Question

a) Two 8kΩ resistors are connected in parallel. Calculate the equivalent
resistance,
b) Two 8kΩ resistors are connected in series. Calculate the equivalent resistance
I need answers with steps correctly.
pls ​

Answer 1

Given

• We here have two cases where in case 1 they are connected in series and in case 2 it is connected in parallel
• R₁ = R₂ = 8 kΩ

To Find

• Equivalent resistance

Solution

☯ The equivalent resistance of a series combination is given by, Rₙₑₜ = R₁ + R₂ ... + Rₙ

☯ The equivalent resistance of a parallel combination is given by, 1/Rₙₑₜ = 1/R₁ + 1/R₂ ... + 1/Rₙ

━━━━━━━━━━━━━━━━━━━━━━━━━

✭ Equivalent Resistance (Series) :

→ Rₙₑₜ = R₁ + R₂ ... + Rₙ

• R₁ = 8 kΩ
• R₂ = 8 kΩ

→ Rₙₑₜ = 8 + 8

→ Rₙₑₜ = 16 kΩ

∴ The equivalent Resistance of the series combination is 16 kΩ

━━━━━━━━━━━━━━━━━━━━━━━━━

✭ Equivalent Resistance (Parallel) :

→ 1/Rₙₑₜ = 1/R₁ + 1/R₂ ... + 1/Rₙ

• R₁ = 8 kΩ
• R₂ = 8 kΩ

→ 1/Rₙₑₜ = 1/8 + 1/8

→ 1/Rₙₑₜ = (1+1)/8

→ 1/Rₙₑₜ = 2/8

→ Rₙₑₜ = 8/2

→ Rₙₑₜ = 4 kΩ

∴ The equivalent Resistance of the Parallel combination is 4 kΩ

Answer 2

$\Large\bf\pink{GiVeN,}$

$\:\:\:\:\:\:\:\red\bullet\:\:\rm{a)\:\:Two\:8k\Omega\:resistors\:are\:connected\:in\:parallel\:.}$

$\:\:\:\:\:\:\:\orange\bullet\:\:\rm{b)\:\:Two\:8k\Omega\:resistors\:are\:connected\:in\:series\:.}$

$\Large\bf\blue{CoNcEpT,}$

☆ When two resistors $\Big[\:\bf{i.e.\:R_1\:\:and\:\:R_2}\:\Big]\:$ are connected in series, then the equivalent resistance is

$\purple\bigstar\:\:\bf{\color{indigo}R_{eq}\:=\:R_1\:+\:R_2}$

$\bf\red{And,}$

☆ When two resistors $\Big[\:\bf{i.e.\:R_1\:\:and\:\:R_2}\:\Big]\:$ are connected in parallel, then the equivalent resistance is

$\blue\bigstar\:\:\bf\purple{\dfrac{1}{R_{eq}}\:=\:\dfrac{1}{R_1}\:+\:\dfrac{1}{R_2}}$

$\Large\bf\orange{CaLcUlAtIoN,}$

a)

• $\bf{\red{R_1}}\:=\:8k\Omega\:$

• $\bf{\red{R_2}}\:=\:8k\Omega\:$

■ According to the concept,

$\longmapsto\:\:\bf{\dfrac{1}{R_{eq}}\:=\:\dfrac{1}{R_1}\:+\:\dfrac{1}{R_2}}$

$\longmapsto\:\:\rm{\dfrac{1}{R_{eq}}\:=\:\dfrac{1}{8}\:+\:\dfrac{1}{8}}$

$\longmapsto\:\:\rm{\dfrac{1}{R_{eq}}\:=\:\dfrac{1\:+\:1}{8}\:}$

$\longmapsto\:\:\rm{\dfrac{1}{R_{eq}}\:=\:\dfrac{2}{8}\:}$

$\longmapsto\:\:\rm{R_{eq}\:=\:\dfrac{8}{2}\:}$

$\longmapsto\:\:\bf\green{R_{eq}\:=\:4k\Omega\:}$

b)

• $\bf{\red{R_1}}\:=\:8k\Omega\:$

• $\bf{\red{R_2}}\:=\:8k\Omega\:$

■ According to the concept,

$\longmapsto\:\:\bf{R_{eq}\:=\:R_1\:+\:R_2}$

$\longmapsto\:\:\rm{R_{eq}\:=\:8\:+\:8}$

$\longmapsto\:\:\bf\green{R_{eq}\:=\:16k\Omega}$