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Correct OPTION is (C) 11

Easy explanation: Multiplying FACTOR = m = \(\FRAC{I}{I_1} \)

Now, \(\frac{I_1}{I_2} = \frac{100}{1000} \)

∴ \(\frac{I_1}{100} = \frac{I_2}{1000} = \frac{I}{1000} \)

∴ \(\frac{I}{I_1}\) = 11.

Correct OPTION is (d) 3.4 %

For explanation I would SAY: Ammeter error ∆I = ± 1%

VOLTMETER error ∆V = ± 2.4%

We know that \(\frac{∆R}{R} = \frac{∆V}{V} + \frac{∆I}{I}\)

∴ MAXIMUM percentage error = 2.4% + 1% = 3.4%.

The correct ANSWER is (C) 25.0 Ω

For explanation: Ish Rsh = Im Rm

Ish = I – Im or, \(\frac{I}{I_m} – 1 = \frac{R_m}{R_{sh}}\)

Now, m = \(\frac{I}{I_m}\)

Or, m – 1 = \(\frac{R_m}{R_{sh}}\)

∴ Rsh = 25 Ω.

The correct option is (C) 0.025 Ω

Easy explanation: Rsh = \( \FRAC{R_m}{m-1}\)

Where, m is the MULTIPLICATION FACTOR = 500/100 = 5

∴ Rsh = 0.1/4 = 0.025 Ω.

Correct option is (c) 1000 rev/kW-h

The BEST explanation: Meter constant is 14.1 A-s/rev

\(\frac{14.1}{3600}\) A-h/rev = \(\frac{14.4 × 250}{3600}\)

So, W = 1 W-h/rev

Hence, 1 rev/W-h = 1000 rev/kW-h.

Correct option is (d) 2.4 kΩ

To EXPLAIN: Voltage across R1, V1 = 35 – 5 = 30 V

Current in the CIRCUIT, I = \(\displaystyle\frac{5}{\frac{600 ×1200}{600+1200}} = \frac{5}{400}\) A

∴ R = \(\frac{30 × 400}{5}\) = 2.4 kΩ.

Correct choice is (d) 4.71 %

Easy explanation: Current through the voltmeter Iv = \(\frac{180}{2000}\)

Current through R, IR = 2 – 9/100 = 1.91 A

Since, 1.91 R = 180

∴ R = 94.24

∴ PERCENTAGE ERROR = \(\frac{94.24-90}{90}\) × 100 = 4.71 %.

The CORRECT answer is (b) -10.56%

Best EXPLANATION: IM = 250/I = 250 A

Ip

Or, R = \(\frac{V^2}{R} \)

∴ R = ± (2 × 1.5 + 5) = ± 8%.

Right ANSWER is (B) Ra = Rb Rc/Rd

The explanation is: At balance condition, POTENTIAL at B = Potential at D

∴ VA – IA Ra = VA – Ib Rb

Or, \( \FRAC{I_a}{I_b}=\frac{R_b}{R_a}\)

Similarly,\( \frac{I_a}{I_b}=\frac{R_d}{R_c}\)

∴ Ra = \( \frac{R_b R_c}{R_d}\).

Right choice is (B) 6.67%

The explanation is: RT = \( \FRAC{V_T}{I_T}\)

VT = 200 V, IT = 10 A

So, RT = 20 kΩ

Resistance of voltmeter,

RV = 1000 × 300 = 300 kΩ

Voltmeter is in parallel with UNKNOWN resistor,

RX = \( \frac{R_T R_V}{R_T – R_V} = \frac{20 × 300}{280}\) = 21.43 kΩ

Percentage error = \(\frac{Actual-Apparent}{Actual}\) × 100

= \(\frac{21.43-20}{21.43}\) × 100 = 6.67%.

Correct choice is (a) 20 kΩ

Easiest explanation: RT = \( \frac{V_T}{I_T}\)

VT = 200 V,IT = 10 A

So, 20 kΩ.

The correct option is (a) 6.67%

Easy explanation: R1 = 10 ± 5%

R2 = 5 ± 10%

R1 = 10 ± \( \FRAC{5}{100}\) × 10 = 10 ± 0.5Ω

R2 = 5 ± \( \frac{5}{100}\) × 5 = 5 ± 0.5Ω

The limiting VALUE of RESULTANT resistance = 15 ± 1

Percentage limiting error of series combination of resistance = \( \frac{1}{15}\) × 100 = 6.67%.

