Question

The values of two resistors are (5.0+0.2) and (10.0+0.1). What is the percentage error in the equivalence resistance when connected in parallel?

Answer 1

Answer:

The percentage error in the equivalence resistance when connected in a parallel combination is 7%.

Explanation:

When the resistors are connected in parallel their equivalent resistance is given as,

$R=\frac{R_1R_2}{R_1+R_2}$          (1)

Where,

R=equivalent resistance

R₁ and R₂ are individual resistance

And the percentage error in the parallel combination is given as,

$\frac{\Delta R}{R}=(\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}+\frac{\Delta R_1+\Delta R_2}{R_1+R_2})\times 100$        (2)

And from the question we have,

$R_1=(5\pm0.2)\Omega$         (3)

$R_2=(10\pm0.1)\Omega$       (4)

By substituting the required values in equation (2) we get;

$\frac{\Delta R}{R}=(\frac{0.2}{5}+\frac{0.1}{10}+\frac{0.2+0.1}{10+5})\times 100$

$\frac{\Delta R}{R}=(0.04+0.01+0.02)\times 100$

$\frac{\Delta R}{R}=0.07\times 100=7\%$

Hence, the percentage error in the equivalence resistance when connected in a parallel combination is 7%.