Question

In a circuit, potential difference across resistance is $V=(4\pm 0.25)V$ and current in resistance, $I=(1\pm 0.1)A.$ What is the value of resistance with its percentage error?

• A. $4\Omega +16.25\%$
• B. $4\Omega +22.25\%$
• C. $(4\pm 0.4)\Omega$
• D. $4\Omega +18.25\%$

Answer 1

The correct option is A

$4\Omega +16.25\%$

Given:
potential difference$\left(V\right)=(4\pm 0.25)V$
Current $\left(I\right)=(1\pm 0.1)V$
As we know,
$R_0=\frac{V_0}{I_0}=\frac{4}{1}=4\Omega$
Percentage Error in resistance is given by,
$\frac{\Delta R}{R}\times 100=(\frac{\Delta V}{V}+\frac{\Delta I}{I})\times 100\frac{\Delta R}{R}\times 100=(\frac{0.25}{4}+\frac{0.1}{1.0})\times 100=0.1625\times 100\frac{\Delta R}{R}\%=16.25\%$
Therefore,
$R=4\Omega +16.25\%$