# DC Ammeter MCQ || Basic DC Ammeter Questions and Answers

1. An ammeter is connected in ________ with the circuit.

1. Parallel
2. Series
3. Both parallel and series
4. None of the above

Explanation:

In a series connection, the current flowing through all the components of the circuit is the same. An ammeter measures the current in the circuit, hence it is connected in series so that the same current that is there in the circuit flows through it and gets measured.

Ammeter is a very low resistance(nearly zero) device. If it will be connected in parallel, it would draw most of the current and would get damaged.

2. A moving coil instrument has full-scale deflection at 50 mV and 10 mA. The value of shunt resistance required to be connected to convert it into a (0-5A) ammeter is:

1. 0.005 Ω
2. 0.01 Ω
3. 0.001 Ω
4. 1 Ω

Explanation:

Given that,

Full-scale deflection voltage (Vm) = 50 mV

Full scale deflection current (Im) = 10 mA

Meter resistance (Rm) = 50/10 = 5 Ω

Required full scale reading (I) = 5 A

As the two resistances, Rm and Rsh are in parallel, the voltage drop across the resistance is equal.

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

where

Rm = internal resistance of the coil

Rsh = Shunt resistance

I = Required full-scale range

Im = Full scale deflection of current

${R_{sh}} = \frac{5}{{\left( {\frac{5}{{0.01}} – 1} \right)}} = 0.01\:{\rm{\Omega }}$

3. A moving coil galvanometer can be converted into a voltmeter by introducing a resistance of a _________.

1. Large value in parallel
2. Small value in series
3. Small value in parallel
4. Large value in series

Explanation:

A galvanometer can be converted into a voltmeter by connecting a high resistance in series connection within it. The scale is calibrated in volt. The value of the resistance connected in series decides the range of the voltmeter. The resistance is calculated by this equation which is connected in series.

4. A 10A DC Ammeter has a resistance of 0.1 Ω is to be extended to 50 A, the required shunt wire is

1. Manganin wire of 20 m Ω
2. Constantan wire of 20 m Ω
3. Manganin wire of 25 m Ω
4. Constantan wire of 25 m Ω

Answer.3. Manganin wire of 25 m Ω

Explanation:

A shunt is a low-value resistance having minimum temperature co-efficient and is made up of manganin. Because manganin has a very low value of temperature coefficient.

Given

Rm = 0.1 Ω

M = 50/10 = 5

Rsh = 0.1/(5 − 1) = 0.025Ω

Rsh = 0.025 Ω = 25 m Ω

Therefore the shunt is made up of 25 mΩ and manganin

5. A moving coil galvanometer can be converted into an Ammeter by introducing a resistance of a _________.

1. Large value in parallel
2. Small value in series
3. Small value in parallel
4. Large value in series

Explanation:

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.

6. If an ammeter is to be used in place of a voltmeter, we must connect with the ammeter a

1. High resistance in parallel
2. High resistance in series
3. Low resistance in parallel
4. Low resistance in series

Explanation:

An ammeter is a low resistance device and is always connected in series with the circuit. So, to use an ammeter in place of a voltmeter a low resistance must be connected in parallel with the ammeter to make its resistance small.

7. What happens when the ammeter is connected in parallel?

1. Open circuited
2. Closed circuited
3. Short circuited
4. None of the above

Explanation:

Ammeter consists of a wire of low resistance (ideally zero). When connected in parallel, a large amount of current passes through it due to the low resistance of the ammeter. Hence gets burned i.e. short-circuited.

8. An ammeter has a current range of 0-5 A, and its internal resistance is 0.2 Ω. In order to change the range to 0-25 A, we need to add a resistance of

1. 0.8 Ω in series with the meter
2. 1.0 Ω in series with the meter
3. 0.04 Ω in parallel with the meter
4. 0.05 Ω in parallel with the meter

Answer.3. 0.05 Ω in parallel with the meter

Explanation:

Given –

Rm = 0.2 Ω

M= Multiplying factor = 25/5 = 5

Rsh = Rm/(m − 1)

Rsh = 0.2/(5 −1)

Rsh = 0.05Ω

9. A 1 mA ammeter has a resistance of 100 Ω. It is to be converted to a 1A ammeter. The value of shunt resistance is

1. 0.001 Ω
2. 0.1001 Ω
3. 1000000 Ω
4. 100 Ω

Explanation:

Given that,

Full scale deflection current (Im) = 1 mA

Meter resistance (Rm) =  100 Ω

Required full-scale reading (I) = 1 A

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

${R_{sh}} = \frac{{100}}{{\left( {\frac{{1000}}{1} – 1} \right)}} = 0.1\:{\rm{\Omega }}$

10. Identify the main reason why an Ammeter should have very low internal resistance

1. It should provide damping
2. It should decrease the circuit resistance.
3. It should not affect the circuit resistance
4. It should not burn

Answer.3. It should not affect the circuit resistance

Explanation:

• For reading a higher range of current ammeter shunt is used.
• When high current is to be measured most of the current passed through a low resistance called a shunt
• A high value of internal resistance will vary the current considerably in an Ammeter.
• The internal resistance of a milli-ammeter must be very low for minimal effect on the current in the circuit

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