Construction of CRO MCQ Quiz in తెలుగు - Objective Question with Answer for Construction of CRO - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Deflection sensitivity: The deflection sensitivity of a cathode ray tube is defined as the deflection of the screen per unit deflection voltage.

Deflection sensitivity,

\(S = \frac{D}{{{E_d}}} = \frac{{L{l_d}}}{{2d{E_a}}}m/V\)

Where D = deflection of an electron beam on the screen in the Y direction in m

Ed = potential between deflecting plates in V

L = distance between the screen and the center of the deflecting plates in m

ld = length of the deflecting plates in m

d = distance between deflecting plates in m

Ea = accelerating voltage in V

Deflection factor: The deflection factor of a CRT is defined as the reciprocal of sensitivity.

Deflection factor \(D = \frac{1}{S} = \frac{{2d{E_a}}}{{L{l_d}}}V/m\)

Calculation:

Given that:  D₁  =  10 mm,  \({E_{{a_2}}} = \frac{{{E_{{a_1}}}}}{2}\) and E_d2 = 2 × E_d1  

\(D \propto \frac{{{E_d}}}{{{E_a}}}\; \Rightarrow \;\frac{{{D_2}}}{{{D_1}}} = \frac{{{E_d}_2}}{{{E_{{d_1}}}}}\frac{{{E_{{a_1}}}}}{{{E_a}_2}}\)

\({D_2} = \left( {\frac{{2 \times {E_{{d_1}}}}}{{{E_{{d_1}}}}} \times \frac{{{E_a}_1}}{{\frac{{{E_{{a_1}}}}}{2}}}} \right){D_1}\)

\(D_2\) = 4 × \(D_1\)

\(D_2\) = 4 × 10 = 40 mm

Now the deflection of the electron beam D₂ = 40 mm     

Concept: Dual beam oscilloscope:

• The dual beam oscilloscope emits two electron beams that are displayed simultaneously on a single scope, which could be individually or jointly controlled.

• There are two individual vertical input channels for two electron beams coming from different sources.
• Each channel has its own attenuator and pre-amplifier. Therefore, the amplitude of each channel can be controlled eventually.
• The two channels may have common or independent time base circuits which allow different sweep rates.
• Each beam passes through different channels for separate vertical deflection before it crosses a single set of horizontal plate.
• The horizontal amplifier is compiled by sweep generator to drive the plate which gives common horizontal deflection.
• The horizontal plates allow both the electron beams across the screen at the same time.

Note: The dual beam oscilloscope has two different electron gun which passes through two completely separate vertical channels, whereas dual trace oscilloscope has single electron beam which get split into two and passes through two separate channels.

Explanation:

To calculate the maximum velocity of the beam of electrons in a Cathode Ray Tube (CRT) with a given cathode anode voltage, we can use the principles of energy conservation and the known properties of electrons.

Given Data:

• Cathode anode voltage (V) = 182 V
• Charge of electron (e) = 1.6 × 10^-19 C
• Mass of electron (m) = 9.1 × 10^-31 kg

Assumptions:

• Electrons leave the cathode with zero velocity.

When an electron is accelerated from rest through a potential difference (V), it gains kinetic energy equal to the work done on it by the electric field. The work done on the electron by the electric field is given by:

Work Done (W) = eV

Since the electron starts from rest, its initial kinetic energy is zero. The final kinetic energy (K.E.) of the electron is given by:

K.E. = 1/2 mv²

By the principle of energy conservation, the work done on the electron is equal to its kinetic energy. Therefore, we can write:

eV = 1/2 mv²

Solving for the velocity (v) of the electron:

v = \(\sqrt{(2eV/ m)}\)

Substituting the given values into the equation:

v = \(\sqrt{((2 \times 1.6 × 10^{-19} C \times 182 V) / (9.1 × 10^{-31})}\)

Therefore, the maximum velocity of the beam of electrons in the CRT is:

v = 8 × 10⁶ m/s

The correct option is: Option 2: 8 × 10⁶ m/s

Given that, a compensated 20 : 1 attenuator is obtained.

\(\begin{array}{l} \frac{{{R_2}}}{{{R_1} + {R_2}}} = \frac{1}{{20}}\\ \Rightarrow \frac{1}{{1 + {R_2}}} = \frac{1}{{20}} \end{array}\)

⇒ R₂ = 19 MΩ

\(\begin{array}{l} \frac{{{C_1}}}{{{C_1} + {C_2}}} = \frac{1}{{20}}\\ \Rightarrow \frac{{{C_1}}}{{{C_1} + 100}} = \frac{1}{{20}} \end{array}\)

⇒ 20 C₁ = C₁ + 100

Explanation:

The CRO stands for a cathode ray oscilloscope.

