Explanation:

The formula connecting Potential difference(v), Current (i) and resistance (R) is given by Ohm's law,.

which is V = i * R

Applying log on both sides ,

=> log(V) - log(i) = log(R)

[Since, Log(a) - log(b) = log(a/b)]

Representing error with ∆,.

So, Relative errors will be of this form,

(∆R/R) = (∆i/i) + (∆V/V)

Why (+)? , It's because in error calculation you always add the error to get the max error possible.

Give in the question that,

V = 100

∆V = ±5 = 5

i = 10

∆i = 0.2

=> Using the relation above,

(5/100) + (0.2/10) = (∆R/R)

=> 0.07 = ∆R/R

=> ∆R/R = 0.07

=> Percentage error = ∆R/R% = 7%

But if you want the accurate error or minimum error possible, it will be

0.05 - 0.02 = 0.03 => %Error = 3%

Everytime when someone asks for an error, One must report the maximum possible.

Resistance = V/i = 100/10 = 10,

=> Resistance will be 10±∆R

∆R = 0.07*10 = 0.7

=> Resistance = 10±0.7

Therefore,

Percentage error = 7%

Resistance = 10±0.7

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