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Question: The resistance \(R = \dfrac{V}{I}\) where \(V = \left( {100 \pm 5} \right)V\) and \(I = \left( {10 \...

The resistance R=VIR = \dfrac{V}{I} where V=(100±5)VV = \left( {100 \pm 5} \right)V and I=(10±0.2)AI = \left( {10 \pm 0.2} \right)A. What is the total error in RR?

Explanation

Solution

Here it is asked to find the total error in the resistance. So first using the resistance formula we can find the constant value of the resistance. Now using the division formula of error calculation putting the all known values in the formula we can find the absolute error value of the resistance and from that we can write the total error as evaluated value ±\pm absolute error in resistance.

Formula used:
Dividing formula of error calculation:
ΔRR=ΔVV+ΔII\dfrac{{\Delta R}}{R} = \dfrac{{\Delta V}}{V} + \dfrac{{\Delta I}}{I}
Where, ΔR\Delta R, ΔV\Delta V, ΔI\Delta I are the absolute error values while R,V,IR, V, I are the evaluated values.

Complete step by step answer:
As per the problem we have the resistance R=VIR = \dfrac{V}{I} where V=(100±5)VV = \left( {100 \pm 5} \right)V and I=(10±0.2)AI = \left( {10 \pm 0.2} \right)A.We need to calculate the total error in the resistance R.We know,
V=(100±5)VV = \left( {100 \pm 5} \right)V
I=(10±0.2)A\Rightarrow I = \left( {10 \pm 0.2} \right)A
Now from the above value evaluated value of the voltage and current respectively will be,
V=100VV = 100V and I=10I = 10

Now putting this in the resistance formula we will get the evaluated value of the resistance as,
R=VI10010R = \dfrac{V}{I} \Rightarrow \dfrac{{100}}{{10}}
Hence,
R=10ΩR = 10\Omega
Now using the division formula of error calculation we will get,
ΔRR=ΔVV+ΔII\dfrac{{\Delta R}}{R} = \dfrac{{\Delta V}}{V} + \dfrac{{\Delta I}}{I}
Now putting the given values in the above equation we will get,
ΔR10=5100+0.210 ΔR10=5100+2100\dfrac{{\Delta R}}{{10}} = \dfrac{5}{{100}} + \dfrac{{0.2}}{{10}} \\\ \Rightarrow \dfrac{{\Delta R}}{{10}} = \dfrac{5}{{100}} + \dfrac{2}{{100}}

Now on further solving we will get,
ΔR10=7100 ΔR=710 ΔR=0.7Ω\dfrac{{\Delta R}}{{10}} = \dfrac{7}{{100}} \\\ \Rightarrow \Delta R = \dfrac{7}{{10}} \\\ \Rightarrow \Delta R = 0.7\Omega
Now we can write the total error in the resistance as,
R=(10±0.7)Ω\therefore R = \left( {10 \pm 0.7} \right)\Omega

Hence, the total error in R is R=(10±0.7)ΩR = \left( {10 \pm 0.7} \right)\Omega .

Note: Remember that total error value of a parameter is represented as,
TE=EV±AETE = EV \pm AE
Where, TE = Total Error, EV = Expected value and AE = Absolute error.
An error is an action which is inaccurate or we can say incorrect as it is synonymous with a mistake while measuring. When we are given a physical quantity that is related to two other physical quantities through multiplication or division, then the fractional errors of these two quantities get added up to give us the error in the original physical quantity.