 and 10

To explain: PMMC instrument reads only DC VALUE and since it is a CENTRE zero type, so it will give – 8 values.

So, rms = \(\sqrt{8^2 +(\frac{6\sqrt{2}}{\sqrt{2}})^2}\) = 10 A

Moving iron also reads rms value, so its READING will also be 10 A.

Right OPTION is (d) Zero

Easy explanation: The SPRING gives the controlling torque. It is connected in SERIES with the coil. If the spring is cut OPEN, there will be no deflection.

The correct choice is (c) An ammeter and a voltmeter for measuring DC as well as AC

The explanation is: When the instrument is CONNECTED in the circuit, the current flows through the coil. These currents SET up a magnetic field in the coil. The result is that the pointer attached to the moving system moves from zero POSITION. If the current in the coil is reversed, the direction of deflecting torque REMAINS unchanged. Therefore, these instruments can be used for both DC as well as AC measurements.

Right option is (c) 0.707 A

Easy explanation: We know,TD = Cθ

Also, TD = KI^2

∴ I^2 = \(\FRAC{C}{K}\) θ = C’θ

Or, \(\frac{I^2}{I} = \frac{C’ π}{C’ \frac{\PI}{2}} = \frac{1}{2}\)

So, I = 0.707 A.

Right option is (d) 5 mA

To EXPLAIN I would say: The lower current Il will DECIDE the total current.

∴ T × \(\frac{25}{125}\) = 1 mA

Or, T = 5 mA.

Right OPTION is (c) 99.60 kΩ

The BEST I can explain: (RS + 400) × 10^-3 × 4 = 400

Or, RS = 100000 – 400 = 99.6 kΩ.

The correct option is (b) 0.16 Ω

To explain: VOLTAGE across the meter = 4 × 10^-3 × 4 × 10^2 = 1.6 V

Current through the SHUNT = 10 – 0.004 = 9.996 A

∴ Shunt resistance = \(\FRAC{16}{10 × 9.996} \) = 0.16 Ω.

Correct option is (b) All coils in series

For explanation I would say: SINCE the CT SUPPLIES the current to the current coil of a WATTMETER, THEREFORE the coils are CONNECTED in series so that the current remains the same. If they were connected in parallel then the voltages would have been the same but the currents would not be the same and thus efficiency would decrease.

The correct option is (C) Electromagnetically coupled

The best explanation: A POTENTIAL TRANSFORMER cannot be electrostatically coupled since CRO are electrostatically coupled. ALSO, they cannot be conductively coupled. But since they are KIND of electrically coupled hence electromagnetically coupled is the only correct option.

Right answer is (B) The large voltage may act as a SAFETY hazard for the operators and MANY even raptures the insulation

Easy explanation: Never open the circuit of the SECONDARY winding of a current transformer while to the primary winding is energized. Failure to observe this precaution may lead to serious consequences both to the operating PERSONNEL and to the transformer.

Correct OPTION is (B) -28.57

To explain I would SAY: Im = 350/1 = 350 A

Ip = \(((nI_s^2)^2 + (I_m^2)^2)^{0.5}\) = 490.05

n = 350/7 = 50

∴ R = \(\FRAC{I_P}{I_S} = \frac{490.05}{7}\) = 70

∴ Percentage ratio error = \(\frac{50-70}{70}\) × 100 = -28.57%.

Correct choice is (c) -1.0

To EXPLAIN: Turn COMPENSATION only ALTERS ratio error n=400

Ratio error = -0.5% = – \(\frac{0.5}{100}\) × 400 = -2

So, Actual ratio = R = n+1 = 401

Nominal Ratio KN = 400/1 = 400

Now, if the number of TURNS are reduced by one, n = 399, R = 400

Ratio error = \(\frac{K_N-R}{R} = \frac{200-200}{200}\) = 0.

Right choice is (B) 5.7%

Best explanation: E = \(\frac{10}{200}\) V

I = 60 A

I = \(\frac{mmf}{N_P}\) = 250 A

Actual rating R = 200 + \(\frac{60}{5}\) = 212

So, percentage ratio ERROR = \(\frac{K_n-R}{R}\) × 100 = \(\frac{200-212}{212}\) × 100 = 5.7%.

