Electrical and Electronic Measurements Test 2

Full scale deflection (I_m) = 2 A

Voltage drop at FSD (V_m) = 100 mV

Internal resistance \(\left( {{R_m}} \right) = \frac{{{V_m}}}{{{I_m}}} = \frac{{100 \times {{10}^{ - 3}}}}{2} = 50\;{\rm{m\Omega }}\)

Required full scale deflection (I) = 10 A

To extend the range of an ammeter we need to connect a resistor in parallel with the meter.

Concept:

The deflection angle in a moving iron instrument is given by

\(\theta = \frac{1}{{{K_c}}}\frac{{{I^2}}}{2}\frac{{dL}}{{d\theta }}\)

I is the current flows through the coil

K_c is the spring constant

\(\frac{{dL}}{{d\theta }}\) is the change in self-inductance of a coil with change in deflection angle.

Calculation:

Given that, I = 1 A

K_c = 10 μNm/degree

Change in self-inductance (dL) = 2.51 mH

Change in deflection (dθ) = 90°

\( \Rightarrow \theta = \frac{1}{{10 \times {{10}^{ - 6}}}} \times \frac{{{{\left( 1 \right)}^2}}}{2} \times \frac{{2.51 \times {{10}^{ - 3}}}}{{90}}\)

Concept:

The deflection angle in an electrodynamic instrument is given by

\(\theta = \frac{1}{{{K_c}}}{I_1}{I_2}\frac{{dM}}{{d\theta }}\)

I₁ is the current flows through fixed coil

I₂ is the current flows through moving coil

K_c is the spring constant

M is the change mutual inductance between both the coils with change in deflection angle

θ ∝ I₁ I₂

Calculation:

Case 1:

θ₁ = 60°, I₁₁ = 5 mA, I₂₁ = 4 mA

Case 2:

I₁₂ = I₂₂ = 3 mA

\( \Rightarrow \frac{{{\theta _1}}}{{{\theta _2}}} = \frac{{{I_{11}}{I_{21}}}}{{{I_{12}}{I_{22}}}}\)

\( \Rightarrow \frac{{60}}{{{\theta _2}}} = \frac{{5 \times 4}}{{3 \times 3}}\) 

\(Total\;circuit\;measured\;resistance\;\left( {{R_m}} \right) = \frac{{75}}{{2.5\;mA}} = 30\;k{\rm{\Omega }}\)

Resistance of voltmeter, R_v = 100 × 500 = 50 kΩ

As the voltmeter is in parallel with the unknown resistance, we have

\({R_m} = \frac{{{R_T}\;{R_V}}}{{{R_T} + {R_V}}}\)

\(\Rightarrow {R_T} = \frac{{{R_m}\;{R_v}}}{{{R_V} - {R_m}}} = \frac{{50 \times 30}}{{50 - 30}} = 75\;k{\rm{\Omega }}\)

\(Percentage\;error = \frac{{measured\;value - true\;value}}{{true\;value}} \times 100\)

\(= \frac{{30 - 75}}{{75}} \times 100 = - 60\;\% \)

Concept:

The deflection angle in a PMMC instrument is given by

\(\theta = \frac{1}{{{K_c}}}BINA\)

K_c is the spring constant

B is the flux density

N is the number of turns

A is the area

Calculation:

Given that, K_c = 0.5 μ Nm/degree

N = 100

B = 0.8 wb/m²

I = 2 mA

Length = 1.25 cm

Width = 1 cm

Area (A) = 1.25 × 1 = 1.25 cm²

Deflection angle,

\(\theta = \frac{1}{{0.5 \times {{10}^{ - 6}}}} \times 0.8 \times 2 \times {10^{ - 3}} \times 100 \times 1.25 \times {10^{ - 4}}\)

⇒ θ = 40°

Concept:

The deflection in an electrostatic instrument is given by

\(\theta = \frac{1}{2}\frac{{{V^2}}}{{{K_c}}}\frac{{dC}}{{d\theta }}\)

V is the voltage

K_c is the spring constant

\(\frac{{dC}}{{d\theta }}\) is the change in capacitance with change in deflection angle.

Calculation:

Given that,

Change in capacitance (dC) = 95.7 – 80 = 15.7 pF

Change in angle (dθ) = 104 – 95 = 9°

Spring constant (K_c) = 2 × 10^-6 Nm/degree

Deflection angle (θ) = 100°

\(\theta = \frac{1}{2}\frac{{{V^2}}}{{{K_c}}}\frac{{dC}}{{d\theta }}\)

\( \Rightarrow \frac{{100 \times \pi }}{{180}} = \frac{1}{2} \times \frac{{{V^2}}}{{2 \times {{10}^{ - 6}}}} \times \frac{{15.7 \times {{10}^{ - 12}}}}{9}\)

Concept:

The deflection in an electrodynamic instrument is given by

\(\theta = \frac{{{I^2}}}{{{K_c}}}\frac{{dM}}{{d\theta }}\)

I is the current

K_c is the spring constant

M is the change mutual inductance between both the coils with change in deflection angle.

Calculation:

Given that,

θ = 100° \( = \frac{{5\pi }}{9}\;rad\)

I = 200 mA

\(M = 500\sin \left( {\theta - \frac{{2\pi }}{9}} \right)\)

\(\frac{{dM}}{{d\theta }} = 500\cos \left( {\theta - \frac{{2\pi }}{9}} \right)\)

At \(\theta = \frac{{5\pi }}{9},\frac{{dM}}{{d\theta }} = 500\cos \left( {\frac{{5\pi }}{9} - \frac{{2\pi }}{9}} \right)\)

\(= 500\cos \left( {\frac{\pi }{3}} \right) = 250\;\;\mu H/radian\)

\(\frac{{5\pi }}{9} = \frac{{{{\left( {200 \times {{10}^{ - 3}}} \right)}^2}}}{{{K_c}}} \times 250 \times {10^{ - 6}}\)

⇒ K_c = 5.73 μ N-m/radian