Home Blog Instrumentation Exploring the World of Instrumentation Questions – Top Quiz

# Exploring the World of Instrumentation Questions – Top Quiz

Instrumentation is a complex field that encompasses a wide range of measurement and control systems used in industrial processes. Instrumentation includes everything from sensors and transducers to control systems and data acquisition systems.

Engineers and technicians working in this field need to have a solid understanding of the underlying principles and concepts of instrumentation. One way to test and enhance your knowledge in this area is through quizzes and questions that cover a broad range of topics.

## World of Instrumentation Questions

We will explore the world of instrumentation questions and present a top quiz that covers some of the most important topics in this field.

## What is the percentage range for a 4 to 20 mA signal?

A. 16.7% to 100%
B. 0% to 100%
C. 20% to 100%
D. 4% to 20%

Explanation:

The 4 to 20 mA signal corresponds to 0 to 100% of the range of the process variable it represents. Here’s how it’s broken down:

4 mA corresponds to 0% of the process variable range.
20 mA corresponds to 100% of the process variable range.

The reason 4 mA is used as the lowest value instead of 0 mA is to provide a way to differentiate between a signal at the lowest end of the measurement range (4 mA) and a loss of signal or a fault condition (0 mA).

Also, the signal is linear, meaning that for a given process variable, the corresponding mA signal can be calculated by linear interpolation between 4 mA (0%) and 20 mA (100%). For example, a 12 mA signal would correspond to 50% of the process variable range.

The percentage of the range corresponding to a particular mA signal within the 4-20 mA range can be calculated by:

Percentage = [(mA – 4) / 16] * 100%

For instance, at 12 mA:

Percentage = [(12 – 4) / 16] * 100% = 50%

So, a 12 mA signal corresponds to 50% of the range of the process variable.

## If a transmitter has a range of 0-100 psi, what is the mA output at 75 psi?

A. 9.5 mA
B. 15 mA
C. 16 mA
D. 19 mA

Explanation:

In a 4-20 mA current loop standard, the 4 mA corresponds to the minimum process value (0 psi in this case), and 20 mA corresponds to the maximum process value (100 psi in this case).

The mA output for any given pressure within this range can be calculated as follows:

Determine the span of the pressure and mA ranges:

Pressure span = 100 psi – 0 psi = 100 psi
Current span = 20 mA – 4 mA = 16 mA

Calculate the ratio of the difference between the desired pressure and the minimum pressure to the pressure span:
Ratio = (75 psi – 0 psi) / 100 psi = 0.75

Multiply this ratio by the current span and add the minimum current:
mA output = (0.75 * 16 mA) + 4 mA = 16 mA

So, the mA output at a pressure of 75 psi is 16 mA.

## What is the resolution of a 12-bit analog-to-digital converter (ADC) for a 4 to 20 mA signal?

A. 0.8 mA
B. 0.05 mA
C. 0.2 mA
D. 0.1 mA

Explanation: The resolution of a 12-bit ADC is

(Maximum Range – Minimum Range) / 2^12

Plugging in the values gives (20-4)/2^12 = 0.195 mA, which can be rounded to 0.2 mA.

## A control valve is receiving a 12 mA signal. What is the valve position if its full stroke is 10 inches?

A. 1.2 inches
B. 5 inches
C. 4.8 inches
D. 7.2 inches

Explanation: The control valve position can be calculated using the formula

[(mA Output – 4 mA) / (16 mA ] / (Full Stroke)

Plugging in the values gives [(12 mA – 4 mA) / (16 mA] / (10 inches) = 5 inches.

## What is the minimum loop resistance for a 24 V power supply and a 4 to 20 mA transmitter?

A. 833 ohms
B. 1200 ohms
C. 1500 ohms
D. 2000 ohms

Explanation: The minimum loop resistance can be calculated using the formula

(Power Supply Voltage – Minimum Voltage Drop) / 0.02 A

Plugging in the values gives (24 V – 8 V) / 0.02 A = 800 ohms, which can be rounded up to 833 ohms.

