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Question

Question: In the experiment of verification of Ohm's law the error in the current measurement is \( 1\% \) , w...

In the experiment of verification of Ohm's law the error in the current measurement is 1%1\% , while that in the voltage measurement is 2%2\% . The error in the resistance has a maximum value of
A. 1%1\%
B. 2%2\%
C. 3%3\%
D. None of these

Explanation

Solution

Hint
In the question, the value of error in current and voltage is given in terms of percentage. By using the equation of the ohm’s law we calculate the value of the error in the resistance. In the equation all the values will be given in terms of the error, so we wrote the values in terms of percentage.
The expression for finding the error in the resistance is -
R=VIR = \dfrac{V}{I}
Where, RR be the resistance, VV be the voltage and II be the current.

Complete step by step answer
Given that,
Error in Current measurement, ΔII×100=1%\dfrac{{\Delta I}}{I} \times 100 = 1\%
Error in voltage measurement, ΔVV×100=2%\dfrac{{\Delta V}}{V} \times 100 = 2\%
Error in resistance measurement, ΔRR×100\dfrac{{\Delta R}}{R} \times 100 .
Now we find the value of error measurement in resistance,
So that
ΔRR×100=ΔII×100+ΔVV×100\dfrac{{\Delta R}}{R} \times 100 = \dfrac{{\Delta I}}{I} \times 100 + \dfrac{{\Delta V}}{V} \times 100
Substitute the value of error in current and the voltage in the above equation, we get,
ΔRR×100=1+2\dfrac{{\Delta R}}{R} \times 100 = 1 + 2
Perform the arithmetic operation in the above equation, we get
ΔRR×100=3%\dfrac{{\Delta R}}{R} \times 100 = 3\%
Therefore, the error in the resistance has a maximum of 3%.3\% .
Hence, from the above option, option (C) is correct.

Note
In the question, error of resistance is unknown so change the equation of ohm’s law in terms of resistance. In this case, Voltage is unknown; we convert the equation of ohm’s law in terms of voltage. According to the ohm’s law, current through the conductor is directly proportional to the voltage across the points.