Question

22-IF THREE RESISTORS 2,4,8 OHM ARE COMBINED IN PARALLEL COMBINATION THEN FIND THEIR NET

REDISTANCE.
​

Answer 1

Answer:

1.14285

Explanation:

1/R = 1/R1+1/R2+1/R3

1/R= 1/2+1/4+1/8

1/R = 0.875

R = 1/0.875

R= 1.14285

Answer 2

» Given:

resistance of three resistors = 2Ω , 4Ω , 8Ω

they are in series connection.

» We know that,

in series connection,

${\pink{\boxed{R_{eq} = R_1+R_2+R_3}}}\\ \\ \\ \bf Solution:\\ \\ \\R_1 = 2\Omega\\ \\ \\ R_2 = 4\Omega\\ \\ \\R_3 = 8\Omega\\ \\ \\ R_{eq} = 2+4+8\\ \\ \\ R_{eq} = 14\Omega$

» For parallel connection -

in parallel connection,

${\pink{\boxed{\frac{1}{R_{eq}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}}}\\ \\ \\ \bf Solution:\\ \\ \\R_1 = 2\Omega\\ \\ \\ R_2 = 4\Omega\\ \\ \\R_3 = 8\Omega\\ \\ \\ \frac{1}{R_{eq}} = \frac{1}{2}+\frac{1}{4}+\frac{1}{8} \\ \\ \\ = \frac{4+2+1}{8}\\ \\ \\ = \frac{7}{8}\\ \\ \\ \therefore\:R_{eq} = \frac{8}{7}\\ \\ \\ {\purple{\boxed{R_{eq} = 1.1428 \Omega}}}$