Question

Two resistances $R_1=(3.0\pm 0.1)\Omega$ and $R_2=(6.0\pm 2)\Omega$ 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
$R_5=R_1+R_2=3.0+6.0=9.0\Omega$
Error in $R_5=\Delta R_5=(9.0\pm 0.3)\Omega$
Thus choice (a) is correct and choice (b) is wrong. The minimum value is obtained when the resistance are joined in parallel.
$\frac{1}{R_F}=\frac{1}{R_1}+\frac{1}{R_2}$
$\Rightarrow R_F=\frac{R_1{R}_2}{R_1+R_2}=\frac{3.0\times 6.0}{3.0+6.0}=2.0Omega$
Now $R_F=\frac{R_1{R}_2}{R_5}$ $(\because R_1+R_2=R_5)$
$\therefore \frac{\Delta R_F}{R_F}=\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}+\frac{\Delta R_5}{R_5}$
$=\frac{0.1}{3.0}+\frac{0.2}{6.0}+\frac{0.3}{9.0}$
$=0.033+0.033+0.033$
$=0.099\sim 0.1$
$\therefore \Delta R_F=0.1\times R_F=0.1\times 2\Omega =0.2\Omega$
$\therefore$ Minimum value is $(R_F\pm \Delta R_F)=(2.0\pm 0.2)\Omega$
Hence choice (c) is wrong and choice (d) is correct