#### To determine

**To show:** The relative error in calculating *I* is approximately the same as the relative error in *R*. That is, ΔII≈−dRR.

#### Explanation

**Given:**

The current *I* passes through the resistor with resistance *R* and *V* is constant and *R* is measured certain error.

**Proof:**

The ohms law state that the voltage drop is, V=RI.

That is, the current I=VR.

The differential is dI=f′(R)dR.

The derivative of the function f(R)=VR is computed as follows,

f′(R)=ddR(VR)=VddR(1R)=V(−1R−1−1)=−VR2

Substitute f′(R)=−VR2 in dI=f′(R)dR,

dI=(−VR2)dR

The relative error ΔII is computed as follows,

ΔII≈dII=(−VR2)dRVR=(−1R)dR=−dRR

Hence, the relative error in calculating *I* is approximately the same (in magnitude) as the relative error in *R*.