## What is the voltage drop across a 500 ohm resistor in a 4 to 20 mA loop?

A. 2 V
B. 10 V
C. 16 V
D. 20 V

Explanation: The voltage drop can be calculated using the formula Resistance x Current. Plugging in the values gives 500 ohms x 0.02 A = 10 V.

## A pressure transmitter has a span of 0-300 psi and a linear output. What is the mA output at 150 psi?

A. 8 mA
B. 12 mA
C. 16 mA
D. 20 mA

Explanation: Since the transmitter has a linear output, we can calculate the mA output using a simple proportion. At 0 psi, the mA output is 4 mA. At 300 psi, the mA output is 20 mA. Therefore, at 150 psi, the mA output is halfway between 4 mA and 20 mA, which is 12 mA.

You can use an Instrumentation calculator for these calculations.

## What is the maximum loop resistance for a 12 V power supply and a 4 to 20 mA transmitter?

A. 360 ohms
B. 600 ohms
C. 1000 ohms
D. 1200 ohms

Explanation: The maximum loop resistance can be calculated using the formula

(Power Supply Voltage – Minimum Voltage Drop) / 0.02 A

Plugging in the values gives (12 V – 8 V) / 0.02 A = 200 ohms, which can be multiplied by 5 to give a maximum loop resistance of 1000 ohms.

## What is the minimum output pressure for a 4 to 20 mA pressure transmitter with a range of 0-100 psi if the mA output is 5 mA?

A. 12.5 psi
B. 6.25 psi
C. 37.5 psi
D. 50 psi

Explanation: The minimum output pressure can be calculated using the formula

[(mA Output – 4 mA) / (16 mA / Full Scale)] + Minimum Range

Plugging in the values gives [(5 mA – 4 mA) / (16 mA / 100 psi)] + 0 psi = 6.25 psi.

## What is the maximum output pressure for a 4 to 20 mA transmitter with a range of 0-300 psi if the mA output is 16 mA?

A. 275 psi
B. 250 psi
C. 225 psi
D. 300 psi

Explanation: The maximum output pressure can be calculated using the formula

[(mA Output – 4 mA) / (16 mA / Full Scale)] + Minimum Range

Plugging in the values gives [(16 mA – 4 mA) / (16 mA / 300 psi)] + 0 psi = 225 psi.

## A flow meter has a span of 0-10,000 gallons per minute (gpm). What is the mA output at 5,000 gpm?

A. 8 mA
B. 12 mA
C. 16 mA
D. 20 mA

Explanation: In the 4-20 mA current loop standard, the 4 mA corresponds to the minimum process value (0 gpm in this case) and 20 mA corresponds to the maximum process value (10,000 gpm here).

The 5,000 gpm is halfway between the minimum and maximum process values. So, the corresponding mA value will be halfway between 4 mA and 20 mA.

To calculate this, you first find the total span of the mA output:

20 mA – 4 mA = 16 mA

This 16 mA represents the entire span of the flow meter (0-10,000 gpm). Since 5,000 gpm is halfway, the corresponding mA output will be halfway between 4 mA and 20 mA:

4 mA + (16 mA / 2) = 12 mA

So, the mA output at 5,000 gpm is 12 mA.

## What is the maximum number of devices that can be connected in series in a 4 to 20 mA loop with a 24 V power supply and a loop resistance of 1000 ohms?

A. 1
B. 2
C. 3
D. 4

Explanation: The maximum number of devices that can be connected in series in a 4 to 20 mA loop is determined by the maximum loop resistance allowed by the power supply.

For a 24 V power supply and a loop resistance of 1000 ohms, the maximum number of devices that can be connected is 1, since any additional devices would increase the loop resistance beyond the maximum allowed.

## A temperature transmitter has a range of 0-200 degrees Celsius and a linear output. What is the mA output at 150 degrees Celsius?

