# SSC JE Measurement & Instrumentation Solved Questions (2018-2009)

Ques.21. Determine the value of a shunt resistance (in Ohms) required to convert a galvanometer into an ammeter of reading up to15 A. The internal resistance of the galvanometer is 30 Ohms and the value of the current for a full-scale deflection is 0.3A. (SSC-2018 Set-3,5)

1. 0.84
2. 0.74
3. 0.61
4. 0.52

Explanation:-

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

I = Total current = 15A
Im = full-scale deflection current of ammeter = 0.3
Ish = shunt current
Rm = resistance of the ammeter = 30Ω
Rsh = shunt resistance = ?

The full-scale deflection current Im is given as

$\begin{array}{l}{I_m} = \dfrac{{I \times {R_{sh}}}}{{{R_m} + {R_{sh}}}}\\\\{0.3} = \dfrac{{15 \times R_{sh}}}{{(30 + R_{sh})}}\\\\{R_{sh}} = 0.61ohms\end{array}$

Ques.22. Determine the bandwidth (in Hz) of a sine wave that is reproduced by the CRO having a rise time of 70 ms.(SSC-2018 Set-3)

1. 5
2. 10
3. 15
4. 20

Explanation:-

The time required by the waveform to rise from 10 % to 90 % of its peak value is called rise time, denoted as tr

The bandwidth is inversely proportional to rise time and the relation is given by,

Bandwidth = 0.35 ⁄ tr

Where tr = rise time = 70 ms = 70 x 10−3

Bandwidth = 0.35 ⁄  70 x 10−3

Bandwidth = 5 Hz

Ques.23. Determine the quality factor for Maxwell’s inductance-capacitance bridge is supplied by a frequency of 50 Hz.(SSC-2018 Set-3)

1. 0.16
2. 0.27
3. 0.36
4. 0.47

Explanation:-

Note:- You can solve this question in two ways and both the ways are explained in details

In this bridge, the unknown inductance is measured by comparison with a standard variable capacitance.

The unknown inductor L1 of effective resistance R1 in the branch AB is compared with the standard known variable capacitor C4 on arm CD. The other resistances R2, R3, and R4 are known as non-inductive resistors.

Under the balance condition, the unknown quantities i.e resistor R1 and C1 can be determined by the

R1 = R2 × R3 ⁄ R4

&

L1 = C4 × R2 × R3

Here

R2 = 15 Ω

R3 = 35 Ω

R4 = 60 Ω

C4 = 25 μF

Hence the value of R1 will be

R1 = 15 × 35 ⁄ 60

R1 = 8.75 Ω

&

L1 = 25 × 10−6 × 15 × 35

L1 = 13.125 × 10−3

In the Maxwell’s Inductance-Capacitance Bridge, Q-factor or the quality factor of the inductor under measurement can be found at balance condition to be

Q = ωL1/R1

or

Q = ωC4 R4

where ω = 2πf = 2 × 3.14 × 50 = 314

Q = 314 × 13.125 × 10−3 ⁄ 8.75

Q = 0.471

or

Q = 314 × 25 × 10−6 × 60

Q = 0.471

Ques.24. What will be the deflection (in rad) of a moving iron instrument, when the inductance of the moving iron instrument is (20 + 4θ) μH, where θ is the deflection in radian from zero position and the deflection current is 5A? Assuming spring constant k = 10 × 10−6 Nm/rad.(SSC-2018 Set-3)

1. 5
2. 10
3. 15
4. 20

Explanation:-

The deflection in the moving iron instrument is given as

$\theta = \dfrac{1}{2}\dfrac{{{I^2}}}{K}\dfrac{{dL}}{{d\theta }}$

Where I = current = 5 A

K = Spring constant = 10 × 10−6 Nm/rad

dl/dθ = change in Inductance

Change in the inductance

dl/dθ = d(20 + 4θ) ⁄ dθ μH

dl/dθ = 4 × 10−6 H

$\begin{array}{l}\theta = \dfrac{1}{2} \times \dfrac{{{5^2}}}{{10 \times {{10}^{ - 6}}}} \times 4 \times {10^{ - 6}}\\\\\theta {\rm{ }} = {\text{ 5 Radian}}\end{array}$

Ques.25. Determine the value of the unknown capacitance C( in μF) for the circuit given below, when no current flows through the detector (D). (SSC-2018 Set-3)

1. 20
2. 40
3. 30
4. 80

The bridge given in the above question is De Sauty’s Bridge. This is the simplest method of finding out the value of an unknown capacitor in terms of a known standard capacitor. Configuration and phasor diagram of a De Sauty’s bridge under-balanced condition is shown in Figure

The unknown capacitor C1 in the branch AB is compared with the standard known standard capacitor C2 on arm AC. The bridge can be balanced by varying either of the non-inductive resistors R3 or R4.

