Electrical and Electronic Measurements Test 1

We have resistance:

\(R = \frac{V}{I} = V{I^{ - 1}}\)

Weighted probable error in the resistance due to voltage is:

\({r_{RV}} = \frac{{\partial R}}{{\partial V}}{r_V} \)

\(= {I^{ - 1}}{r_V} = \frac{{{r_V}}}{I} \)

\(= \frac{{ \pm 12}}{{10}} = \pm 1.2\;{\text{Ω}}\)

Weighted probable error in the resistance due to current is:

\({r_{RI}}\frac{{\partial R}}{{\partial V}}{r_V} = \frac{V}{{{I^2}}}{r_I}\)

\( = \frac{{100}}{{{{\left( {10} \right)}^2}}} \times \left( { \pm 2} \right) = \pm 2\;{\text{Ω }}\) 
Probable error in computed resistance is:

\({r_R} = \;\sqrt {{{\left( {{r_{RV}}} \right)}^2} + {{\left( {{r_{RI}}} \right)}^2}} \)

 \(\sqrt {{{\left( {1.2} \right)}^2} + {{\left( 2 \right)}^2}} = 2.33\;{\text{Ω}}\)

Measurement by direct methods are not always possible, feasible and practicable. In most of the cases direct method of measurements are inaccurate because they involve human factors. Moreover, they are less sensitive. Due to these reasons use of direct method of measurement is limited while indirect methods are commonly used.

Electrodynamometer voltmeter reads RMS value

E = 20 sin ωt + 40 sin (3ωt + 30°) + 50 sin (5ωt + 60°)

\({E_{rms}} = \sqrt {{{\left( {\frac{{20}}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{{40}}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{{50}}{{\sqrt 2 }}} \right)}^2}}\)

Since one of the resistance (i.e. R₁) is accurate only to three significant figures, therefore the result of R _total has to be reduced to three significant figures. Hence, total resistance

• The deflection factor (scale factor) relates the change in deflection to the change in resistance.
• Given: A change of 8 Ω produces a deflection change of 3 mm.
• Deflection factor = Change in Resistance / Change in Deflection = 8 Ω / 3 mm.
• Calculate: 8 / 3 = 2.67 Ω/mm.
• Thus, the deflection factor is 2.67 Ω/mm, which matches option B.

Concept:

In a two-wattmeter method,

Total power (P) = P₁ + P₂

The power factor of the system = cos ϕ

\(\tan \phi = \sqrt 3 \frac{{[{P_1} - {P_2}]}}{{\left[ {{P_1} + {P_2}} \right]}}\)

Where P₁ & P₂ are power of wattmeters

Calculation:

P = 30 W, cos ϕ = 0.4

If P₁ and P₂ are two individual wattmeter readings then according to the problem,

P = P₁ + P₂ = 30 kW

Power factor angle ϕ = cos^-1 (0.4) = 66.4°

So, \(\phi = {\tan ^{ - 1}}\left( {\sqrt 3 \frac{{\left( {{P_1} - {P_2}} \right)}}{{\left( {{P_1} + {P_2}} \right)}}} \right)\)

\(66.4^\circ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{P_1} - {P_2}} \right)}}{{30}}} \right)\)

\(\frac{{2.288\; \times \;30}}{{\sqrt 3 }} = {P_1} - {P_2}\)  

39.69 kW = P₁ – P₂

E = VI cos   × t × 10^-3 kWh

\( = \frac{{230 \times 50 \times 1 \times 37}}{{3600}} \times {10^{ - 3}}\)

= 0.118 kWh

It takes 61 revolutions for 0.118 kWh

⇒ for 520 revolutions, it will be 1.007 kWh.

E_meas = 1.0075 kWh

E_true = 1 kWh

\(\% Error = \frac{{1.0075 - 1}}{1} \times 100 = 0.75\% \)

In a thermocouple ammeter,

Deflection (θ) ∝ I²

\(\Rightarrow \frac{{{\theta _1}}}{{{\theta _2}}} = {\left( {\frac{{{I_1}}}{{{I_2}}}} \right)^2}\)

\(\Rightarrow \frac{{{\theta _1}}}{{\left( {\frac{{{\theta _1}}}{4}} \right)}} = {\left( {\frac{{10}}{{{I_2}}}} \right)^2}\)

The circuit diagram for the measurement of voltage using multimeter is shown below:

Resistance at 15° C = 5 kΩ

Voltmeter reading = 200 V

Current at 15° C = 200/5000 = 0.04 A

Resistance at 50° C is, R₅₀ = R₁₅ (1 + α Δt)

= 5 (1 + 0.004 (50 – 15)) = 5.7 kΩ

Current at 50° C = 200/5,700

Reading at 50° C \( = \frac{{200 \times \frac{{200}}{{5700}}}}{{0.04}} = 175.4\;V\)

I_m = 0.1 mA, R_m = 99 Ω, R_sh = 1Ω

At 0.5 FSD, I_m = 0.57 mA

At full scale deflection,

Meter voltage, V_m = I_m R_m

= 0.1 mA × 99 = 9.9 mV

And I_sh R_sh = V_m

So, \({I_{sh}} = \frac{{{V_m}}}{{{R_{sh}}}} = \frac{{9.9\;m}}{{1\;{\text{Ω }}}} = 9.9\;mA\)

Total current, I = I_sh + I_m

= 9.9 mA + 0.1 mA = 10 mA.

