Question

How can three resistors of resistances 2 ohm, 3 ohm, and 6 ohm be connected to give a total resistance of A. 4 ohm
B. 1 ohm

Answer 1

$\huge\green{\mathfrak{solution}}$

Given:

R₁=2Ω

R₂=3Ω

R₃=6Ω

Answer:

(a)R₂ and R₃ in parallel with R₁ in series

1/R=1/R₂ + 1/R₃

$\small{⇒1/R=1/3 + 1/6}$

$\small{⇒1/R= 1/2}$

$\small{⇒<strong>R</strong><strong>=</strong><strong>2</strong><strong>Ω</strong>}$

Resistance in series = R + R₁

$\small{⇒2 + 2 Ω}$

$\small{⇒<strong>4</strong><strong>Ω</strong>}$

Hence the net equivalent resistance in the circuit is 4Ω.

(b) When R₁, R₂, R₃ are connected in parallel

1/R=1/R₁ + 1/R₂ + 1/R₃

$\small{⇒1/R=1/2 + 1/3 + 1/6}$

$\small{⇒1/R= 1/2 + 1/2}$

$\small{⇒1/R=1Ω}$

$\small{⇒<strong>R=1Ω</strong>}$

Hence the net equivalent resistance in the circuit is 1Ω

Answer 2

Referred to the attachments.......