The CORRECT ANSWER is (b) 180 μWb

The EXPLANATION: Turn ratio = 1000/5 = 200

NP = 1

So, NS = 200

Secondary impedance = 1.6Ω

Secondary induced voltage, ES = 5 × 1.6 = 8 V

∴ ES = 4.44 F N ∅

So,∅ = \(\frac{E_S}{4.44 \,f \,N} = \frac{8}{4.44 \,f \,N}\) = 180 μWb.

Right ANSWER is (B) 45 μWb

The explanation: Voltage induced in the secondary,

ES = IS × ZS = 5 V

ES = 4.44 f ∅ N

∴ ∅ = \(\frac{E_S}{4.44 \,f \,N} = \frac{5}{4.44 \,f \,N}\) = 45 μWb.

Correct choice is (a) 3.1°

The EXPLANATION is: Secondary burden is PURELY RESISTIVE and the resistance of burden is equal to the resistance of the secondary winding; the resistance of secondary winding = 1Ω. The voltage induced in secondary × resistance of secondary winding = 7 × 2 = 14V. Secondary power factor is unity as the load is purely resistive. The LOSS component of no-load current is to be neglected i.e. Ie = 0. IM = 300 A.

Secondary winding current IS = 7 A

Reflected secondary winding current = n IS = 5600 A

∴ tan θ = \(\frac{I_M}{nI_S}\). So, θ = 3.1°.

The CORRECT option is (a) Ip shifts towards Io

Best explanation: When the power factor of the secondary burden is REDUCED, Ip shifts towards Io. VOLTAGES Vp and VS COME closer to Ep and Es.

Correct option is (a) True

Easy EXPLANATION: As the secondary burden is increased, WINDING voltage drop increases. Voltage Vp is leading in phase while Vs is LAGGING in phase. As a result the phase ANGLE increases with the secondary burden.

Correct answer is (c) ratio error INCREASES with secondary burden

For explanation I would say: As the secondary burden is INCREASED, the ratio error also increases and BECOMES more NEGATIVE. Ratio error varies linearly with respect to the change in the secondary burden.

Right option is (a) increases Is

The EXPLANATION: When the SECONDARY BURDEN of a potential TRANSFORMER increases, it leads to an increase in the secondary CURRENT. As a result the primary current also increases.

The correct answer is (a) by reducing secondary turns

Best explanation: In a P.T., at no LOAD, we get

where, R is the RATIO error

n is the turns ratio

Is is the secondary winding current

Ie is the iron loss component

Im is the magnetising component

From the above equation it is SEEN that to REDUCE the ratio error, actual ratio and nominal ratio MUST be made equal. This can be done by reducing the secondary turns.

The correct answer is (d) using a GOOD core MATERIAL

The explanation: By making use of a good quality core material, low value of flux density and following REQUIRED PRECAUTIONS in the core assembly we can MINIMISE the value of the ratio error.

Right OPTION is (B) False

Best explanation: In a C.T. the various components of current such as magnetising current, iron loss COMPONENT of current are ALMOST comparable in magnitude with the value of the load current.

The CORRECT answer is (c) secondary current

The best I can explain: In a P.T., the difference between actual RATIO and TURNS ratio is GIVEN by the relation,

where, R is the ratio error

n is the turns ratio

Is is the secondary winding current

Ie is the IRON loss component

Im is the magnetising component

It is seen from the above equation that the ratio error in a P.T. depends on the secondary current, magnetising and iron loss components of current.

The correct answer is (a) True

To ELABORATE: In a P.T., a HIGH FLUX density in the core, gives rise to a less number of turns. This in turn results in a LOWER leakage reactance.

Correct choice is (b) False

To EXPLAIN: In a P.T., FULL voltage APPEARS across the primary winding when it is connected across the LINE. While when a C.T. is connected in series with a line, a very small voltage appears across the primary winding.