It consists of a set of blocks which are, shown in the following block diagram.

Additional Information

Subparts of Cathode Ray Tube are

• Electron gun: It emits electrons and forms them into a beam with the help of a heater, cathode, grid, and pre-accelerating, accelerating, and focusing anode.
• Vertical deflection plates: VDP is applied with a test signal. Whose waveform has to be observed on the screen VDP is kept in a horizontal position.
• Horizontal deflection plates:  HDP is used for shifting the electron beam horizontally i.e. horizontal time scale is adjusted by changing HDP potential manually.
• Screen: A standard screen size is 8 cm by 10 cm. The screen is coated with a phosphor that emits light when struck by the electron beam

Hence, the Time-base generator is a part of CRO but not a part of CRT.

Uses of CRO:

• Used to measure the voltage, current, frequency, inductance, admittance, resistance, and power factor.
• used to monitor the signal properties as well as characteristics and also
• Used to control the analog signals.

Important Points

• Vertical deflection plates are mounted horizontally.
• Horizontal plates are mounted vertically as shown in the block diagram. 

Given that 10 to 1 attenuation probe.

\(\left( {{C_i}||C} \right) = \frac{C}{{10}}\)

\(\Rightarrow \frac{{{C_i} \times C}}{{{C_i} + C}} = \frac{C}{{10}}\)

\(\Rightarrow 9{C_i} = C \Rightarrow {C_i} = \frac{C}{9} = \frac{{45\;pF}}{9} = 5\;pF\)

 Effective input capacitance = (45 || 5)

= 4.5 pF

In channel I,

peak to peak voltage is 5 V

peak to peak  divisions of Upper trace voltage = 2

for one division voltage is 2.5 V

In channel 2, the no. of divisions for unknown voltage = 3

Voltage = 2.5 × 3 = 7.5 V

frequency of upper trace is 1 KHz

time period \(T = \frac{1}{f} = 1\ msec\)

For four divisions time period = 1 msec

Explanation:

In a Cathode Ray Oscilloscope (CRO), the measurement of time delay is achieved by using the time base control and measuring the horizontal shift. This method leverages the fundamental operation of the CRO, where the horizontal deflection is proportional to time, allowing for accurate time measurements of electronic signals.

Important Information:

Analyzing the other options:

1. By varying the intensity of the electron beam: This option is incorrect because varying the intensity (brightness) of the electron beam does not affect the time measurement. The intensity control adjusts the brightness of the displayed waveform but does not influence the horizontal deflection or time base settings.
2. By adjusting the brightness of the display: Similar to the first option, this method is not related to measuring time delay. Adjusting the brightness only changes how visible the waveform is on the screen and does not impact the time measurement.
3. By using the time base control and measuring the horizontal shift: This is the correct method for measuring time delay in a CRO. The time base control sets the horizontal deflection rate, and by measuring the horizontal shift of the waveform, the time delay can be accurately determined.
4. By changing the vertical deflection voltage: This option is incorrect as well. Changing the vertical deflection voltage affects the amplitude of the displayed signal but does not influence the horizontal deflection or time measurement.

In conclusion, the correct method for measuring time delay in a Cathode Ray Oscilloscope (CRO) is by using the time base control and measuring the horizontal shift, as this method directly relates to the time interval represented on the screen. The other options do not affect the time measurement and are thus not suitable for this purpose.

The correct answer is: 1) The propagation speed of the signal in the line

Explanation:

In an electrical delay line used in oscilloscopes, the delay time is determined by:

• The physical length of the delay line

• The propagation speed of the signal through the line (which depends on the medium's properties, such as its dielectric constant and inductance/capacitance per unit length).

Option Analysis

2. The resistance of the delay line → Affects signal attenuation, not the delay time.

3. The frequency of the input signal → The delay line is typically designed to be frequency-independent for a wide range of signals.

4. The speed of the electron beam → This relates to the CRT display, not the delay line.

Thus, the correct choice is 1) The propagation speed of the signal in the line.

Velocity of electron \( = \sqrt {\frac{{2e{E_0}}}{m}}\)

\(= \sqrt {\frac{{2 \times 1.6 \times {{10}^{ - 19}} \times 800}}{{9.1 \times {{10}^{ - 31}}}}} \) 

= 16.8 × 10⁶ m/sec