Question

You're given 4 resistances as $\sf {4\Omega ,8\Omega ,12\Omega ,24\Omega.}$
If we combined all 4 resistances.Then find:-

$\sf{:}\longrightarrow Highest \:Resistance$
$\sf{:}\longrightarrow Lowest\: Resistance$

Note:-
Briefly Explain your answer in 60+words.

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Answer 1

$\huge\fbox\green{Given}$

$\sf R_{1}$ = 4 Ω

$\sf R_{2}$ = 8 Ω

$\sf R_{3}$ = 12 Ω

$\sf R_{4}$ = 24 Ω

$\huge\fbox\green{To\: find}$

• Highest resistance
• Lowest resistance

$\huge\fbox\green{Concept}$

• When resistors are connected in series the overall resistance is the highest

• when the resistors are connected in parallel the overall resistance is the lowest

$\huge\fbox\green{Formula}$

For highest resistance:-

:$\implies$ $\underline{\boxed{\sf R = R_{1} + R_{2} + R_{3} + R_{4} ... }}$

for lowest resistance:-

:$\implies$ $\underline{\boxed{\sf \dfrac{1}{R} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + \dfrac{1}{R_{4}} ... }}$

$\huge\fbox\green{Solution}$

highest resistance:-

:$\implies$ $\sf R = 4+8+12+24$

:$\implies$ $\sf R = 48 Ω$

Lowest resistance:-

:$\implies$ R = $\sf\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{24}$

:$\implies$ $\sf\dfrac{6+3+2+1}{24}$

:$\implies$ $\sf\cancel\dfrac{12}{24}$ = $\sf\dfrac{1}{2}$

:$\implies$ R = 2 Ω

___________________________

Answer 2

$\huge \mathbb {ANSWER:-}$

$\huge\colorbox{goldenrod}{given}$

$\sf R_{1} =4ᘯ$

$\sf R_{2} =8ᘯ$

$\sf R_{3} =12ᘯ$

$\sf R_{4} =24ᘯ$

$\huge \colorbox{goldenrod}{to \: find}$

• Highest Resistance
• Lowest Resistance

$\huge \colorbox{goldenrod}{formula}$

Highest Resistance:-

$\implies \boxed {R = R_{1} + R_{2} + R_{3} + R_{4}}$

Lowest Resistance:-

$\implies \boxed{\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} }$

Finding Highest Resistance:-

$\leadsto \sf{R= 4 + 8 + 12 + 24}$

$\leadsto \sf{R = 48ᘯ}$

Finding Lowest Resistance:-

$\leadsto \sf{R = \frac{1}{4} + \frac{1}{8} } + \frac{1}{12} + \frac{1}{24}$

$\leadsto \sf{R = \frac{6 + 6 + 2 + 1}{24} }$

$\leadsto \sf{R= \frac{12}{24} = \frac{1}{2} }$

$\leadsto \sf{R = 2ᘯ}$