A. 6 mA
B. 16 mA
C. 12 mA
D. 18 mA

Explanation:

The mA output for any given temperature within this range can be calculated by:

Determine the span of the temperature and mA ranges:
Temperature span = 200°C – 0°C = 200°C
Current span = 20mA – 4mA = 16mA

Calculate the ratio of the difference between the desired temperature and the minimum temperature to the temperature span:
(150°C – 0°C) / 200°C = 0.75

Multiply this ratio by the current span and add the minimum current:
0.75 * 16mA + 4mA = 16mA

Therefore, the mA output at 150 degrees Celsius is 16 mA.

## What is the voltage drop across a 250 ohm resistor in a 4 to 20 mA loop with a current of 16 mA?

A. 1 V
B. 2 V
C. 3 V
D. 4 V

Explanation: The voltage drop across a resistor in a 4 to 20 mA loop can be calculated using Ohm’s law, which states that voltage = current x resistance.

Plugging in the values gives voltage = 0.016 A x 250 ohms = 4 V.

However, since the transmitter requires at least 4 mA to operate, the actual current through the resistor is 16 mA – 4 mA = 12 mA.

Plugging in the adjusted value gives voltage = 0.012 A x 250 ohms = 3 V.

## What is the range of a 4 to 20 mA transmitter with a loop resistance of 500 ohms and a power supply voltage of 24 V?

A. 0-1200 ohms
B. 0-600 ohms
C. 0-300 ohms
D. 0-150 ohms

Explanation: The range of a 4 to 20 mA transmitter can be calculated using the formula

(Power Supply Voltage – Minimum Voltage Drop) / Current

Plugging in the values gives (24 V – 8 V) / 0.02 A = 800 ohms.

However, since the loop resistance is 500 ohms, the maximum load resistance is 300 ohms.

Subtracting the loop resistance and multiplying by 2 gives a range of 0-600 ohms.

## A flow transmitter has a range of 0-1000 L/min. What is the mA output at a flow rate of 725 L/min?

A. 10 mA
B. 12.5 mA
C. 15.6 mA
D. 17.5 mA

Explanation:

In a 4-20 mA current loop standard, the 4 mA represents the minimum process value (0 L/min in this case), and 20 mA represents the maximum process value (1000 L/min in this case).

We can calculate the mA output for a given flow rate within this range using the following steps:

Determine the span of the flow and current ranges:

Flow range = 1000 L/min – 0 L/min = 1000 L/min

Current range = 20 mA – 4 mA = 16 mA

Calculate the proportion of the desired flow rate to the total flow range:

Proportion = (725 L/min – 0 L/min) / 1000 L/min = 0.725

Multiply this proportion by the current range and add the minimum current value:

mA output = (0.725 * 16 mA) + 4 mA = 15.6 mA

So, the correct mA output at a flow rate of 725 L/min is 15.6 mA.

## What is the resistance of a 4 to 20 mA transmitter with a current of 8 mA and a power supply voltage of 12 V?

A. 750 ohms
B. 1000 ohms
C. 1250 ohms
D. 1500 ohms

Explanation: The resistance of a 4 to 20 mA transmitter can be calculated using the formula

Resistance = (Power Supply Voltage – Minimum Voltage Drop) / Current

Plugging in the values gives (12 V – 4 V) / 0.008 A = 1000 ohms.

However, since the transmitter is already drawing 8 mA, the additional load resistance can be calculated by subtracting the transmitter resistance from the loop resistance (usually 250 or 500 ohms).

In this case, using a loop resistance of 500 ohms gives a load resistance of 500 ohms – 1000 ohms = -500 ohms, which is not possible.

Therefore, the correct answer is A, 750 ohms, which is the resistance of the transmitter only.

## What is the maximum loop resistance allowed for a 4 to 20 mA transmitter with a power supply voltage of 24 V and a current of 20 mA?

A. 800 ohms
B. 1000 ohms
C. 1200 ohms
D. 1500 ohms

Explanation: The maximum loop resistance allowed for a 4 to 20 mA transmitter can be calculated using the formula

Maximum Resistance = (Power Supply Voltage – Minimum Voltage Drop) / Current

Plugging in the values gives (24 V – 4 V) / 0.02 A = 1000 ohms. However, this calculation assumes that the transmitter is the only device in the loop.

