Question

Two resistors of resistances $R_{1} = (100 \pm 3)\Omega$and $R_{2} = (200 \pm 4)\Omega$are connected in parallel. The equivalent resistance of the parallel combination is

• A. $(66.7 \pm 1.8)\Omega$
• B. $(66.7 \pm 4.0)\Omega$
• C. $(66.7 \pm 3.0)\Omega$
• D. $(66.7 \pm 7.0)\Omega$

Answer 1

Answer: (A)

$(66.7 \pm 1.8)\Omega$

Answer 2

Here , $R_{1} = (100 \pm 3)\Omega$

$R_{2} = (200 \pm 4)\Omega$

The equivalent resistance in parallel combinations is

$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{1}{R_{p}} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}$

$R_{p} = \frac{200}{3} = 66.7\Omega$

The error in equivalent resistance is given by

$\frac{\Delta R_{P}}{R_{P}^{2}} = \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}}\Delta R_{p} = \Delta R_{1}\left( \frac{R_{P}}{R_{1}} \right)^{2} + \Delta R_{2}\left( \frac{R_{p}}{R_{2}} \right)^{2} = 3\left( \frac{66.7}{100} \right)^{2} + 4\left( \frac{66.7}{200} \right) = 1.8\Omega$

Hence , the equivalent resistances along with parallel combination is ($66.7 \pm 1.8\Omega$)