Question

Given that two resistance $R_1=(4\pm 0.2)\Omega$ and $R_2=(4\pm 0.3)\Omega$. What will be their equivalent resistance when connected in series and in parallel respectively?

• A. $(8\pm 0.5)\Omega ;(6\pm 18.75$%)
• B. $(8\pm 0.1)\Omega ;(2\pm 18.75$%)
• C. $(8\pm 0.1)\Omega ;(6\pm 18.75$%)
• D. $(8\pm 0.5)\Omega ;(2\pm 18.75$%)

Answer 1

Answer: (D)

$(8\pm 0.5)\Omega ;(2\pm 18.75$%)

$R_s=R_1+R_2=(4\pm 0.2)+(4\pm 0.3)=(8\pm 0.5)\Omega$

$R_p=\frac{R_1{R}_2}{R_1+R_2}=\frac{(4\pm 0.2)(4\pm 0.3)}{(4\pm 0.2)+(4\pm 0.3)}=\frac{16\pm 0.5}{8\pm 0.5}=\frac{16\pm 12.5}{2\pm 6.25}=($2$\pm$ $18.75$%)$\Omega$