Answer.1. Megger

Explanation:-

Testing of Insulation Resistance of Complete Installation to Earth

The principal purpose of testing the insulation is to verify that there are no inadvertent connections between live conductors and between live and Earth before the installation is energized. Tests are required between live conductors (e.g. between phases and between phase(s) and neutral) and between all live conductors and Earth.

Insulation resistance of the installation depends on many factors such as atmospheric conditions, humidity, dirt, etc. As such its calculation is not possible, but it can be readily measured. Normally the insulation resistance is quite high and can be measured by an instrument called “megger” usually used for the measurement of high resistance. The main object in performing this test is to ascertain whether the complete wiring is sound enough to avoid leakage current. As per the Indian electricity rules, the leakage current of installation should not exceed 1/5000 of the maximum supply-demand of the consumer. As such the insulation resistance of the complete installation to earth should not be less than 1 MΩ. Insulation resistance test must be conducted on a D.C voltage of value not less than twice the standard supply voltage. However, it should not exceed  500 V for medium voltage circuits.
 Insulation resistance of the installation to earth is measured by a megger generating dc voltage of about 500 V. Insulation resistance is measured under the following conditions.
(i) Main switch in off position.

(ii) The main fuse to be taken out.

(iii) All other fuses to be provided.

(iv) All lamps must be put in the on position.

(v) All switches must be made on.

(vi) Live and neutral wire to be shorted together.

Testing of Insulation Resistance Between Conductors

After testing the insulation resistance of installation to earth, the insulation resistance between line and neutral should be measured. The major object in conducting this test is again to ensure that the insulation is quite sound between the conductors and there are no chances of an appreciable leakage current between them.

This test should be carried out under the following conditions of installation

(i) The Main switch off.

(ii) The main fuse has withdrawn.

(iii) All other fuses in their position.

(iv) All switches in on position.

(v) All lamps from the holders removed.

(vi)Two terminals of the meager are connected to the line and neutral wire.

Insulation resistance measured under the above condition by a megger should not be less than 50 MΩ divided by the number of outlets(points + switches). However, it need not be more than 1 MΩ.

Answer.4. 3.75

Explanation:-

Extension of Ammeter Range

The current range of a DC moving coil ammeter is extended by connecting a shunt resistance Rs (low resistance) across the coil, the circuit as shown in Figure

I = Total current = 5A
I_m = full-scale deflection current of ammeter = ?
I_sh = shunt current
R_m = resistance of the ammeter = 25Ω
R_sh = shunt resistance = 75Ω

The full-scale deflection current I_m is given as

[latex]\begin{array}{l}{I_m} = \dfrac{{I \times {R_{sh}}}}{{{R_m} + {R_{sh}}}}\\\\{I_m} = \dfrac{{5 \times 75}}{{(25 + 75)}}\\\\{I_m} = 3.75A\end{array}[/latex]

Answer.1. 2286.3

Explanation:-

When the load is balanced the Power consumed by the two wattmeters is

P = W1 + W2 = √3 V_L I_L cosφ

Where

V_Ph = Phase voltage = 220 V

Line voltage = √3 × V_ph = √3 × 200

I_Ph = Phase current = Line current I_L = 10 A

Power factor cosφ = 0.8

∴ Total Power consumed

P = √3 × √3 × 220 × 10 × .8

W₁ + W₂ = 5280 watt

Now

Sin²φ = 1 − cos²φ

Sin²φ = 1 − (0.8)²

Sinφ = 0.6

Now tanφ = Sinφ ⁄ Cosφ = 0.6 ⁄ 0.8 = 0.75

The power factor of the two wattmeters is given as

[latex]\begin{array}{l}\tan \Phi = \sqrt 3 \left( {\dfrac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)\\\\0.75 = \sqrt 3 \left( {\dfrac{{{W_1} – {W_2}}}{{5280}}} \right)\\\\{W_1} – {W_2} = \dfrac{{3960}}{{\sqrt 3 }}\\\\{{\rm{W}}_{\rm{1}}}{\rm{ – }}{{\rm{W}}_{\rm{2}}}{\rm{ = 2286}}{\text{.3 watt}}\end{array}[/latex]

Answer.2. 2000

Voltmeter sensitivity (ohm per volt ratings)

The sensitivity of a voltmeter is given in ohms per volt. It is determined by dividing the sum of the resistance of the meter (R_m) plus the series resistance (R_s), by the full-scale reading in volts. In equation form, sensitivity is expressed as follows:

Sensitivity (S) = (R_m + R_s) ⁄ V_fld

Now as per the given question

R_m = 400 KΩ

R_s = 100 KΩ

V_fld = 250 V

S = (400 + 100) ⁄ 250

S = (500 × 10³) ⁄ 250

S = 2000 ohm/volt

Answer.1. 7.92

Explanation:-

System voltage i.e Nominal voltage = 11000 V

V₁/V₂ = N₁/N₂

98/V₂ = 104

Primary = Turn ratio × Secondary voltage = 98 × 104 = 10192 V

Percentage of error is the Potential error

= (Nominal voltage − Primary voltage)/Primary voltage

(11000 − 10192)/10192

= 7.92%

Answer.1. 51

Explanation:-

Reading of Wattmeter 1 W₁ = 200 W

Reading of Wattmeter 2 W₂ = 1200 W

The power factor of the two wattmeters is given as

[latex]\begin{array}{l}\tan \Phi = \sqrt 3 \left( {\left| {\dfrac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right|} \right)\\\\\tan \Phi = \sqrt 3 \left( {\left| {\dfrac{{200 – 1200}}{{200 + 1200}}} \right|} \right)\\\\\tan \Phi = \sqrt 3 \left| {\dfrac{{ – 1000}}{{1400}}} \right|\\\\\tan \Phi = \dfrac{{\sqrt 3 }}{{1.4}}\\\\\tan \Phi = 1.237\\\\\Phi = {\tan ^{ – 1}}1.237\\\\\Phi = {51^ \circ }\end{array}[/latex]

Answer.2. These instruments are suitable for both AC and DC currents

Explanation:-

The principle on which a Permanent Magnet Moving Coil (PMMC) instrument operates is, “the torque is exerted on a current-carrying coil placed in the field of a permanent magnet”.

PMMC type of instrument can be operated in direct current only. In alternating current, the instrument does not operate because in the positive half the pointer experiences a force in one direction and in the negative half, the pointer experiences the force in the opposite direction. Due to the inertia of the pointer, it remains in its zero position.

Answer.3. Multimeter

Explanation:-

A multimeter is a very useful electronic instrument that can be used for the measurement of three quantities, namely voltage, current, and resistance. This instrument can also be used for both dc and ac voltages and currents. Multimeters are available in both analog and digital form. Although analog multimeters are being replaced by digital multimeters.

Answer.3. 0.5

Explanation:-

The reading of two wattmeters can be expressed as

W₁ =  V_LI_Lcos(30 + φ)
W₂ =  V_LI_Lcos(30 − φ)

(i) When PF is unity ( φ = 0°)

W₁ =  V_LI_Lcos30°
W₂ =  V_LI_Lcos30°

Both wattmeters read equal and positive reading i.e upscale reading

(ii) When PF is 0.5 (φ = 60°)

W₁ =  V_LI_Lcos90° = 0
W₂ =  V_LI_Lcos30°

Hence total power is measured by wattmeter W₂ alone

(iii) When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)

W₁ = Negative
W₂ = positive (since cos(−φ) = cosφ)

The wattmeter W₂ reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°. However, in wattmeter W₁, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives negative (i.e. downscale) reading.

Wattmeter cannot show a negative reading as it has an only positive scale. An indication of negative reading is that the pointer tries to deflect in a negative direction i.e. to the left of zero. In such a case, reading can be converted to positive by interchanging either pressure coil connections or by interchanging current coil connections. Remember that interchanging connections of both the coils will have no effect on wattmeter reading.

 (iv) When P.F reads zero (φ = 90°)

Such a case occurs when the load consist of pure inductance or pure capacitance

W₁ =  V_LI_Lcos(30 + 90°) = – V_LI_Lsin30°
W₂ =  V_LI_Lcos(30 − 90°) = V_LI_Lsin30°

In this condition, the two wattmeter reads equal and opposite i.e W₁ + W₂ = 0

Answer.1. The frictional error is low

Explanation:-

The permanent magnet moving coil (PMMC)-type instrument is the basic dc measuring instrument. In the instruments, a permanent magnet, generally of horseshoe type, creates a magnetic field in which a coil of fine wire of the number of turns is placed. The coil is wound on a very light aluminum drum and is pivots on jewel bearings so that the coil is free to move when current flows through it. The current-carrying coil placed in the magnetic field experiences a torque and tries to turn. its free turning is restricted by spring tension attached to its shaft. The moving coil produces a defecting torque which is opposed by the control torque produced by the spring action. A simplified diagram of a PMMC type instrument is shown in the figure.

Its operating principle is the same as that of the D’Arsonval galvanometer. When a current flow through the coil, a magnetic field is produced which reacts with the magnetic field produced by the permanent magnet.

The current to be measured or a definite fraction of it proportional to the voltage to be measured is passed through the coil. A deflecting torque is produced on account of the reaction of the permanent magnetic field with the magnetic field produced by the coil. The direction of deflecting torque can be found by applying Flemming’s left-hand rule.

Advantages of PMMC Type instruments

• Uniform scale
• Since driving power is small, power consumption is low.
• Due to aluminum or copper former, hysteresis loss is absent.
• Very effective and reliable eddy current damping.
• The torque-weight ratio is very high.
• Due to the application of intense polarized or unidirectional field, the stray magnetic field has no effect.
• Using shunts or multipliers, ranges can be extended.

Disadvantages of PMMC Type Instrument

• PMMC type of instrument can be operated in direct current only. In alternating current, the instrument does not operate because in the positive half the pointer experiences a force in one direction and in the negative half, the pointer experiences the force in the opposite direction. Due to the inertia of the pointer, it remains in its zero position.
• Compared to moving iron instruments, these instruments are costlier,
• Friction error due to Jewel and pivot suspension, temperature, aging of control springs, and permanent magnets might introduce errors.