At Under-balanced condition, since no current flows through the detector, nodes B and C are at the same potentials i.e. V1 = V2 and V3 = V4.

C1 = C2 × R4 ⁄ R3

C1 = ?

C2 = 20μF = 20 × 10−6 F

R3 = 40 ohms

R4 = 60 ohms

Now putting the value of capacitor and Resistor we get

C1 = 20 × 10−6 × 60 ⁄ 40

C1 = 30 μF

Ques.26. Which of the following is a type of recording instrument? (SSC-2018 Set-4)

1. Ammeter
2. Megger
3. X-Y Plotter
4. Voltmeter

Explanation:-

Recording instruments: These instruments give a continuous record of the given electrical quantity which is being measured over a specific period.

In such a recording Instruments, the readings are recorded by drawing the graph. The pointer of such instruments is provided with a marker i.e. pen or pencil, which moves on graph paper as per the reading. The X-Y plotter is the best example of such an instrument.

X-Y recorders are used for many measurements and tests as follows:-

• Speed/torque measurements of electric motors.
• Lift/drag wind-tunnel tests.
• Plotting of characteristic curves of vacuum tubes, transistors, rectifiers, zoner diodes, etc.
• Measurement of physical quantities, such as force, pressure, temperature, etc.
• Regulation curves of the power supply.
• Electrical characteristics of materials such as resistance vs. temperature
• Characteristics of governors, such as speed vs. load.

Note:- Ammeter, Voltmeter, Meggar all are Indicating Instrument

Ques.27. In the ‘two-wattmeter method’ of power calculation of a 3-phase balanced star connected system, what is the power factor of the system if both the wattmeters show a positive reading? (SSC-2018 Set-4)

1.  0
2. Greater than 0 but less than 0.5
3. 0.5
4. Greater than 0.5 but less than 1

Answer.4. Greater than 0.5 but less than 1

Explanation:-

The reading of two wattmeters can be expressed as

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

(i) When PF is unity ( φ = 0°)

W1 =  VLILcos30°
W2 =  VLILcos30°

(ii) When PF is 0.5 (φ = 60°)

W1 =  VLILcos90° = 0
W2 =  VLILcos30°

Hence total power is measured by wattmeter W2 alone

(iii) Power Factor Is More Than 0.5 But Less Than One (i.e.,1 > cosφ > 0.5) or 60° > φ > 0°

Under such conditions, the readings of the wattmeter will be

W1 =  VLILcos(30 + φ) (positive, larger)
W2 =  VLILcos(30 − φ) (negative, smaller)

Hence when the power factor is more than 0.5 but less than one, both me wattmeters give positive (upscale) readings. However, wattmeter W1 gives larger reading than wattmeter W2.

Now, total power, P = W1 + W2 = larger reading + smaller readings

(iv) 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 only a 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 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.

(v) When P.F reads zero (φ = 90°)

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

W1 =  VLILcos(30 + 90°) = – VLILsin30°
W2 =  VLILcos(30 − 90°) = VLILsin30°

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

Ques.28.  A resistance of 75 Ohms is connected in the shunt of a galvanometer, having an internal resistance of 25 Ohms, to convert it into an ammeter. What is the value of current (in A) flowing through the galvanometer, if the total current in the circuit is 5 A? (SSC-2018 Set-4)

1. 2
2. 2.5
3. 3.65
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
Im = full-scale deflection current of ammeter = ?
Ish = shunt current
Rm = resistance of the ammeter = 25Ω
Rsh = shunt resistance = 75Ω

The full-scale deflection current Im is given as

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

Ques.29. Which of the following is TRUE about analog multimeter? (SSC-2018 Set-4)

1. It is a type of absolute instrument
2. It is a type of indicating instrument
3. It is a type of recording instrument.
4. It is a type of integrating instrument.

Answer.2. It is a type of indicating instrument

Explanation:-

A multimeter is an indicating 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.

Ques.30. The circuit given below is of a circuit _________ bridge. (SSC-2018 Set-4)