Question 294: Current Electricity

Question

Two resistors R
1
= (4 $\pm$ 0.8) $\Omega$ and R
2
= (4 $\pm$ 0.4) $\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be :

Accepted Answer

Given,
R
1
= (4 $\pm$ 0.8) $\Omega$
R
2
= (4 $\pm$ 0.4) $\Omega$
Equivalent resistance when the resistors are connected in parallel is given by
${1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} \Rightarrow {1 \over {{R_{eq}}}} = {1 \over 4} + {1 \over 4}$
${R_{eq}} = 2\,\Omega $
Now, ${{\Delta {R_{eq}}} \over {R_{eq}^2}} = {{\Delta {R_1}} \over {R_1^2}} + {{\Delta {R_2}} \over {R_2^2}}$
Substituting the values in the above equation, we get
${{\Delta {R_{eq}}} \over 4} = {{0.8} \over {16}} + {{0.4} \over {16}} \Rightarrow \Delta {R_{eq}} = 0.3\,\Omega $
$	herefore$ The equivalent resistance in parallel combination is ${R_{eq}} = (2 \pm 0.3)\Omega $.

Answer

(4 $\pm$ 0.4) $\Omega$

Answer

(2 $\pm$ 0.4) $\Omega$

Answer

(2 $\pm$ 0.3) $\Omega$

Answer

(4 $\pm$ 0.3) $\Omega$

Current Electricity

Two resistors R 1 = (4 \pm 0.8) \Omega and R 2 = (4 \pm 0.4) \Omega are connected in parallel. The equivalent resistance of their parallel combination will be :

Choose the correct answer: