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:

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

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

Electrodynamometer voltmeter reads RMS value

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

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

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

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

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

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

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

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

Analog instrument:

Voltage error = ± 2% of 25V = ± 0.5 V

\(Error = \pm \frac{{0.5V}}{{20V}} \times = \pm 2.5\% \)

Digital instrument:

For 20 V displayed = on a \(3\frac{1}{2}\) digit display.

1 Digit = 0.1 V

Voltage error = ± (0.6% of reading + 1 digit)

= ± (1.2V + 0.1V) = ± 0.22 V

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

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

= 160 V