Correct CHOICE is (C) ±15.8 Ω

For EXPLANATION: Given,R1 = 100 ± 5 Ω

R2 = 150 ± 15 Ω

Now, R = R1 + R2

The probable errors in this CASE, R = \( ± (R_1^2+R_2^2 )^{0.5}\) = ± 15.8 Ω.

The correct ANSWER is (a) 4%

For explanation I would say: The magnitude of limiting ERROR of the instrument

ρA = 0.01 × 400 = 4 V

The magnitude of voltage being measured = 250 V

The RELATIVE at this voltage Er = \( \frac{4}{250}\) = 0.016

∴ Voltage measured is between the limits

Aa = As(1± Er)

= 250(1 ± 0.016)

= 250 ± 4 V.

The correct option is (b) 101.5%

To EXPLAIN: TRUE CURRENT = 25(1 + 0.01) = 25.25 A

True resistance R = 1(1 – 0.005) = 0.995Ω

∴ True power = I^2R = 634.37 W

Measured power = (25)^2 × 1 = 625 W

∴ \( \frac{True \,power}{Measured \,power}\) × 100 = \(\frac{634.37}{625}\) × 100 = 101.5%.

Right option is (c) 8.33%

Easy explanation: Here, R1 and R2 are in parallel.

Then, \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)

Or, R = \(\frac{50}{15}\) kΩ

∴\( \frac{△R}{R} = \frac{△R_1}{R_1^2} + \frac{△R_2}{R_2^2}\)

And △R1 = 0.5×10^3,△R2 = 0.5×10^3

∴\( \frac{△R}{R} = \frac{10 × 10^3}{3 × 10 × 10^3} × \frac{0.5 × 10^3}{10 × 10^3} + \frac{10}{3} × \frac{10^3}{5 × 10^3} × \frac{0.5 × 10^3}{5 × 10^3}\)

= \( \frac{0.5}{30} + \frac{1}{15} = \frac{2.5}{30}\) = 8.33%.

Correct ANSWER is (b) ZERO

Best explanation: In Wheatstone bridge, balance condition is

R1R3 = R2R4

Here, R1 = 5, R2 = 10, R3 = 16, R4 = 8

And when the Wheatstone bridge is balanced then, at Vout voltage will be Zero.

The CORRECT option is (B) Carey-Foster bridge METHOD

Easy EXPLANATION: Kelvin’s double bridge method is used for measurement of Low Resistance, Anderson Bridge is not used for measurement of Resistance and Direct-Deflection method is used for Measurement of High Resistance.

The correct option is (b) 10 mA

Explanation: Total LENGTH of the slide wire = 10 × 200

Total resistance of slide wire = 2000 Ω

∴ Resistance per CM = 1 Ω

Resistance of 104.5 cm = 104.5 Ω

This corresponds to a voltage of 1.045 V

∴ Current = \(\frac{1.045}{104.5}\) = 10 mA.

The correct ANSWER is (b) Bridge method

Explanation: Bridge method applies the concept of null point or bridge balance CONDITION. MULTIMETER and Megger are used for measuring very high RESISTANCES and Voltmeter-Ammeter method is used for Low resistances. A null type instrument has higher accuracy as compared to a deflection type instrument.

Right CHOICE is (c) Kelvin double bridge

Easiest explanation: De-Sauty’s bridge is USED for measurement of Capacitance; MAXWELL’s bridge is used for measurement of INDUCTANCE and Wein Bridge for FREQUENCY. Kelvin double bridge is used for measurement of Low resistance.

Right option is (c) 14.95 A

To elaborate: Voltage drop per unit length = \(\frac{1.53}{42}\) = 0.036 V/CM

Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V

∴ Current through resistor, I = \(\frac{2.99}{0.2}\) = 14.95 A.

The CORRECT option is (c) 2.54 Ω

Explanation: Here, V = 31.5∠38.4°

I = 12.4∠27.5°

Z = \(\frac{31.5∠38.4°}{12.4∠27.5°}\) = 2.54∠10.9°

But Z = R + JX = 2.49 + j0.48

∴ REACTANCE X= 2.54 Ω.

