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:

Current transformer:

• A current transformer is a device that is used to measure high alternating current in a conductor.
• In the figure shown below the conductor act as the primary winding of a single turn that passed through the circular laminated iron core.
• The secondary winding consists of a large number of turns of fine wire wrapped around the core.
• Due to transformer action, the secondary current transforms into a lower value.

• Therefore, the current transformer changes the current into a lower value that can easily be measured by the measuring instrument.
• Along with a Potential transformer (PT), a current transformer (CT) can measure Power and energy also.
• Hence, CT used with Ammeter, Wattmeter, and Watt-hour meter.

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

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

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

Current Transformer: The secondary side of the current transformer is always kept short-circuited in order to avoid core saturation and high voltage induction so that the current transformer can be used to measure high values of currents.

• The current transformer works on the principle of shorted secondary.
• It means that the burden on the system Zb is equal to 0.
• Thus, the current transformer produces a current in its secondary which is proportional to the current in its primary.

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

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

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

The secondary side of the current transformer is always kept short-circuited in order to avoid core saturation and high voltage induction so that the current transformer can be used to measure high values of currents.

• Current transformer works on the principle of shorted secondary
• It means that the burden on the system Zb is equal to 0
• Thus, the current transformer produces a current in its secondary which is proportional to the current in its primary

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

• To reduce the error in standard resistance due to stray inductance, the adjacent wires should carry current in opposite direction so as to reduce the resultant magnetic field.
• To reduce the error due to contact (lead) resistance, a four terminal construction is used.
• To reduce the skin effect when used at high frequency a.c., small diameter wires are used.

At high and medium frequencies, a resistor can be represented as shown in figure below.

The equivalent impedance of the above shown resistor at an angular frequency ω rad/s is

If the resistance is to remain independent of frequency i.e. to have the same value of resistance at medium frequencies as with d.c, we have:

CR² = 2 L

so that, R_eq = R

At medium frequencies, we have:

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

Let the the required capacitance be C₁.

Now, when an external capacitance C₁ is connected in parallel, then new phase angle

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

Let the resistance R be changed to R + ΔR.

The Thevenin's equivalent circuit across terminals bd is shown below.