DC Ammeter MCQ || Basic DC Ammeter Questions and Answers - Page 2 of 2

11. A 0.5 Ω resistance is required to be connected in parallel to a moving coil instrument whose full-scale deflection is 1 mA; so that this instrument can measure 10 mA current. The internal resistance of this instrument is

5.0 Ω
4.5 Ω
2.25 Ω
0.45 Ω

Answer.2. 4.5 Ω

Explanation:

Given that, I = 10 mA, I_m = 1 mA

⇒ m = 10

Shunt resistance (Rsh) = 0.5 Ω

R_sh = R_m/(m − 1)

0.5 = R_m/9

R_m = 4.5 Ω

12. Two milli-ammeters with full-scale currents of 1 mA and 10 mA are connected in parallel and they read 0.5 mA and 2.5 mA respectively. Their internal resistances are in the ratio of

1∶10
10:1
1:5
5:1

Answer.2. 10:1

Explanation:

Since ammeters are connected in parallel and hence their voltage must be the same:

i.e. V₁ = V₂

I_FS1 × R_m1 = I_FS2 × R_m2

R_m1/R_m2 = I_FS2/I_FS1 = 10/1

13. A moving-coil instrument gives a full-scale deflection of 10 mA when a potential difference of 10 mV is applied across its terminals. To measure currents up to 100 A, the same instrument can be used

With shunt resistance of 0.0001 Ω
With series resistance of 0.01 Ω
With shunt resistance of 0.01 Ω
With series resistance of 0.0001 Ω

Answer.1. With shunt resistance of 0.0001 Ω

Explanation:

We can extend the range of the ammeter by placing a shunt resistance.

Given that,

Full scale deflection current (Im) = 10 mA

Full-scale deflection voltage (Vm) = 10 mV

Meter resistance (Rm) = V_m/I_m= 1 Ω

Required full scale reading (I) = 100 A

${R_{sh}} = \frac{{{R_m}}}{{\left[ {\frac{I}{{{I_m}}} – 1} \right]}}$

${R_{sh}} = \frac{1}{{\left( {\frac{{100}}{{0.01}} – 1} \right)}} = 0.0001\:{\rm{\Omega }}$

14. Internal resistance of an ideal ammeter is

Zero
Infinite
Small
Big

Answer.1. Zero

Explanation:

An ammeter is an instrument that is used to measure the current flowing through the circuit

It has low resistance, ideally zero but practically it has a very low internal resistance

The connecting ammeter in series allows all of the circuit current to pass through it.

15. A 4.0-ampere ammeter has a resistance of 0.01 Ω. Determine the efficiency of the instrument.

25 A/W
25.01 A/W
16 A/W
1.6 A/W

Answer.1. 25 A/W

Explanation:

The given instrument is ammeter,

I = 4 A

R = 0.01 Ω

Hence, P = I²R = 4² × 0.01 = 0.16 W

Hence, efficiency = I/P = 4/0.16 = 25 A/W

16. Series resistance required to read 0-250 V with a moving coil instrument of internal resistance 2 ohm and full-scale deflection of 50 mA is

49998 Ω
4998 Ω
498 Ω
49.8 Ω

Answer.2. 4998 Ω

Explanation:

Given- A Voltmeter with Rm = 2Ω

Vnew = 250 V

I_fs= 50mA

Vm = Ifs × Rm = 50 × 2

V_m= 0.1V

m = V_new/V_m

m = 250/0.1 = 2500

R_se= 2(2500−1) = 4998Ω

17. A shunt resistance of 50 Ω is required for extending the range of an ammeter from 100 μA to 500 μA. The value of internal resistance of the ammeter is:

400 Ω
200 Ω
800 Ω
100 Ω

Answer.2. 200 Ω

Explanation:

Given

Rsh = 50 Ω

M = 500/100 = 5

R_m = R_sh(m – 1)

R_m = 50(5 -1)

R_m = 200 Ω

18. Internal resistance of a micro-ammeter is 500 ohm. Shunt resistance required to increase its range from 0-100 μA to 0-10 A will be approximately

0.05 Ω
0.005 Ω
0.5 Ω
5.0 Ω

Answer.2. 0.005 Ω

Explanation:

Given- An ammeter with R_m= 500 Ω

I_m= 100 μA,

I_new= 10 A

m = I_new/I_m= 10/100 µA

m = 10⁵

R_sh = 500/(10⁵ − 1)

R_sh = 0.005 Ω

19. A (0-50)A moving coil ammeter has a voltage drop of 0.1V across its terminals at full-scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0 – 500 A) is

0.25 mΩ
0.16 mΩ
0.22 mΩ
0.30 mΩ

Answer.3. 0.22 mΩ

Explanation:

Given,

The voltage drop across the ammeter is 0.1 V for full-scale deflection

Current flowing through the meter at full-scale deflection is 50 A

Now the range is extended to 500 A,

Current flowing through shunt I_sh = 500 – 50 = 450 A

As the shunt and meter are in parallel, therefore, the voltage drop across them will be equal

The voltage drop across shunt = Shunt resistance × shunt current

⇒ 0.1V = R_sh × 450 A

⇒ R_sh = 0.22 mΩ

20. A 1 mA galvanometer with an internal resistance of 50 Ω is to be converted to measure 5 A (full-scale). What is the value of the shunt resistance required for this conversion?