The CORRECT choice is (c) \(– \frac{R_x}{R_x + R_v}\) × 100

Easiest EXPLANATION: PERCENTAGE ERROR = \(\frac{R_m– R_x}{R_x}\) × 100

\(= \frac{R_x R_v – R_x(R_x-R_v)}{R_x (R_x+R_v)}\) × 100

∴ Percentage Error = \(– \frac{R_x}{R_x + R_v}\) × 100.

Right choice is (d) very low resistance

Best explanation: Once an electrical apparatus is GROUNDED, most of its COMPONENTS are at ground potential. When the DIFFERENT parts of electrical MACHINERY are connected to the ground, they POSSESS very low resistance.

The correct answer is (c) zero potential

Best explanation: After earthing, the VARIOUS PARTS of electrical machinery such as casing, ARMORING of CABLES, ETC are at zero potential.

The correct ANSWER is (a) True

For explanation I would say: The complications involved in LAYING SEPARATE telephone cables and the cast used for TRACTION work is minimised. As a result earthing is used as a return conductor in telephone lines and for traction work.

The correct choice is (a) earthed

To EXPLAIN: Earthing can be used to maintain a CONSTANT line voltage in a THREE phase SYSTEM. This is achieved by earthing the NEUTRAL.

The correct choice is (d) by MEASURING resistance

For EXPLANATION I would SAY: The resistance of the earth electrode is measured in order to check whether it is in a good condition or not.

The CORRECT option is (c) LOW resistance

For EXPLANATION: In the case of occurrence of any leakage currents due to poor shielding of the apparatus, the earth electrode is USED to provide a very low resistance path from the electrical appliances to the earth.

Correct answer is (b) material used for earthing

The best I can explain: ELECTRODE connected to the MAIN is basically a source of e.m.f. CONDUCTING material that is used for connecting electrical machinery to the earth is known as an earth electrode.

Correct option is (a) connecting electrical machines to earth

To explain: Connecting electrical machines to the general MASS of the earth by MAKING use of a CONDUCTING material with very LOW resistance is known as earthing.

Right ANSWER is (c) used for MEASURING high RESISTANCE with respect to ground

The best explanation: TEMPERATURE protection can be provided to a bridge circuit by making use of heat sinks. Megger is used for determining very high resistances between the conducting part of a circuit and ground.

Right answer is (d) reversely calibrated based on the POINTER position

The EXPLANATION is: The scale in a megger is calibrated such that normal position of a pointer indicates infinity while full scale corresponds to ZERO RESISTANCE.

The CORRECT answer is (a) megohmmeter

To explain: A GALVANOMETER is used to DETECT the balance condition. A multimeter can be used for measurement of voltage, CURRENT, resistance and power. A megger is also known as a megohmmeter as it is used for the measurement of RESISTANCES in the order of Mῼ.

Correct option is (d) pointer shows deflection

Explanation: VOLTAGE drop across the circuit is measured by using a VOLTMETER. When the pointer DISPLAYS full deflection, then there is an electrical CONTINUITY between the two POINTS.

The CORRECT option is (c) megger

Easiest explanation: A galvanometer is used for detecting the BALANCE condition in a bridge circuit. An ammeter measures the current flow through the circuit while the VOLTMETER is used for determining the VOLTAGE across the bridge circuit. A megger circuit can be used to determine the continuity between any TWO points.

The correct OPTION is (B) devices such as MOTOR, cable, etc

Explanation: SOLID state devices consist of power electronic components and drives. One of the applications of a Megger circuit is in verifying the ELECTRICAL insulation levels of devices such as motor, cable, generator, etc.

The correct OPTION is (d) scaling

The EXPLANATION is: TEMPERATURE is controlled by minimising the flow of current through the circuit. Better scaling PROPORTIONS can be achieved in a MEGGER by making use of a compensating coil.

Right answer is (d) electromagnetic induction

Best EXPLANATION: OHM’s law is applicable to only purely resistive circuits which are BASED on the LINEARITY PRINCIPLE. Megger basically works on the principle of electromagnetic induction.

Correct option is (c) 3 COILS

For explanation I would say: The moving element of an OHMMETER in a megger consists of three coils, NAMELY current or deflection COIL, pressure or control coil and compensating coil.