Basic DC Ammeter MCQ Quiz in मल्याळम - Objective Question with Answer for Basic DC Ammeter - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given data,

Voltage = 30 volts

Resistance of meter (R_m) = 300 Ω

The inductive reactance of meter (X_m) = 400 Ω

Case 1: With DC Supply:

With DC Supply inductance offers an impedance of 0 Ω or acts as a short circuit.

Hence, the current through the meter (I_DC) is only due to resistance and it will be,

\(I_{DC} =\frac{V}{R_m}=\frac{30}{300}=0.1\ A\)

Since the meter reads correctly on DC, it means that a current of 0.1 A flowing through the instrument will give a reading of 30 V.

Case 2: With AC Supply:

With DC supply, the current through the meter (I_AC) is due to both resistance as well as reactance or by means of impedance,

\(I_{AC} =\frac{V}{Z_m}\)

Here, \(Z_m=\sqrt{R_M^2+X_M^2}=\sqrt{300^2+400^2}=500\ \Omega\)

Now, \(I_{AC} =\frac{30}{500}=0.06\ A\)

Now, meter reading due to due to both resistance as well as reactance will be,

Reading = R_m × I_AC = 300 × 0.06 = 18 V

Conclusion:

Error due to AC Measurement = 30 - 18 = 12 V

Hence, % Error = \(\frac{12}{30}\times 100=40%\)

Concept: 

• Shunts are used for the range extension of ammeters.
• 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.
• 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.
• 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.

In the figure

I is total current flowing in the circuit

Ish is the current through the shunt resistor

Rm is the ammeter resistance

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

Calculation:

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

Concept:

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

Here Rm = internal resistance of the coil

Rsh = Shunt resistance

I = Required full-scale range

Im = Full scale deflection of current

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

\({I_m}{R_m} = \left( {I - {I_m}} \right){R_{sh}}\)

\({R_m} = \left( {\frac{I}{{{I_m}}} - 1} \right){R_{sh}}\)

\(⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}\)

\(⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {m - 1} \right)}}\)

Where \(m = \frac{I}{{{I_m}}}\)

‘m’ is called multiplying power

Calculation:

Given that,

Full-scale deflection voltage (Vm) = 100 mV

Full scale reading (I) = 10 mA =  0.01 A.

Let's consider

Meter resistance = Rm

Full scale deflection current = Im

m = 200/(0.01) = 20000
R_m = 100 mV / 10mA = 10 Ω

\(⇒ {R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}\)

= \(\frac{10}{20000 - 1}\)

= 500.02 µΩ

Concept:

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

Here Rm = internal resistance of the coil

Rsh = Shunt resistance

I = Required full-scale range

Im = Full scale deflection of current

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

\({I_m}{R_m} = \left( {I - {I_m}} \right){R_{sh}}\)

\({R_m} = \left( {\frac{I}{{{I_m}}} - 1} \right){R_{sh}}\)

\(\Rightarrow {R_{sh}} = \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}\)

\(\Rightarrow {R_{sh}} = \frac{{{R_m}}}{{\left( {m - 1} \right)}}\)

Where \(m = \frac{I}{{{I_m}}}\)

‘m’ is called multiplying power

Calculation:

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

\({R_{sh}} = \frac{{{R_m}}}{{\left[ {\frac{I}{{{I_m}}} - 1} \right]}}\)

\({R_{sh}} = \frac{{1}}{{\left( {\frac{{100}}{{0.01}} - 1} \right)}} = 0.0001\;{\rm{\Omega }}\)

Note:

To increase the ranges of ammeter, we need to connect a small shunt resistance in parallel with ammeters.

To increase the ranges of a voltmeter, we need to connect a high series of multiplier resistance in series with voltmeters.

Concept: 

• Shunts are used for the range extension of ammeters.
• 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.
• 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.
• 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.
• The ratio of maximum current (with shunt) to the full-scale deflection current (without shunt) is known as the ‘multiplying power’ or ‘multiplying factor’ of the shunt.
   