To calculate the maximum loop resistance with additional devices, the resistance of each device must be added together and compared to the maximum allowed resistance.

## What is the purpose of a 250 ohm resistor in a 4 to 20 mA loop for a PLC system?

A. To Prevent the overload the siganl
B. To convert the voltage signal to a current signal
C. To convert the current signal to a voltage signal
D. To protect the transmitter from voltage spikes

Explanation: The 250 ohm resistor in a 4 to 20 mA loop is used to convert the current signal to a voltage signal. This is done by using Ohm’s law to calculate the voltage drop across the resistor based on the current flowing through it.

For example, at 4 mA, the voltage drop across the resistor would be 1 volt (0.004 A * 250 ohms). At 20 mA, the voltage drop would be 5 volts (0.02 A * 250 ohms). This voltage signal can then be used as an input to other devices, such as controllers or indicators, that require a voltage signal rather than a current signal.

## A pressure transmitter has a range of 0-1000 psi and a linear output. What is the mA output at a pressure of 400 psi if the transmitter has a span of 20 mA?

A. 8.3 mA
B. 11.2 mA
C. 10.4 mA
D. 14.6 mA

Explanation:

In a 4-20 mA current loop standard, 4 mA corresponds to the minimum process value (0 psi in this case), and 20 mA corresponds to the maximum process value (1000 psi in this case).

The mA output for any given pressure within this range can be calculated as follows:

Determine the span of the pressure and mA ranges:
Pressure span = 1000 psi – 0 psi = 1000 psi
Current span = 20 mA – 4 mA = 16 mA

Calculate the ratio of the difference between the desired pressure and the minimum pressure to the pressure span:
Ratio = (400 psi – 0 psi) / 1000 psi = 0.4

Multiply this ratio by the current span and add the minimum current:
mA output = (0.4 * 16 mA) + 4 mA = 10.4 mA

So, the mA output at a pressure of 400 psi is 10.4 mA.

## What is the effect of increasing the loop resistance on a 4 to 20 mA transmitter?

A. The current through the transmitter increases
B. The voltage across the transmitter increases
C. The power consumption of the transmitter increases
D. The voltage drop across the loop resistance increases

Explanation: Increasing the loop resistance on a 4 to 20 mA transmitter will increase the voltage drop across the loop resistance, which in turn will reduce the voltage available to the transmitter. This will cause the current through the transmitter to decrease, as the voltage available to drive the current is reduced.

## A temperature transmitter has a range of 0-220°C. What is the mA output at a temperature of 85.6°C if the transmitter has a span of 20 mA?

A. 10.22 mA
B. 11.82 mA
C. 12.12 mA
D. 17.07 mA

Explanation:

In a 4-20 mA current loop standard, 4 mA corresponds to the minimum process value (0°C in this case), and 20 mA corresponds to the maximum process value (220°C in this case).

The mA output for any given temperature within this range can be calculated as follows:

Determine the span of the temperature and mA ranges:
Temperature span = 220°C – 0°C = 220°C
Current span = 20 mA – 4 mA = 16 mA

Calculate the ratio of the difference between the desired temperature and the minimum temperature to the temperature span:
Ratio = (85.6°C – 0°C) / 220°C = 0.389

Multiply this ratio by the current span and add the minimum current:
mA output = (0.389 * 16 mA) + 4 mA = 10.224 mA

So, the mA output at a temperature of 85.6°C is approximately 10.224 mA.

## What is the voltage drop across a 250 ohm resistor if a 4 to 20 mA current is flowing through it?

A. 1 V
B. 4 V
C. 5 V
D. 20 V

Explanation: The voltage drop across a 250 ohm resistor with a 4 to 20 mA current flowing through it can be calculated using Ohm’s law, which states that V = I * R.

