Electronic AC Voltmeters MCQ Quiz in বাংলা - Objective Question with Answer for Electronic AC Voltmeters - বিনামূল্যে ডাউনলোড করুন [PDF]

In a thermocouple ammeter,

Deflection (θ) ∝ I²

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

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

Concept

The RMS value of an alternating current is the steady (D.C) current that produces the same amount of heat as the A.C. 

The RMS value of any signal is given by:

\(RMS=\sqrt{{1\over T}\int_{-\infty}^{\infty}x^2(t)\space dt}\)

where, T = Time period of the signal

Explanation

The RMS value of a purely resistive AC circuit is given by:

\(\rm V_{rrms}=\frac{V_m}{\sqrt2}\)

where, Vm = Peak value of the signal

From the above expression, we observe that rms value is directly proportional to the peak value of the signal.

So if the peak value of the output signal increases, its rms value of the ac component increases.

Explanation:

Ripple Factor Calculation:

The ripple factor is a measure of the residual AC component present in the output of a rectifier circuit in relation to the DC component. It quantifies the effectiveness of the rectifier in converting AC to DC. The formula for the ripple factor (ρ) is given by:

Ripple Factor (ρ) = (RMS value of the AC component) / (DC value)

To express the ripple factor in percentage:

Ripple Factor (%) = [(RMS value of the AC component) / (DC value)] × 100

From the problem statement:

• DC voltage (V_DC) = 25 V
• RMS voltage of the AC component (V_AC) = 1.5 V

Substituting the given values into the formula:

Ripple Factor (%) = [(1.5) / (25)] × 100

Ripple Factor (%) = (0.06) × 100

Ripple Factor (%) = 6%

Thus, the ripple factor is 6%, which matches Option 4.

Important Information:

The ripple factor is a crucial parameter in rectifier design and operation. It provides insights into the quality of the DC output and the extent of filtering required in the circuit. A lower ripple factor indicates a smoother and more stable DC output.

Analysis of Other Options

To further understand the analysis, let’s evaluate the other options:

• Option 1 (0.17): This value is significantly lower than the calculated ripple factor of 6%. It may have resulted from an incorrect calculation or misunderstanding of the formula.
• Option 2 (0.34): This value is also incorrect and does not match the calculated ripple factor. It suggests a calculation error or incorrect substitution of values.
• Option 3 (0.25): This value is closer to 25% and does not correspond to the calculated ripple factor of 6%. Again, it might result from an incorrect computation.
• Option 4 (0.06): This value is correct. When expressed as a percentage (multiplied by 100), it gives 6%, which aligns with the accurate calculation.

Conclusion:

The ripple factor is a key parameter for evaluating the performance of a rectifier circuit. For the given voltmeter readings, the ripple factor was calculated to be 6%, corresponding to Option 4. This value indicates the quality of the rectified output and helps in designing the appropriate filtering mechanism for further smoothing of the DC signal.

The Correct Answer is : The amount of oxidn

Key Points

• Amount of Peroxides: The peroxide value represents the amount of peroxide, specifically peroxides of fatty acids, in an oil or fat. This is an indicator of the extent of oxidation that the substance has undergone.
• Indicator of Rancidity: The peroxide value is often used as an early indicator of the rancidity of an oil or fat. Higher values typically indicate increased levels of spoilage. This is because the oxidation of fats and oils leads to the formation of products that have undesirable flavors and smells.
• Quality Check: Peroxide values are often used as part of a quality check in the food industry. Keeping an eye on the peroxide value helps in maintaining the quality and increasing the shelf life of oil-based products.
• Monitoring Storage and Transportation: Peroxide values can change with the storage and handling conditions. For example, if an oil is stored in high heat or exposed to light, it can lead to an increase in peroxide values. Thus, frequent monitoring of peroxide values can help to optimize storage and transportation conditions.
• Relation to Unsaturated Fats: The peroxide value is particularly relevant for oils and fats containing unsaturated fatty acids, as these are more prone to oxidation.
• Health Implications: Extremely high peroxide values could indicate a health risk as the consumption of highly oxidized oils can lead to several health issues. For example, it can contribute to an increase in free radicals in the body.

Rms voltage measured by meter = 5v

Actual voltage value sensed by meter \( = \frac{5}{{1.15}} = 4.35\;V\)

The voltage measured by meter is average value.

\({V_{avg}} = \frac{{2{V_m}}}{\pi }\) 

\(\Rightarrow {V_m} = \frac{\pi }{2} \times {V_{avg}} = \frac{\pi }{2} \times 4.35 = 6.83\;V\) 

\({V_{rms}} = \frac{{{V_m}}}{{\sqrt 2 }} = 4.83\;V\) 

\(\% \;error = \frac{{5 - 4.83}}{{4.83}} \times 100 = 3.52\% \)

Full scale deflection of PMMC instrument (I_FSD) = 2 mA

\(DC\;Sensitivity\;\left( {{S_{DC}}} \right) = \frac{1}{{{I_{FSD}}}} = \frac{1}{{2 \times {{10}^{ - 3}}}} = 0.5\;k{\text{Ω}}/V.\)

The given voltmeter is half wave rectifier type.

For this instrument,

AC sensitivity = 0.45 × dc sensitivity

= 0.45 × 0.5 kΩ /V = 0.225 kΩ/V

Resistance value of series resistor

R_s = S_AC × V - R_m – R_d

= (0.225 × 10³ × 20) – 100 – 50

The dc sensitivity \(= \frac{1}{{100 \times {{10}^{ - 6}}}} = 10k{\rm{\Omega }}/v\)

For a halfwave rectifier,

Ac sensitivity = 0.45 × dc sensitivity

= 0.45 × 10 = 4.5 kΩ/v

Series multiplier = (S_ac) (V) - R_m - R_d

= (4.5)(20) – 2 – 0.5

We known that the form factor for full wave rectifier sinusoidal waveform 1.11

\(\therefore Average\;value\;of\;voltage\;is\;{V_{avg}} = \frac{{R.M.S\;value}}{{Form\;factor}}\) 

\(= \frac{{4.44}}{{1.11}} = 4\;Volt\) 

For triangular waveform, peak value of voltage is

V_m = 2 V_avg = 8 volt

R.M.S value of voltage is

\(V = \frac{{{V_m}}}{{\sqrt 3 }} = \frac{8}{{\sqrt 3 }} = 4.62\;volt\) 

The d.c sensitivity is \({S_{dc}} = \frac{1}{{{I_{fs}}}} = 500\;{\rm{\Omega }}/V\)

For full wave rectifier circuit, a.c sensitivity

S_ac = 0.9 × 500 = 450 Ω/v

Resistance of multiplier R_s = S_acV – R_m – 2R_d

The deflection of a thermo-electric ammeter is θ_F = kI²

\(\frac{{{\theta _f}}}{k} = {\left( {20} \right)^2} = 400\)

The deflection at half scale

\(\frac{{{\theta _f}}}{2} = k{I^2}\)