Electrical and Electronic Measurements Test 2

Given that, R = 820 ± 10% Ω

I = 10 mA

Full scale reading of ammeter = 25 mA ± 2%

error \(= \pm \frac{2}{{100}} \times 25 = \pm 0.5mA\)

Power dissipated in the resistor, P = I²R

By applying logarithmic on both sides,

⇒ ln P = ln I²R

⇒ ln P = 2 ln I + ln R

Differentiate w.r.t. ‘P’,

\(\frac{1}{P} = \frac{2}{I}\frac{{dI}}{{dP}} + \frac{1}{R}\frac{{dR}}{{dP}}\) 

\(\frac{{{\rm{\Delta }}P}}{P} \times 100 = \frac{{2{\rm{\Delta }}I}}{I} \times 100 + \frac{{{\rm{\Delta }}R}}{R} \times 100\) 

Sensitivity of meter 1, (S₁) = 10 kΩ/V

Internal resistance of meter 1 (RM₁) = 50 × 10 = 500 kΩ

Sensitivity of meter 2, (S₂) = 25 kΩ/V

Internal resistance of meter 2 (Rm₂) = 25 × 50 = 1250 kΩ

\({I_{FSD1}} = \frac{1}{{{S_1}}} = 0.1\;mA\) 

\({I_{FSD2}} = \frac{1}{{{S_2}}} = 0.04\;mA\) 

As these two meters are connected in series the maximum current that can pass through both the meters = min (I_FSD1, I_FSD2) = 0.04 mA

• Moving coil instruments are used as voltmeters and ammeters.
• Moving iron instruments can be used as ammeter and voltmeter but can’t be used as wattmeter. So, moving iron wattmeter does not exist.
• Induction type instruments can be used as voltmeter. It is normally used as wattmeter or energy meter.
• Electrodynamometer type instruments can be used as ammeter, voltmeter and wattmeter.

Concept:

Energy consumption is, E = V × I × cos ϕ × t × 10^-3 kWh

And error A_m - A_t

Where A_m = measured value

A_t = true value

Calculation:

V = 230, I = 10, cos ϕ = 0.8, t = 30 min

(Energy consumption)_t = V × I × cos ϕ × t × 10^-3 kW

\(= 230 \times 10 \times 0.8 \times \left( {\frac{{30}}{{60}}} \right) \times {10^{ - 3}}\)

(E.C)_t = 0.92 kWh

(E.C)_m = 58.25 – 57.35 = 0.90 kWh

Error = A_m - A_t = 0.90 – 0.92 = -0.02 kWh

PMMC voltmeter is used to measure DC voltages. A moving iron instrument is used to measure both AC and DC voltages.

Rectifier type instruments are used to measure both AC and DC voltages.

Electrodynamometer instruments are used to measure both AC and DC voltages.

Given that, Voltage (V) = 300 V

Current (I) = 16 A

Current coil resistance (R_CC) = 0.2 Ω

Pressure coil resistance (R_PC) = 20 kΩ

Power factor = cos ϕ = 0.4

Power (P) = VI cos ϕ = 300 × 16 × 0.4 = 1.92 kW

As the pressure coil is connected across load side, the error is due to pressure coil resistance.

Error \( = \frac{{{V^2}}}{{{R_{PC}}}} = \frac{{{{300}^2}}}{{20 \times {{10}^3}}} = 4.5\;W\)

Nominal ratio, \({K_n} = \frac{{200}}{1} = 200\)

Actual transformation ratio,

\(R = \frac{{{I_p}}}{{{I_s}}} = \frac{{100}}{{0.495}} = 202.02\)

\(Ratio\;error = \frac{{Nominal\;ratio - Actual\;ratio}}{{Actual\;ratio}} \times 100\)

\(= \frac{{200 - 202.02}}{{202.02}} \times 100\)

Given that = C = 300 pF

V₁ = 300 V

V₂ = 100 V

t = 120 S

\({R_{ins}} = \frac{t}{{C\;ln\left( {\frac{{{v_1}}}{{{v_2}}}} \right)}}\)