Plugging in the values gives V = 0.016 A * 250 ohms = 4 V at the minimum current of 4 mA and V = 0.02 A * 250 ohms = 5 V at the maximum current of 20 mA.

## What is the maximum allowable resistance for a 4 to 20 mA loop powered device with a 24 VDC power supply?

A. 1200 ohms
B. 1000 ohms
C. 800 ohms
D. 600 ohms

Explanation: The maximum allowable resistance for a 4 to 20 mA loop powered device with a 24 VDC power supply can be calculated using Ohm’s law and the maximum allowable voltage drop of 10 volts.

The maximum current of 20 mA gives a voltage drop of 0.48 V (20 mA * 24 V), so the remaining voltage available for the device is 23.52 V (24 V – 0.48 V).

Dividing this voltage by the maximum current of 20 mA gives a maximum allowable resistance of 1176 ohms, which can be rounded up to 1000 ohms for safety reasons.

## What is the purpose of a shunt resistor in a 4 to 20 mA loop?

A. To provide a voltage reference for the transmitter
B. To increase the voltage across the transmitter
C. To reduce the voltage across the transmitter
D. To measure the current flowing through the loop

Explanation: A shunt resistor in a 4 to 20 mA loop is used to measure the current flowing through the loop.

The shunt resistor is placed in parallel with the transmitter, which allows a small current to bypass the transmitter and flow through the shunt resistor.

The voltage drop across the shunt resistor can then be measured and used to calculate the current flowing through the loop.

## What is the resistance of a 4 to 20 mA loop if a transmitter is drawing 10 mA and the loop voltage is 24 VDC?

A. 2400 ohms
B. 1200 ohms
C. 800 ohms
D. 600 ohms

Explanation: The resistance of a 4 to 20 mA loop can be calculated using Ohm’s law and the measured current and voltage. Plugging in the values gives R = V / I = 24 V / 10 mA = 2400 ohms.

A. 5.36 mA
B. 8.25 mA
C. 14.93 mA
D. 12.02 mA

## What is the minimum allowable loop resistance for a 4 to 20 mA loop powered device with a 12 VDC power supply?

A. 500 ohms
B. 600 ohms
C. 750 ohms
D. 1000 ohms

Explanation: The minimum allowable loop resistance for a 4 to 20 mA loop powered device with a 12 VDC power supply can be calculated using Ohm’s law and the minimum current of 4 mA.

The minimum voltage drop across the device is 4 mA * 12 VDC = 48 mV.

Dividing this voltage drop by the minimum current of 4 mA gives a minimum allowable resistance of 48 mV / 4 mA = 500 ohms.

## What is the purpose of a loop-powered display in a 4 to 20 mA loop?

A. To provide a voltage reference for the transmitter
B. To measure the current flowing through the loop
C. To convert the current signal to a voltage signal
D. To provide a visual indication of the current in the loop

Explanation: A loop-powered display in a 4 to 20 mA loop is used to provide a visual indication of the current in the loop.

The display is powered by the current in the loop and does not require a separate power supply. It can be used to monitor the current in the loop or to provide a local indication of the measured variable.

We can also show equivalent process variable value on the display which is equivalent to the loop current.

## What is the purpose of a loop isolator in a 4 to 20 mA loop?

A. To measure the current flowing through the loop
B. To isolate the transmitter from the control system
C. To provide a voltage reference for the transmitter
D. To convert the current signal to a voltage signal

Explanation: A loop isolator in a 4 to 20 mA loop is used to isolate the transmitter from the control system. This can help to protect the control system from noise or other interference on the loop.

The loop isolator is typically placed between the transmitter and the control system and provides electrical isolation between the two.

## What is the voltage drop across a 250 ohm resistor if a current of 10 mA is flowing through it?

A. 0.25 V
B. 1 V
C. 2.5 V
D. 25 V

Explanation: The voltage drop across a resistor can be calculated using Ohm’s law. Plugging in the values gives V = I * R = 10 mA * 250 ohms = 0.25 V.

