Physics, asked by nikita197465, 1 year ago

The current flowing through a resistor connected in a electrical circuit and the potential difference developed across its ends are shown in the given ammeter and voltmeter find the least count of both what is the voltage and current across the given resistor

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Answered by BrainlyBotv100
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Answer: First of all we must find the least count of the ammeter and voltmeter. Least count is the smallest value that can be measured by the measuring instrument.


Least count = Range/Number of n divisions


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Least count of ammeter = 1/50 = 0.02A Least count of voltmeter = 3/30 = 0.1V

The next part of the question asks us to find the current and potential difference/voltage passing through the resistor.

In this question,

Current = Value shown by ammeter × Least count = 15 × 0.02A = 0.3A

Voltage/Potential Difference = Value shown by voltmeter × Least count = 21 × 0.1V = 2.1V


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wikremsinghp4v470: amazingly explained the first part but i could not understand how you found out the current and voltage
wikremsinghp4v470: why dont we directly write the readings in the ammeter
wikremsinghp4v470: as current
BrainlyBotv100: Yes, that is the way we usually do it, but when we look at the readings, it shows the range and we cannot directly conclude the value shown by the measuring instrument. So to find the values and prove it, we find it numerically.
khokharakg: Thanks a lot...... It was really helpful......amazingly explained..
sakshisharma75344: PLZ explain the second part of ques
sakshisharma75344: plz answer fast.....
Answered by abhi178
42

Least count : the smallest reading which can be accurately measured with a Vernier caliper, screw guage or any device is called its least count.

Least can be measured by using formula,

\textbf{Least count}=\frac{\textbf{Range}}{\textbf{number of divisions}}

for Ammeter,

least count of Ammeter = Range/number of divisions

Range = 0.4 - 0.2 = 0.2A

number of divisions = 10

so, least count of Ammeter = 0.2/10 = 0.02

here, main scale = 0.2A , n = 5 and least count = 0.02

reading of Ammeter = main scale + n × least count of Ammeter

= 0.2 + 5 × 0.02 = 0.2 + 0.1 = 0.3A

for voltmeter,

range = 2 - 1 = 1V

number of divisions = 10

so, least count of voltmeter = 1/10 = 0.1

here, main scale = 2V, n = 1 and least count of voltmeter = 0.1

reading of voltmeter = main scale + n × least count of voltmeter

= 2 + 1 × 0.1 = 2.1V

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