Full Wave Bridge Rectifier Type EVM MCQ Quiz - Objective Question with Answer for Full Wave Bridge Rectifier Type EVM - Download Free PDF

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.

Concept

For triangular wave shape, the peak value of voltage is:

\(V_m=2V_{avg}\)

If the meter is calibrated in terms of RMS value, then the average value is given by:

\(V_{avg}={V_o\over FF}\)

The form factor of the bridge rectifier is 1.11

Calculation

The meter uses a full wave rectifier circuit which indicates the value of 3.33 V. The form factor for full wave rectified sinusoidal waveform is 1.11 

\(V_{avg}={3.33\over 1.11}=3\)

Peak voltage, \(V_m=2\times 3=6\space V\)

Concept

For triangular wave shape, the peak value of voltage is:

\(V_m=2V_{avg}\)

If the meter is calibrated in terms of RMS value, then the average value is given by:

\(V_{avg}={V_o\over FF}\)

The form factor of the bridge rectifier is 1.11

Calculation

The meter uses a full wave rectifier circuit which indicates the value of 3.33 V. The form factor for full wave rectified sinusoidal waveform is 1.11 

\(V_{avg}={3.33\over 1.11}=3\)

Peak voltage, \(V_m=2\times 3=6\space V\)

The correct answer is option 2): (3 kΩ)

Concept:

The series resistor  Rs  is called a “multiplier” resistor because it multiplies the working range of the meter movement as it proportionately divides the measured voltage across it.

Series “multiplier” resistors are used to give voltmeter movements higher range, and parallel “shunt” resistors are used to allow ammeter movements to measure currents beyond their natural range.

The given Circuit of AC voltmeter is full wave rectifier based  AC voltmeter

For full wave rectifier DC sensitivity

Sdc = \(1\over I_{fs}\) 

I_fs is Current required to produce full-scale deflection

For full wave rectifier AC sensitivity Sac

Sac = 0.9 × Sdc

Series multiplier resistance value

 Rs =( Sac × V) - Rm - 2Rd

where 

V is the  Voltage RMS value of  sinusoidal AC range of the voltmeter

R_d is the forward resistance of the diode.

Calculation:

Given

R_m = 500 Ω

V_rms = 10 V

I_fs = 2 × 10 ^-3 A

R_d  =  500 Ω

S_dc = \(1\over I_{fs}\)

= \(1\over 2 × 10 ^{-3}\)

= 0.5 × 10 3 Ω\V

S_ac = 0.9 × Sdc

= 0.9 × 0.5 × 10 3

= 0.45 × 10 3  Ω\V

Series multiplier resistance value

 R_s =( S_ac × V) - R_m - 2R_d

= ( 0.45 × 10 3 ×  10 ) - 500 - 1000

= 3000

= 3 kΩ 

\({V_{rm{s_{\left( {ind} \right)}}}} = form~factor \times {V_{dc}}\)

The form factor of square wave = 1

⇒ V_rms = 1 × V_dc = V_dc

The output of full-wave rectifier for

Sinusoidal input \(= \frac{{2{V_m}}}{\pi }\)

Given that, V_m = 2 V

\(\Rightarrow {V_{dc}} = \frac{4}{\pi }\)

Concept

For triangular wave shape, the peak value of voltage is:

\(V_m=2V_{avg}\)

If the meter is calibrated in terms of RMS value, then the average value is given by:

\(V_{avg}={V_o\over FF}\)

The form factor of the bridge rectifier is 1.11

Calculation

The meter uses a full wave rectifier circuit which indicates the value of 3.33 V. The form factor for full wave rectified sinusoidal waveform is 1.11 

\(V_{avg}={3.33\over 1.11}=3\)

Peak voltage, \(V_m=2\times 3=6\space V\)

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.

Concept

For triangular wave shape, the peak value of voltage is:

\(V_m=2V_{avg}\)

If the meter is calibrated in terms of RMS value, then the average value is given by:

\(V_{avg}={V_o\over FF}\)

The form factor of the bridge rectifier is 1.11

Calculation

The meter uses a full wave rectifier circuit which indicates the value of 3.33 V. The form factor for full wave rectified sinusoidal waveform is 1.11 

\(V_{avg}={3.33\over 1.11}=3\)

Peak voltage, \(V_m=2\times 3=6\space V\)

The correct answer is option 2): (3 kΩ)

Concept:

The series resistor  Rs  is called a “multiplier” resistor because it multiplies the working range of the meter movement as it proportionately divides the measured voltage across it.

Series “multiplier” resistors are used to give voltmeter movements higher range, and parallel “shunt” resistors are used to allow ammeter movements to measure currents beyond their natural range.

The given Circuit of AC voltmeter is full wave rectifier based  AC voltmeter

For full wave rectifier DC sensitivity

Sdc = \(1\over I_{fs}\) 

I_fs is Current required to produce full-scale deflection

For full wave rectifier AC sensitivity Sac

Sac = 0.9 × Sdc

Series multiplier resistance value

 Rs =( Sac × V) - Rm - 2Rd

where 

V is the  Voltage RMS value of  sinusoidal AC range of the voltmeter

R_d is the forward resistance of the diode.

Calculation:

Given

R_m = 500 Ω

V_rms = 10 V

I_fs = 2 × 10 ^-3 A

R_d  =  500 Ω

S_dc = \(1\over I_{fs}\)

= \(1\over 2 × 10 ^{-3}\)

= 0.5 × 10 3 Ω\V

S_ac = 0.9 × Sdc

= 0.9 × 0.5 × 10 3

= 0.45 × 10 3  Ω\V

Series multiplier resistance value

 R_s =( S_ac × V) - R_m - 2R_d

= ( 0.45 × 10 3 ×  10 ) - 500 - 1000

= 3000

= 3 kΩ 

The meter uses a full wave rectifier circuit it indicates value of 2.77 V

The form factor for full wave rectified sinusoidal waveform is 1.11

The average value of applied voltage, V_avg = 2.49 V

For triangular wave shape, peak value of voltage V_m = 2 V_avg = 4.98 V

\(\begin{array}{l} {V_{rms}} = \frac{{{V_m}}}{{\sqrt 3 }} = \frac{{4.98}}{{\sqrt 3 }} = 2.87\;V\\ \% \;error = \frac{{2.77 - 2.87}}{{2.87}} \times 100 = - 3.48\;\% \end{array}\)

Concept

For triangular wave shape, the peak value of voltage is:

\(V_m=2V_{avg}\)

If the meter is calibrated in terms of RMS value, then the average value is given by:

\(V_{avg}={V_o\over FF}\)

The form factor of the bridge rectifier is 1.11

Calculation

The meter uses a full wave rectifier circuit which indicates the value of 3.33 V. The form factor for full wave rectified sinusoidal waveform is 1.11 

\(V_{avg}={3.33\over 1.11}=3\)

Peak voltage, \(V_m=2\times 3=6\space V\)

\({V_{rm{s_{\left( {ind} \right)}}}} = form~factor \times {V_{dc}}\)

The form factor of square wave = 1

⇒ V_rms = 1 × V_dc = V_dc

The output of full-wave rectifier for

Sinusoidal input \(= \frac{{2{V_m}}}{\pi }\)

Given that, V_m = 2 V

\(\Rightarrow {V_{dc}} = \frac{4}{\pi }\)

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.

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.

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

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

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

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