Question

A current of $(2.5\pm 0.05)$ A flows through a wire and develops a potential difference of $(10\pm 0.1)$ volt. Resistance of the wire in ohm, is

• A. $4\pm 0.012$
• B. $4\pm 0.4$
• C. $4\pm 0.08$
• D. $4\pm 0.02$

Answer 1

Answer: (A)

,Step 1: Given that:

Current($i$)= $(2.5\pm 0.05)A$

Potential difference($V$)= $(10\pm 0.1)V$

Step 2: Calculation of resistance of the wire:

According to Ohm's law;

Potential difference= Resistance $\times$ Current

That is; $V=iR$

Hence, $R=\frac{V}{i}$

As the given values of current and potential difference consist of the measured values as well as the error in the measurements.

Therefore using only the measured values,

The value of resistance in the wire will be;

$R=\frac{10V}{2.5A}$

$R=4\text{Ω}$

Step 3: Calculation of the error in the value of resistance:

As $R=\frac{V}{i}$ ,

Taking log on both sides we get,

$logR=log\frac{V}{i}$

$logR=logV-logi$

Differentiating both sides , we get,

$\frac{\Delta R}{R}=\frac{\Delta V}{V}-\frac{\Delta i}{i}$

${\left(\frac{\Delta R}{R}\right)}_{max}=\frac{\Delta V}{V}+\frac{\Delta i}{i}$

$\Delta R=R(\frac{\Delta V}{V}+\frac{\Delta i}{i})$

Where, ∆R is the error in the value of R.

Thus, putting the values, we get

$\Delta R=4\text{Ω}(\frac{0.1}{10V}+\frac{0.05}{2.5A})$

$\Delta R=4\times (\frac{1}{100}+\frac{5}{250})$

$\Delta R=4\times (\frac{1}{100}+\frac{1}{50})$

$\Delta R=4\times \left(\frac{1+2}{100}\right)$

$\Delta R=4\times \left(\frac{3}{100}\right)$

$\Delta R=\left(\frac{3}{25}\right)$

$\Delta R=0.12\text{Ω}$

Step 4: Calculation of the net value of resistance:

Thus, the resistance of the wire will be:

$R+\Delta R=(4\pm 0.12)\text{Ω}$

Thus,

Option A) $(4\pm 0.12)ohm$ is the correct option.