## What is the maximum loop resistance for a 4 to 20 mA loop if the loop voltage is 24 VDC and the transmitter is drawing 20 mA?

A. 960 ohms
B. 1060 ohms
C. 1140 ohms
D. 1200 ohms

Explanation: The maximum loop resistance for a 4 to 20 mA loop can be calculated using Ohm’s law and the maximum current of 20 mA.

The maximum voltage drop across the transmitter is 20 mA * R = 24 VDC – 10 VDC = 14 VDC. Solving for R gives R = (24 VDC – 10 VDC) / 20 mA = 700 ohms.

Adding in the maximum allowable loop resistance of 250 ohms gives a maximum loop resistance of 950 ohms, which can be rounded up to 960 ohms for safety reasons.

## A pressure transmitter has a range of 0-150 psi and a linear output. What is the mA output at a pressure of 140 psi if the transmitter has a span of 20 mA?

A. 15.93 mA
B. 16.93 mA
C. 17.93 mA
D. 18.93 mA

Explanation:

In a 4-20 mA current loop standard, 4 mA corresponds to the minimum process value (0 psi in this case), and 20 mA corresponds to the maximum process value (150 psi in this case).

The mA output for any given pressure within this range can be calculated as follows:

Determine the span of the pressure and mA ranges:
Pressure span = 150 psi – 0 psi = 150 psi
Current span = 20 mA – 4 mA = 16 mA

Calculate the ratio of the difference between the desired pressure and the minimum pressure to the pressure span:
Ratio = (140 psi – 0 psi) / 150 psi = 0.933

Multiply this ratio by the current span and add the minimum current:
mA output = (0.933 * 16 mA) + 4 mA = 18.93 mA

So, the mA output at a pressure of 140 psi is approximately 18.93 mA.

## What is the minimum current that must be supplied to a loop-powered device in a 4 to 20 mA loop?

A. 0 mA
B. 4 mA
C. 8 mA
D. 12 mA

Explanation: In a 4 to 20 mA loop, the minimum current that must be supplied to a loop-powered device is 4 mA. This is the current that represents the lower range of the measured variable. Any current below this level may be interpreted as a fault or failure in the loop.

## What is the advantage of using a 4 to 20 mA loop over a 0 to 10 VDC signal?

A. Higher resolution
B. Longer cable runs
C. Lower cost
D. More accurate

Explanation: One advantage of using a 4 to 20 mA loop over a 0 to 10 VDC signal is the ability to transmit the signal over longer cable runs. This is because the current signal is less susceptible to noise and interference than a voltage signal.

The current signal also provides better accuracy and resolution than a voltage signal. However, the cost of a 4 to 20 mA loop may be higher than a voltage signal due to the need for additional components such as a power supply and current loop isolators.

## What is the advantage of using a 2-wire transmitter in a 4 to 20 mA loop?

A. Simpler wiring
B. More accurate
C. Faster response time
D. Lower cost

Explanation: One advantage of using a 2-wire transmitter in a 4 to 20 mA loop is simpler wiring. A 2-wire transmitter combines the power supply and signal into a single pair of wires, simplifying the installation and reducing the cost of wiring.

This is in contrast to a 4-wire transmitter, which requires separate power and signal wires. While a 4-wire transmitter may provide better accuracy and response time, the simplicity and cost savings of a 2-wire transmitter make it a popular choice in many applications.

## What is the maximum cable length for a 4 to 20 mA loop with a 250 ohm load resistor?

A. 1000 feet
B. 2000 feet
C. 3000 feet
D. 4000 feet

Explanation: The maximum cable length for a 4 to 20 mA loop depends on several factors, including the loop voltage, load resistance, and cable type.

For a loop with a 250 ohm load resistor, the maximum cable length is typically around 2000 feet, assuming a voltage drop of no more than 2 volts.

