Question

Two resistors $R_1=(3.0\pm 0.1)\Omega$ and $R_2=(6\pm 0.5)\Omega$ are connected in parallel. The resistance of the combination is :

• A. $(2.0\pm 0.4)\Omega$
• B. $(2.0\pm 0.1)\Omega$
• C. $(2.0\pm 0.2)\Omega$
• D. $(2.00\pm 0.4)\Omega$

Answer 1

Answer: (B)

$(2.0\pm 0.1)\Omega$

Let $R$ is the equivalent resistance for parallel combination and $\Delta R$ is the error(in its measurement respective to the error in the values of resistors)

Hence,

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R}$

$\frac{1}{R}=\frac{R_1{R}_2}{R_1+R_2}$

$R=\frac{3\times 6}{3+6}$

$\therefore R=2\Omega$

Now,

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

$\frac{\Delta R}{R^2}=\frac{0.1}{(3{)}^2}+\frac{0.5}{(6{)}^2}$

$\frac{\Delta R}{R^2}=\frac{0.1}{4}$

$\Delta R=0.1$

Hence the resistance of the combination is

$R=(2.00\pm 0.1)\Omega$