\(= \frac{{120}}{{\left( {300 \times {{10}^{ - 12}}} \right)\ln \left( {\frac{{300}}{{100}}} \right)}}\)

Given that, V_ph = 200 V

⇒ V_L = 200 √3 V

I_ph = I_L = 10 A

Power factor = cos ϕ = 0.866

⇒ ϕ = cos^-1 (0.866) = 30°

P₁ = V_LI_L cos (30 + ϕ)

= 200 √3 × 10 × cos 60°

= 1732 W

P₂ = V_LI_L cos (30 - ϕ)

= √3 × 200 × 10 × cos 0°

= 3464.1 W

Concept:

% error due to pressure coil inductance in a dynamometer type wattmeter

% error = (tan ϕ tan β)

β: impedance angle of pressure coil

ϕ: power factor angle

Error = tan ϕ tan β × true power

Calculation:

Pressure coil:

Inductance = 6 mH

Resistance = 2000 Ω

Impedance angle \( = \beta = {\tan ^{ - 1}}\left( {\frac{{\omega L}}{R}} \right)\)

\(\beta = {\tan ^{ - 1}}\left( {\frac{{2\pi \times 50 \times 6 \times {{10}^{ - 3}}}}{{2000}}} \right)\)

tan β = 0.94 × 10^-3

Given cos ϕ = 0; sin ϕ = 1

Error = tan ϕ tan β × true power

\( = \frac{{\sin \phi }}{{\cos \phi }}\tan \beta \times VI\cos \phi = VI\sin \phi \tan \beta \)

Error = 400 × 10 × 1 × 0.94 × 10^-3

The impedance of the meter on 60 Hz is

Z = 2400 + j0.6 × 2π × 60

= 2400 + j72 π

\(\left| Z \right| = \sqrt {{{\left( {2400} \right)}^2} + {{\left( {72\pi } \right)}^2}} = 2410.63\;V\) 

Current drawn by the meter for full-scale deflection \(= \frac{{120}}{{2410.63}}A\)

When it is connected on 600 V, it requires \(\frac{{120}}{{2410}}A\) for full-scale deflection.

Impedance \(= \frac{{600}}{{120}} \times 2410.63 = 12,053.15{\rm{\Omega }}\)

The resistance of the meter \(= \sqrt {{{\left( {12053.15} \right)}^2} - {{\left( {72\pi } \right)}^2}}\)

= 12051 Ω

The resistance to be inserted to increase the range to 600 V = 12051 – 2400 = 9651 Ω

Given that, reference voltage (V_ref) = 75 mV

First integration time (T₁) = 75 ms

Input voltage = 60 + 20 sin (50 πt)

DVM measures only average value.

The measured value = 60 V.

V_in = 60 V

V_in T₁ = V_ref T₂

⇒ (60)(75) = (75)(T₂)

⇒ T₂ = 60 ms

Total conversion time (T) = T₁ + T₂

= 75 + 60 = 135 ms

Best resolution \( = \frac{1}{{{{10}^3}}} \times 100\) mv = 0.1 mV

T_d = 18 × 10^-6 Nm,

I = 60 mA, M = - cos (θ + 30°) mH

The deflection torque is, \({T_d} = {I^2}\frac{{dM}}{{d\theta }}\)

\(18 \times {10^{ - 6}}N.m = {\left( {60 \times {{10}^{ - 5}}} \right)^2}\frac{{dM}}{{d\theta }}\)

\(\frac{{dM}}{{d\theta }} = \frac{{18 \times {{10}^{ - 6}}}}{{{{\left( {60 \times {{10}^{ - 3}}} \right)}^2}}} = 5\;mH/rad\)

\(\frac{{dM}}{{d\theta }} = \frac{d}{{d\theta }}\;\left( { - 5\cos \left( {\theta + 30^\circ } \right)} \right)\)

\(\frac{{dM}}{{d\theta }} = 5\sin \left( {\theta + 30^\circ } \right) \times {10^{ - 3}}\)

We know that \(\frac{{dM}}{{d\theta }} = 5\;mH/rad\)

5 × 10^-3 = 5 sin (θ + 30°) × 10^-3

sin (θ + 30°) = 1