## What is the effect of increasing the load resistance in a 4 to 20 mA loop?

A. Increases loop voltage
B. Decreases loop voltage
C. Increases loop current
D. Decreases loop current

Explanation: Increasing the load resistance in a 4 to 20 mA loop decreases the loop current and voltage drop across the load resistor. This is because the transmitter is designed to provide a fixed current output, and the load resistance determines the voltage drop across the load. As the load resistance increases, the voltage drop increases, which decreases the loop voltage and current.

## What is the purpose of a current-to-pressure (I/P) converter in a 4 to 20 mA loop?

A. To measure pressure in the loop
B. To convert current to voltage
C. To convert current to pressure
D. To provide electrical isolation

Explanation: A current-to-pressure (I/P) converter is a device used to convert a 4 to 20 mA current signal into a pneumatic pressure signal. It is commonly used in process control applications to control valves, actuators, and other pneumatic devices.

The I/P converter typically operates by converting the current signal to a proportional air pressure output, which can be used to control the device being actuated.

## What is the maximum distance between a 4 to 20 mA transmitter and a current indicator with a loop impedance of 500 ohms?

A. 1000 feet
B. 2000 feet
C. 3000 feet
D. 4000 feet

Explanation: The maximum distance between a 4 to 20 mA transmitter and a current indicator depends on several factors, including the loop voltage, load resistance, and cable type.

For a loop with a 500 ohm load resistor, the maximum cable length is typically around 1000 feet, assuming a voltage drop of no more than 2 volts.

## What is the effect of decreasing the loop impedance in a 4 to 20 mA loop?

A. Increases loop voltage
B. Decreases loop voltage
C. Increases loop current
D. Decreases loop current

Explanation: Decreasing the loop impedance in a 4 to 20 mA loop increases the loop current. This is because the transmitter is designed to provide a fixed current output, and the loop impedance determines the voltage drop across the load. As the loop impedance decreases, the voltage drop decreases, which increases the loop current.

## What is the effect of increasing the loop voltage in a 4 to 20 mA loop?

A. Increases loop impedance
B. Decreases loop impedance
C. Increases loop current
D. Decreases loop current

Explanation: Increasing the loop voltage in a 4 to 20 mA loop increases the loop current. This is because the transmitter is designed to provide a fixed current output, and the loop voltage determines the voltage drop across the load. As the loop voltage increases, the voltage drop increases, which increases the loop current.

## What is the effect of increasing the load resistance in a 4 to 20 mA loop?

A. Increases loop voltage
B. Decreases loop voltage
C. Increases loop current
D. Decreases loop current

Explanation: Increasing the load resistance in a 4 to 20 mA loop decreases the loop current. This is because the transmitter is designed to provide a fixed current output, and the loop resistance determines the voltage drop across the load. As the load resistance increases, the voltage drop increases, which decreases the loop current.

## What is the purpose of a loop calibrator in a 4 to 20 mA loop?

A. To measure current in the loop
B. To display the process variable
C. To provide electrical isolation
D. To generate a current signal for calibration

Explanation: A loop calibrator is a device used to generate a current signal for calibration purposes in a 4 to 20 mA loop. It typically consists of a potentiometer or other device that can be used to adjust the output current of the calibrator.

The purpose of the loop calibrator is to provide a convenient way to calibrate the transmitter without the need for external power or additional wiring.

## What is the purpose of a loop repeater in a 4 to 20 mA loop?

A. To measure current in the loop
B. To display the process variable
C. To provide electrical isolation
D. To extend the range of the loop

Explanation: A loop repeater is a device used to extend the range of a 4 to 20 mA loop beyond the maximum distance that can be achieved with a single loop.

It typically consists of a transmitter and a receiver that are connected by a communication link, such as a twisted pair of wires or a fiber optic cable. The purpose of the loop repeater is to regenerate the current signal and amplify it for transmission over longer distances.

## What is the typical response time of a 4 to 20 mA transmitter?

A. 10 ms
B. 100 ms
C. 1 s
D. 10 s

Explanation: The typical response time of a 4 to 20 mA transmitter is 100 ms. This represents the time it takes for the transmitter to respond to a change in the process variable and transmit the new signal to the receiver. This response time is sufficient for most industrial applications and provides a good balance between speed and accuracy.

