250+ TOP MCQs on Advanced Problems on Indicating Instruments – 1 and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Advanced Problems on Indicating Instruments – 1”.

1. In a moving iron meter, the deflection torque is proportional to?
a) Square of the current through the coil
b) Current through the coil
c) Sine of measurand
d) The Square root of the measurand
Answer: a
Clarification: We know that,
T_d = (frac{1}{2} I^2 frac{dl}{dθ} )
∴ The deflection torque is proportional to the square of the current through the coil.

2. The full-scale deflection current of an ammeter is 4 mA and its internal resistance is 400Ω. If this meter is to have a full deflection of 10 A, what is the value of the shunt resistance to be used?
a) 49.99 Ω
b) 0.16 Ω
c) 1.5 Ω
d) 2.6 Ω
Answer: b
Clarification: Voltage across the meter = 4 × 10^-3 × 4 × 10² = 1.6 V
Current through the shunt = 10 – 0.004 = 9.996 A
∴ Shunt resistance = (frac{16}{10 × 9.996} ) = 0.16 Ω.

3. The full-scale deflection current of a meter is 4 mA and its internal resistance is 400Ω. This meter is to have full deflection when 400 V is measured. What is the value of the series resistor to be used?
a) 99.90 kΩ
b) 100 kΩ
c) 99.60 kΩ
d) 100 Ω
Answer: c
Clarification: (R_S + 400) × 10^-3 × 4 = 400
Or, R_S = 100000 – 400 = 99.6 kΩ.

4. Two ammeters, one with a full-scale current of 1 mA and internal resistance of 100 Ω and other a full-scale current of 10 mA and internal resistance of 25 Ω are in parallel. What is the total current, these two meters can carry without the reading out of scale in any meter?
a) 1 mA
b) 10 mA
c) 11 mA
d) 5 mA
Answer: d
Clarification: The lower current I_l will decide the total current.
∴ T × (frac{25}{125}) = 1 mA
Or, T = 5 mA.

5. A meter has a full scale of 90° at a current of 1 A. This meter has a perfect square law response. What is the current when the deflection angle is 45°?
a) 0.5 A
b) 0.25 A
c) 0.707 A
d) 0.67 A
Answer: c
Clarification: We know, T_D = Cθ
Also, T_D = KI²
∴ I² = (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.

6. The scale of a dynamometer type instrument marked in terms of RMS value would be__________
a) Uniform throughout
b) Non-uniform and crowded near the full scale
c) Non-uniform and crowded at the beginning
d) Non-uniform and crowded around mid-scale
Answer: c
Clarification: We have deflection θ ∝ I² for an ammeter and θ ∝ V² for a voltmeter. We have assumed the value for the voltmeter to be constant but it is not true. The value is constant for a radical field but not for a voltmeter.

7. Moving Iron Instrument can be used as ____________
a) An ammeter for measuring DC as well as AC
b) For measuring DC current and voltages only
c) An ammeter and a voltmeter for measuring DC as well as AC
d) For measuring AC current and voltages only
Answer: c
Clarification: 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.

8. A 10 mA PMMC ammeter reads 4 mA in a circuit. Its bottom control spring snaps suddenly. The meter will now show __________
a) 10 mA
b) 8 mA
c) 2 mA
d) Zero
Answer: d
Clarification: 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.

9. The standardization of AC potentiometer is done by ____________
a) Using a DC standard source and d’ Arsonval galvanometer
b) Using AC standard sources and transfer instruments
c) Using a standard AC voltage source
d) Using a DC standard source and transfer instruments
Answer: d
Clarification: Standardization of AC potentiometer is done with the help of standard DC source i.e., a standard cell or a Zener source and a transfer instrument where transfer instrument may be an electrodynamometer or a thermocouple instrument.

10. The inductance of a certain moving- iron ammeter is expressed as L = 10 + 3θ – (frac{θ^2}{4}) μH, where θ is the deflection in radian from the zero position. The control spring torque is 25 × 10^-6 Nm/rad. The meter carries a current of 5 A. What is the deflection?
a) 2.4
b) 2.0
c) 1.2
d) 1.0
Answer: c
Clarification: At equilibrium,
Kθ = (frac{1}{2} I^2 frac{dl}{dθ} )
(25 × 10^-6) θ = (frac{1}{2} I^2 (3 – frac{θ}{2}) ) × 10^-6
∴ 2 θ + (frac{θ}{2}) = 3
Or, θ = 1.2.

.