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The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?
Correct answer is '80'. Can you explain this answer?
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The resistance is measured by the voltmeter ammeter method the voltmet...
V = 50 ± (2% of 100 V)
Im = 50 × 10-3 ± (2% of 100 mA)
Rm = 103 ± 4% = 1000 ± 40
= 960 Ω to 1040 Ω
Limit = 80 Ω
Im = 50 × 10-3 ± (2% of 100 mA)
Rm = 103 ± 4% = 1000 ± 40
= 960 Ω to 1040 Ω
Limit = 80 Ω
Most Upvoted Answer
The resistance is measured by the voltmeter ammeter method the voltmet...
Ohms) can be determined?
To determine the resistance, we can use Ohm's Law: R = V/I, where R is the resistance in ohms, V is the voltage in volts, and I is the current in amperes.
First, we need to convert the voltmeter and ammeter readings to their actual values. We know that the voltmeter reading of 50 V is on a 100 V scale, so the actual voltage is:
50 V / 100 V x Full Scale = 50 V / 100 V x 1000 V = 500 V
Similarly, the ammeter reading of 50 mA is on a 100 mA scale, so the actual current is:
50 mA / 100 mA x Full Scale = 50 mA / 100 mA x 1 A = 0.5 A
Now we can use Ohm's Law to calculate the resistance:
R = V/I = 500 V / 0.5 A = 1000 ohms
To determine the limit of accuracy, we need to consider the errors in the voltmeter and ammeter readings. Since both meters are guaranteed for accuracy within 2% of full scale, the actual readings could be anywhere within ±2% of the full scale values.
For the voltmeter, the possible range of actual readings is:
500 V ± 2% x 1000 V = 500 V ± 20 V
For the ammeter, the possible range of actual readings is:
0.5 A ± 2% x 1 A = 0.5 A ± 0.01 A
Using the extreme values of the actual readings, we can calculate the extreme values of the resistance:
R = (500 V + 20 V) / (0.5 A - 0.01 A) = 1040 ohms
R = (500 V - 20 V) / (0.5 A + 0.01 A) = 960 ohms
Therefore, the limit within which the resistance can be determined is:
960 ohms ≤ R ≤ 1040 ohms
To determine the resistance, we can use Ohm's Law: R = V/I, where R is the resistance in ohms, V is the voltage in volts, and I is the current in amperes.
First, we need to convert the voltmeter and ammeter readings to their actual values. We know that the voltmeter reading of 50 V is on a 100 V scale, so the actual voltage is:
50 V / 100 V x Full Scale = 50 V / 100 V x 1000 V = 500 V
Similarly, the ammeter reading of 50 mA is on a 100 mA scale, so the actual current is:
50 mA / 100 mA x Full Scale = 50 mA / 100 mA x 1 A = 0.5 A
Now we can use Ohm's Law to calculate the resistance:
R = V/I = 500 V / 0.5 A = 1000 ohms
To determine the limit of accuracy, we need to consider the errors in the voltmeter and ammeter readings. Since both meters are guaranteed for accuracy within 2% of full scale, the actual readings could be anywhere within ±2% of the full scale values.
For the voltmeter, the possible range of actual readings is:
500 V ± 2% x 1000 V = 500 V ± 20 V
For the ammeter, the possible range of actual readings is:
0.5 A ± 2% x 1 A = 0.5 A ± 0.01 A
Using the extreme values of the actual readings, we can calculate the extreme values of the resistance:
R = (500 V + 20 V) / (0.5 A - 0.01 A) = 1040 ohms
R = (500 V - 20 V) / (0.5 A + 0.01 A) = 960 ohms
Therefore, the limit within which the resistance can be determined is:
960 ohms ≤ R ≤ 1040 ohms
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The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?Correct answer is '80'. Can you explain this answer?
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The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?Correct answer is '80'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?Correct answer is '80'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?Correct answer is '80'. Can you explain this answer?.
The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?Correct answer is '80'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?Correct answer is '80'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The resistance is measured by the voltmeter ammeter method the voltmeter reding is 50 V on 100 V scale and ammeter reading is 50 mA on 100 mA scale. If both the meters are guaranteed for accuracy with 2% of full scale, what is the limit within which resistance (in Ω ) can be measure?Correct answer is '80'. Can you explain this answer?.
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