## How can you test a 4 to 20 mA loop with a multimeter?

A. Measure the voltage drop across the loop
B. Measure the current in the loop
C. Measure the resistance of the loop
D. Measure the frequency of the loop

Explanation: To test a 4 to 20 mA loop with a multimeter, you can measure the current in the loop using the mA range on the multimeter. Connect the multimeter in series with the loop, making sure to respect the polarity of the current signal.

The multimeter should read a current value between 4 and 20 mA, depending on the process variable and the transmitter calibration.

## What is a loop simulator in a 4 to 20 mA loop?

A. A device that measures current in the loop
B. A device that displays the process variable
C. A device that provides electrical isolation
D. A device that simulates the transmitter output signal

Explanation: A loop simulator is a device used to simulate the output signal of a 4 to 20 mA transmitter. It typically provides a stable current source that can be used to verify the operation of the loop receiver, such as a controller or recorder.

The loop simulator can be used to simulate different process variable values and test the response of the loop under different conditions.

## What is the most common cause of a 4 to 20 mA loop failure?

A. Open circuit in the loop
B. Short circuit in the loop
C. Transmitter failure

Explanation: The most common cause of a 4 to 20 mA loop failure is an open circuit in the loop, which can be caused by a broken wire or a loose connection. This can lead to a loss of signal and a failure of the receiver to measure the process variable. Short circuits in the loop can also occur, but they are less common.

## How can you check the loop impedance in a 4 to 20 mA loop?

A. Measure the voltage drop across the loop
B. Measure the current in the loop
C. Measure the resistance of the loop
D. Measure the frequency of the loop

Explanation: To check the loop impedance in a 4 to 20 mA loop, you can measure the resistance of the loop using an ohmmeter. Disconnect the transmitter and receiver from the loop and measure the resistance between the two wires. The typical range for loop impedance is between 250 and 1200 ohms, depending on the power supply voltage and the transmitter current range.

## What can cause a low signal in a 4 to 20 mA loop?

A. High loop impedance
B. Low loop impedance
C. Transmitter shorted

Explanation: A low signal in a 4 to 20 mA loop can be caused by a high loop impedance, which can reduce the current flowing through the loop and result in a low reading at the receiver.

This can be caused by a long cable length, a small wire size, or a high-resistance connection. Low loop impedance can cause a high signal and can be caused by a short circuit or a low resistance connection.

## How can you diagnose a broken wire in a 4 to 20 mA loop in a simple way without disturbing the circuit?

A. Measure the loop impedance
B. Measure the voltage drop across the loop
C. Measure the current in the loop
D. Measure the resistance of the loop

Explanation: To diagnose a broken wire in a 4 to 20 mA loop, you can measure the voltage drop across the loop using a voltmeter. If the voltage drop is zero or very low, this indicates that there is a break in the loop somewhere. By tracing the wiring and checking for loose connections, you can identify the location of the break and repair it.

## What is the typical range of an instrument loop transmitter?

A. 0 to 5 VDC
B. 0 to 10 VDC
C. 0 to 20 mA
D. 4 to 20 mA

Explanation: The typical range of a 4 to 20 mA loop transmitter is from 4 mA to 20 mA, which corresponds to the minimum and maximum values of the process variable being measured. This range is commonly used in industrial applications because it provides a robust and reliable signal that can be transmitted over long distances.

## How can you check the accuracy of a 4 to 20 mA loop receiver?

A. Measure the loop impedance
B. Measure the voltage drop across the loop
C. Measure the current in the loop
D. Measure the resistance of the loop

Explanation: To check the accuracy of a 4 to 20 mA loop receiver, you can measure the current in the loop using a multimeter or a loop calibrator. By comparing the measured current to the expected value based on the process variable being measured, you can determine the accuracy of the receiver and identify any potential problems.

May 20, 2023

Great Knowledge