DC Ammeter MCQ || Basic DC Ammeter Questions and Answers

Answer.2. Series

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.

Answer.2. 0.01 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

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

Answer.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.

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

R_sh = 0.1/(5 − 1) = 0.025Ω

Rsh = 0.025 Ω = 25 m Ω

Therefore the shunt is made up of 25 mΩ and manganin

Answer.3. Small value in parallel

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.

Answer.3. Low resistance in parallel

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.

Answer.3. Short circuited

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.

Answer.3. 0.05 Ω in parallel with the meter

Explanation:

Given –

Rm = 0.2 Ω

M= Multiplying factor = 25/5 = 5

R_sh = R_m/(m − 1)

R_sh = 0.2/(5 −1)

R_sh = 0.05Ω

Answer.2. 0.1001 Ω

Explanation:

Given that,

Full scale deflection current (Im) = 1 mA

Meter resistance (Rm) =  100 Ω

Required full-scale reading (I) = 1 A

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

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

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