Skip to main content

$R_1 = (4\pm 0.8) \Omega $

$R_2 = (4\pm 0.4) \Omega $

$R_{eq} =? $

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:

(A) $(4 \pm 0.4 ) \Omega $
(B) $(2 \pm 0.3 ) \Omega $
(C) $(4 \pm 0.3 ) \Omega $
(D) $(2 \pm 0.4 ) \Omega $

Solution

We have, $R=\frac {R_1 R_2}{R_1 + R_2 } = \frac {4 \times 4}{4+4} = 2 \Omega $

Further, $\frac {1}{R} = \frac {1}{R_1} + \frac {1}{R_2} $

$\therefore \frac {|\Delta R |}{R^2 }=\frac {|\Delta R_1 |}{R_1 ^2 } + \frac {|\Delta R_2 | }{R_2 ^2 }$

$\therefore \frac {|\Delta R |}{2^2 }=\frac {0.8}{4^2 } + \frac {0.4}{4^2 } = \frac {1.2}{16}$

$\therefore |\Delta R| = 0.3 \Omega $

$R_{eq} = R \pm |\Delta R | = 2 \pm 0.3 \Omega $

Answer: (B)