Similarly, at 0.5 Full-scale deflection,

I_m = 0.5 × 0.1 mA = 0.05 mA

V_m = I_mR_m = 0.05 mA × 99 Ω = 4.95 mV

\({I_{sh}} = \frac{{{V_m}}}{{{R_{sh}}}} = \frac{{4.95\;mA}}{{1\;{\text{Ω }}}} = 4.95\;mA\)

Total current, I = I_s + I_m

Concept:

Force would be \(F = \frac{1}{2}{V^2}\frac{{dc}}{{dx}}\) for electrostatic instruments.

Given:

F = 5 × 10^-3 N, V = 10 kV, dx = 1 mm

Solution:

Force, \(F = \frac{1}{2}{V^2}\frac{{dc}}{{dx}}\)

\( \Rightarrow \frac{{dc}}{{dx}} = \frac{{2F}}{{{V^2}}} = \frac{{2 \times 5 \times {{10}^{ - 3}}}}{{{{\left( {10 \times {{10}^3}} \right)}^2}}}\)

So,\(\frac{{dc}}{{dx}} = 100\;pF/m\).

Given the change in distance = 1 mm

So, dc = 100 × 1 × 10^-3 = 0.1 pF

The given circuit can be reduced to its Thevenin’s equivalent circuit as shown below.

Current Transformer:

• The Current Transformer ( C.T. ), is a type of “instrument transformer” that is designed to produce an alternating current in its secondary winding which is proportional to the current being measured in its primary.
• The transformer's primary winding is physically connected in series with the conductor that carries the measured current flowing in the circuit.
• Generally, current transformers and ammeters are used together as a matched pair in which the design of the current transformer is such as to provide a maximum secondary current corresponding to a full-scale deflection on the ammeter.
• Current transformers “step-down” current levels from thousands of amperes down to standard output of a known ratio to either 5 Amps or 1 Amp for normal operation.

• The potential transformer is a voltage step-down transformer that reduces the voltage of a high voltage circuit to a lower level for the purpose of measurement. 
• The primary winding consists of a large number of turns which is connected across the high voltage side or the line in which measurements have to be taken or to be protected.
• The secondary winding has a lesser number of turns which is connected to the voltmeters.
• The primaries of PT are rated from 400 V to several thousand volts and secondaries are always for 110 V.
• Up to voltages of 5,000, potential transformers are usually of the dry type.
• If the voltage is between 5,000 and 13,800 volts, then the transformer may be either dry type or oil-immersed type.
• For the voltages above 13,800 volts, they are always an oil-immersed type of transformer.

I = 4θⁿ A

L when the meter current is zero is 10 mH i.e. θ = 0

Deflection in radian, \(\theta = \frac{1}{2}\frac{{{I^2}}}{K}\frac{{dL}}{{d\theta }}\)

Because I = 4θⁿ

Integrating the above expression, we have

Where A is constant

Given when meter current is zero

L = 10 mH and Thus θ = 0

∴ 10 × 10^-3 = 0 + A

A = 10 × 10^{-3
\(\therefore L = \frac{K}{{16\left( {1 - n} \right)}}{\theta ^{\left( {2 - 2n} \right)}} + 10 \times {10^{ - 3}}H\)}

Concept:

In a two-wattmeter method,

Total power (P) = P₁ + P₂

The power factor of the system = cos ϕ

\(\tan \phi = \sqrt 3 \frac{{[{P_1} - {P_2}]}}{{\left[ {{P_1} + {P_2}} \right]}}\)

Where P₁ & P₂ are the power of wattmeter A & B

Calculation:

P₁ = 7500 W, P₂ = -1500 W

V = 200 Ω, f = 50 Hz

\(\tan \phi = \sqrt 3 \frac{{\left[ {{P_1} - {P_2}} \right]}}{{\left[ {{P_1} + {P_2}} \right]}} = \frac{{\sqrt 3 \left[ {7500 - \left( { - 1500} \right)} \right]}}{{\left[ {7500 + \left( { - 1500} \right)} \right]}}\)

\(\phi = {\tan ^{ - 1}}\sqrt 3 \left( {\frac{{9000}}{{6000}}} \right)\)   

⇒ ϕ = 68.94°

cos ϕ = cos 68.94° = 0.359

For the whole power measured to appear on wattmeter A, the reading of wattmeter will be zero.