1 Ω
0.01 Ω
1 kΩ
10 Ω

Answer.2. 0.01 Ω

Explanation:

Given that,

Full scale deflection current (Im) = 1 mA

Meter resistance (Rm) = 50 Ω

Required full scale reading (I) = 5 A

${R_{sh}} = \frac{{{R_m}}}{{\left[ {\frac{I}{{{I_m}}} – 1} \right]}}$

${R_{sh}} = \frac{{50}}{{\left( {\frac{5}{{1 \times {{10}^{ – 3}}}} – 1} \right)}} \approx 0.01{\rm{\Omega }}$

21. Range of ammeter can extend by connecting

Shunt parallel to ammeter
Shunt in series to ammeter
Multiplier parallel to ammeter
Multiplier in series to ammeter

Answer.1. Shunt parallel to ammeter

Explanation:

A shunt is a low-value resistance having minimum temperature co-efficient.

It is connected in parallel with the ammeter whose range is to be extended. The combination is connected in series with the circuit whose current is to be measured.

The shunt provides a path for extra current as it is connected across (in parallel with) the instrument.

These shunted instruments can be used to measure currents many times greater than their normal full-scale deflection currents

22. A 1 m Amp, 50 Ω Galvanometer is required to measure 5 Amp (full scale). Find out the value of resistance to be added, across (shunt) the Galvanometer to accomplish this measurement.

10 Ω
0.01 Ω
1.0 Ω
0.001 Ω

Answer.2. 0.01 Ω

Explanation:

Given

I_m = 1 mA

R_m = 50 Ω

I = 5 Amp (full scale)

m = I/I_m

m = 5/(1 × 10^-3)

m = 5000

R_sh = R_m/(m -1)

R_sh = 50/(5000 – 1)

R_sh = 0.01 Ω

23. A galvanometer with a full-scale current of 10 mA has a resistance of 1000 Ω. The multiplying factor (The ratio of measured current to galvanometer current) of a 100 Ω shunt with this galvanometer is

Answer.3. 11

Explanation:

I = 10 mA

Rm = 1000 Ω

Rsh = 100 Ω

R_sh = R_m/(m -1)

m = 1 + R_m/R_sh

m = 1 + 1000/100

m = 11

24. A moving coil galvanometer is made into a DC ammeter by connecting

Low resistance across the meter
High resistance in series with the meter
Pure inductance across the meter
Capacitor in series with the meter

Answer.1. Low resistance across the meter

Explanation:

A galvanometer is a device that detects current in a circuit while an ammeter is a device that measures the current through the circuit. An ammeter is simply a galvanometer to which a small resistance is connected in parallel (shunt resistance).

A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer. This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.

25. A 250:5 CT is used along with an ammeter. If the ammeter reading is 2.7 A, estimate the line current.

135 A
50 A
13.5 A
2.7 A

Answer.1. 135 A

Explanation:

Given, CT ratio = 250/5 = 50

Measured value = 2.7 A

Hence, estimated line current = measured value × CT ratio

⇒ estimated line current = 2.7 × 50 = 135 A

26. A voltmeter has a resistance of 300 Ω and an inductance of 0.12 H. This instrument reads correctly on DC. What is the reading on AC at 100 V when the frequency is 25 Hz?

98.80 V
120.36 V
142.00 V
151.00 V

Answer.1. 98.80 V

Explanation:

Z = R = 300 Ω

I_DC = 100/300 = 0.3333 A

With AC Supply:

$Z = \sqrt {{R^2} + X_L^2}$

= $\sqrt {{{300}^2} + {{(2\pi \times 25 \times 0.12)}^2}}$

Z = 300.59 Ω

I_AC = 100/300.59 = 0.3326 A

Since, in DC 0.3333 A produced by 100 V

It means, in AC 0.3326 A produced by X V.

Hence,

100 ∝ 0.3333 … (1)

X ∝ 0.3326 …. (2)

From equation (1) & (2),

Reading (X) = (100 × 0.3326)/0.3333 = 98.80 V

27. While comparing ammeter and milliammeter the highest resistance will be in

Ammeter
Milliammeter
Both have same resistance
None of the above

Answer.2. Milliammeter

Explanation:

A milliammeter has high resistance than that of a voltmeter

Shunt Resistance

$S = \frac{{{I_g} \times G}}{{I – {I_g}}}$

Where

Ig is the maximum current that can pass through the galvanometer

I is the range of the ammeter.

It is clear that meters of the smaller current range (i.e. milliammeter) must have a greater value of shunt resistance. Now effective resistance of meter = GS/G +S. Since the value of S (shunt resistance) is more for milliammeter, it will have a higher resistance than that of the ammeter.

28. Out of the three ammeter, Voltmeter, Galvanometer which one has the lowest resistance?

Ammeter
Voltmeter
Galvanometer
All have same resistance

Answer.1. Ammeter

Explanation:

Out of the three, the ammeter has the lowest resistance. A galvanometer is converted into an ammeter by connecting a low resistance in parallel with it so that the effective resistance of the ammeter is less than that of the galvanometer (effective resistance is less than the least value of the two resistances).

A galvanometer is converted into a voltmeter by connecting a high resistance in series with it so that a voltmeter has a very high resistance than that of the galvanometer.

29. When an ammeter is put in a circuit, it reads less than the actual current in the circuit because

Ammeter is faulty
Ammeter is not calibrated
Circuit Current Decrease
All of the above

Answer.3. Circuit Current Decrease

Explanation:

When an ammeter is put in a circuit, it reads less than the actual current in the circuit because When an ammeter is put in the circuit, the circuit current decreases slightly due to the resistance of the ammeter. Therefore, it reads less than the actual current in the circuit.

30. Out of the three ammeter, Voltmeter, Galvanometer which one cannot measure alternating current?

Ammeter
Voltmeter
Galvanometer
All have same resistance

Answer.3. Galvanometer

Explanation:

A moving coil galvanometer measures the average value (d.c.) of current. Since the average value of an alternating current over a complete cycle is zero, a moving coil galvanometer cannot detect alternating current in a circuit.

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