In the figure

I is total current flowing in the circuit

Ish is the current through the shunt resistor

Rm is the ammeter resistance

Formula:

\({R_m} = \frac{{{V_m}}}{{{I_m}}}\)

\({R_{sh}} = \frac{{{R_m}}}{{M - 1}}\)

M = Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

Where,

Rsh = Shunt resistance 

Rm = Internal Meter resistance

Vm = Potential difference across the meter

Im = Meter current

Calculation:

Given 

Rsh = 50 Ω

\(M = \frac{{500}}{{100}} = 5\) 
\(R_{m} = 50 \times (5 - 1) = 200 \;Ω \)

Rm = 200 Ω 

Concept:

• For range extension of voltage measurement in moving coil instrument, a resistance is connected in series with coil resistance. 
• Because for a constant value of current, resistance connected in series connection has a higher voltage drop compared to parallel or shunt connection.
• For range extension of current measurement in moving coil instrument, a resistance is connected in parallel or shunt with coil resistance.
• Because for a constant value of voltage, resistance connected in parallel connection has a higher value of current flow compared to series connection.

Formula:

Rse = Rm(M – 1)

\({R_m} = \frac{{{V_m}}}{{{I_m}}}\)

\({R_{sh}} = \frac{{{R_m}}}{{M - 1}}\)

M= Multiplying factor = (Required full scale deflection) / (Initial full scale deflection)

Where,

Rsh = Series resistance

Rm = Meter resistance

Vm = Potential difference across meter

Im = Meter current

Calculation:

Given -

Rm = 0.2 Ω

\(M = \frac{{25}}{{5}} = 5\) 
\(R_{sh} = \frac{{0.2}}{{5 - 1}} = 0.05 \;\Omega \)

Rsh = 0.05 Ω

Clamp-on ammeter:

• Clip-on type current transformer is also known as clamp-on type current transformer.
• Clamp meters rely on the principle of magnetic induction to make non-contact AC current measurements.
• Electric current flowing through a wire produces a magnetic field. Since alternating current frequently reverses polarity, it causes dynamic fluctuations in the magnetic field which are proportional to the current flow.
• A current transformer inside the clamp meter senses the magnetic fluctuations and converts the value to an AC current reading. This type of measurement is convenient for measuring very high AC currents.
• While measuring, the primary is connected to feeder and the secondary is connected to ammeter. The current transformer steps down the high feeder current to low current which can be measured by using ammeter.

Concept Used:

In high voltage A.C. circuits, the measurement cannot be done by using the method of extension of ranges of low range meters by providing suitable shunts.

In such conditions, specially constructed accurate ratio transformers are used. These transformers are used to isolate the instruments from high current and high voltage A.C. circuits.

These are generally classified as

(i) Current transformer - large alternating currents can be measured

(ii) Potential transformer - High voltages can be measured.

Note: The secondary of CT or PT is connected to low range meter.

Application:

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

With DC Supply:

Z = R = 300 Ω

I_DC = \(\frac{100}{300}=0.3333 \ A\)

With AC Supply:

Z = \(\sqrt{R^2+X_L^2}\) = \(\sqrt{300^2+(2\pi\times 25 \times 0.12)^2} = 300.59\ \Omega\)

I_AC = \(\frac{100}{300.59}=0.3326 \ A\)

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) = \(\frac{100\times 0.3326}{0.3333}\) = 98.80 V

Input resistance (R_m) = 50 Ω

Full scale deflection (I_m) = 50 μA

Required full scale deflection (I) = 1 mA

Shunt resistance (R_sh) \(= \frac{{{R_m}}}{{\left( {\frac{I}{{{I_m}}} - 1} \right)}}\)

\(= \frac{{50}}{{\left( {\frac{{1 \times {{10}^{ - 3}}}}{{50 \times {{10}^{ - 6}}}} - 1} \right)}}\) 

= 2.63 Ω

Ammeter input resistance = 50 || 2.63

\(= \frac{{50 \times 2.63}}{{50 + 2.63}}\)