In this case, P₁ = 6000 W, P₂ = 0 W

Power factor at this case = 0.5

Power consumed by each phase \(= \frac{{6000}}{3} = 2000\;W\)

Voltage of each phase \(= \frac{{200}}{{\sqrt 3 }} = 115.47\;V\)

Current in each phase \(= \frac{{2000}}{{115.47\; \times \;0.359}} = 48.246\;A\)

The impedance of each phase \(= \frac{{{V_{ph}}}}{{{I_{ph}}}} = \frac{{115.47}}{{48.246}} = 2.39\;{\rm{\Omega }}\)

The resistance of each phase \(= \frac{{{P_{ph}}}}{{{{\left( {{I_{ph}}} \right)}^2}}} = 0.859\;{\rm{\Omega }}\)

The reactance of each phase \(= \sqrt {{{\left( {2.39} \right)}^2} - {{\left( {0.859} \right)}^2}} = 2.23\;{\rm{\Omega }}\)

∴ cos ϕ = 0.5

⇒ ϕ = cos^-1 (0.5)

tan ϕ = tan (cos^-1 0.5) = √3 = 1.732

Now, \(\tan \phi = \frac{X}{R}\)

∴ Reactance of circuit, X = R tan ϕ

X = 0.859 × 1.73 = 1487 Ω

∴ Capacitive reactance required = 2.23 – 1.487 = 0.743 Ω

And capacitance \(C = \frac{1}{{2\pi\; \times \;50\; \times \;0.743}} = 4284.116\;\mu F\)

The symbols of different equipment used in single line diagram are shown below.

Hence, the given symbol is for the potential transformer.

Voltage (V) = 230 V

Current (I) = 4 A

Power factor = cos ϕ = 1

No. of hours working = 6

Number of revolutions = 2208

Energy (E) = VI cos ϕ t × 10^-3 kWh

= 230 × 4 × 1 × 6 × 10^-3 = 5.52 kWh

Meter constant (k_c) \(= \frac{{number\;of\;revolutions}}{{kWh\;rating}}\)

\(= \frac{{2208}}{{5.52}} = 400\;rev/kWh\) 

Now, revolutions made by meter = 1472

Supply voltage (V) = 230 V

Current (I) = 5 A

No. of hours working = 4

\({K_c} = \frac{{1472}}{{kWh}}\) 

\(\Rightarrow 400 = \frac{{1472}}{{kWh}}\) 

⇒ kWh = 3.68

⇒ VI cos ϕ t = 3.68 × 10³

⇒ 230 × 5 × cos ϕ × 4 = 3.68 × 10³

The ratio of potential transformer \(= \frac{{44000}}{{220}}\)

The ratio of current transformer \(= \frac{{2000}}{{20\;A}}\)

V = 200 V, I = 12.50 A

\(t = \frac{{66}}{{3600}},\cos \phi = 0.91\)

Actual energy consumed during the test period

\(= \sqrt 3 \times ratio\;of\;P.T \times ratio\;of\;C.T \times V \times I \times \cos \phi \times t \times {10^{ - 3}}\)

\(= \sqrt 3 \times \frac{{44000}}{{220}} \times \frac{{2000}}{{20}} \times 200 \times 12.5 \times 0.91 \times {10^{ - 3}} \times \frac{{66}}{{3600}}\)

= 1444.819 kWh

Energy recorded by the meter during the test period,

\(= \frac{{40}}{{0.25}} = 160\;kWh\)

∴ Percentage error \(= \frac{{160 - 1444.819}}{{1444.819}} \times 100\)

= -88.92% low

Ratio error in current transformer:

• In the current transformer, the primary current I_p should be exactly equal to the secondary current multiplied by turns ratio, i.e. K_TIs.
• But there is a difference between primary current I_p should be exactly equal to the secondary current multiplied by the turns ratio.
• This difference is contributed by the core excitation or magnetizing component of no-load current.
• The error in the current transformer introduced due to this difference is called current error or ratio error.

The actual ratio of transformation varies with operating conditions and the error in secondary voltage is defined as Percentage ratio error 

K_n is the nominal ratio
R is the actual ratio
It can be reduced by secondary turns compensation i.e. slightly decreasing the secondary turns.

Phase angle error:
In an ideal voltage transformer, there should not be any phase difference between the primary voltage and the secondary voltage reversed. However, in an actual transformer, there exists a phase difference between Vo­ and V­s reversed.
The phase angle is taken as +ve when secondary voltage reversed leads the primary voltage.
The angle is -ve when the secondary voltage reversed lags the primary voltage.
It can be reduced by keeping the primary and secondary windings are wound as closely as possible.

\(Deflection\;D = \frac{{L{l_d}{E_d}}}{{2d{E_a}}}\)

\({E_d} = \frac{{2d{E_a}D}}{{L{l_d}}} = \frac{{2 \times 7 \times {{10}^{ - 3}} \times 3000 \times 4 \times {{10}^{ - 2}}}}{{.35 \times 3 \times {{10}^{ - 2}}}}\)

= 160 V