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

Answer.3. 0.61

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
I_m = full-scale deflection current of ammeter = 0.3
I_sh = shunt current
R_m = resistance of the ammeter = 30Ω
R_sh = shunt resistance = ?

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}}}}\\\\{0.3} = \dfrac{{15 \times R_{sh}}}{{(30 + R_{sh})}}\\\\{R_{sh}} = 0.61ohms\end{array}[/latex]

Answer.1. 5

Explanation:-

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

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

Bandwidth = 0.35 ⁄ t_r

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

Bandwidth = 0.35 ⁄  70 x 10⁻³

Bandwidth = 5 Hz

Answer.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

Answer.1. 5

Explanation:-

The deflection in the moving iron instrument is given as

[latex display=”true”]\theta = \dfrac{1}{2}\dfrac{{{I^2}}}{K}\dfrac{{dL}}{{d\theta }}[/latex]

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

[latex]\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}[/latex]

Answer.3. 30

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

Answer.3. X-Y Plotter

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

Answer.4. Greater than 0.5 but less than 1

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) 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 =  V_LI_Lcos(30 + φ) (positive, larger)
W₂ =  V_LI_Lcos(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 =  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.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.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.

Answer.3. Maxwell’s inductance

Explanation:-

Maxwell Inductance Bridge

This bridge is used to measure the value of an unknown inductance by comparing it with a variable standard self-inductance. The bridge configuration and phasor diagram under-balanced conditions are shown in Figure

The unknown inductor L₁ of resistance R₁ in the branch AB is compared with the standard known inductor L₂ of resistance R₂ on arm AC. The inductor L₂ is of the same order as the unknown inductor L₁. The resistances R₁, R₂, etc., include, of course, the resistances of contacts and leads in various arms. Branch BD and CD contain known no-inductive resistors R₃ and R₄ respectively.

The bridge is balanced by varying L₂ and one of the resistors R₃ or R₄. Alternatively, R₃ and R₄ can be kept constant, and the resistance of one of the other two arms can be varied by connecting an additional resistor.

Under the balanced condition, no current flows through the detector. Under such conditions, currents in the arms AB and BD are equal (I₁). Similarly, currents in the arms AC and CD are equal (I₂). Under the balanced condition, since nodes B and D are at the same potential, voltage drops across arm BD and CD are equal (V₃ = V₄); similarly, the voltage drop across arms AB and AC are equal (V₁ = V₂).

At the balanced condition 

R₁/R₂ = R₃/R₄ = L₁/L₂