Question

Two resistances $R_1=(3.0\pm 0.1)\Omega$ and $R_2=(6.0\pm 0.2)\Omega$ are to be joined together

• A. The maximum resistance obtainable is $(9.0\pm 0.3)\Omega$
  
• B. The maximum resistance obtainable is $(9.0\pm 0.2)\Omega$
  
• C. The minimum resistance obtainable is $(2.0\pm 0.3)\Omega$
  
• D. The minimum resistance obtainable is $(2.0\pm 0.2)\Omega$

Answer 1

Answer: (A), (D)

The correct options are
A

The maximum resistance obtainable is $(9.0\pm 0.3)\Omega$

D

The minimum resistance obtainable is $(2.0\pm 0.2)\Omega$

The maximum value is obtained when the resistances are joined in series. Therefore, the maximum value is

$R_3=R_1+R_2=3.0+6.0=9.0\Omega$
Error in $R_3=△R_3=(9.0\pm 0.3)\Omega$
Thus choice (a) is correct and choice (b) is wrong. The minimum value is obtained when the resistance are joined in parallel.
$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}\Rightarrow R_p=\frac{R_1{R}_2}{R_1+R_2}=\frac{3.0\times 6.0}{3.0+6.0}=2.0\Omega Now,R_p=\frac{R_1{R}_2}{R_3}(\because R_1+R_2=R_3)\therefore \frac{△R_p}{R_p}=\frac{△R_2}{R_2}+\frac{△R_3}{R_3}=\frac{0.1}{3.0}+\frac{0.2}{6.0}+\frac{0.3}{9.0}=0.033+0.033+0.033=0.099\sim 0.1\therefore △R_p=0.1\times R_p=0.1\times 2\Omega =0.2\Omega \therefore Minimumvalueis(R_p\pm △R_p)=(2.0\pm 0.2)\Omega$
Hence, choice (c) is wrong and choice (d) is correct.