Question

Three resistance of 4,8and 24 Ohms are given.How will they be joined to get a resistance 8 Ohm ?show the line reasoning .​

Answer 1

Answer:

Given:

3 resistors have been given. The value of resistances are

• 4 ohm
• 8 ohm
• 24 ohm

To find:

Combination of the resistances such that equivalent resistance becomes 8 ohm.

Diagram:

Please refer to the attached photo to understand better.

Calculation:

First we shall connect 4 ohm and 8 ohm in series. Then these will be connected is parallel with 24 ohm.

1st combination is in series:

R' = R1 + R2 = 4 + 8 = 12 ohm

2nd combination is in parallel:

1/R eq. = 1/R' + 1/24

=> 1/R eq. = 1/12 + 1/24

=> 1/R eq. = 1/8

=> R eq. = 8 ohm.

So final equivalent resistance is 8 ohm.

Answer 2

$\large{\underline{\underline{\mathfrak{\sf{\blue{Explanation-}}}}}}$

$\begin{lgathered}\bold{Given\:resistances} \begin{cases}\sf{4 \Onega} \\ \sf{8 \Omega}\\ \sf{24 \Omega}\end{cases}\end{lgathered}$

$\orange{\boxed{\pink{\underline{\red{\mathfrak{To\:find-}}}}}}$

• We have to combine the given resistances such that the equivalent resistances is of 8 Ω.

$\orange{\boxed{\pink{\underline{\red{\mathfrak{Concept\:used-}}}}}}$

• To find net resistance in series combination :

Formula : $R_t$ = $R_1\:+\:R_2$ ...

$\begin{lgathered}\bold{Where} \begin{cases}\sf{\green{R_t\:refers\:to\:net\:resistance}} \\ \sf{\purple{R_1\:refers\:to\:resistance\:of\:first\:resistor}}\\ \sf{\color{pink}{R_2\:refers\:to\:resistance\:of\:second\:resistor}}\end{cases}\end{lgathered}$

• To find net resistance in parallel combination:

Formula : $\implies$ $\dfrac{1}{R_t}$ = $\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\begin{lgathered}\bold{Where} \begin{cases}\sf{\green{R_t\:refers\:to\:net\:resistance}} \\ \sf{\purple{R_1\:refers\:to\:resistance\:of\:first\:resistor}}\\ \sf{\color{pink}{R_2\:refers\:to\:resistance\:of\:second\:resistor}}\end{cases}\end{lgathered}$

$\orange{\boxed{\pink{\underline{\red{\mathfrak{Solution-}}}}}}$

$\huge{\underline{\boxed{\sf{\pink{Case-1}}}}}$

$R_t$ = $R_1\:+\:R_2$

Let the resistor of resistance 4 Ω and 8 Ω in series, we get,

$\implies$ $R_t$ = $4\:+\:8$

$\implies$ $R_t$ = $\sf{12 \Omega}$

$\rule{200}2$

$\huge{\underline{\boxed{\sf{\pink{Case-2}}}}}$

$\dfrac{1}{R_t}$ = $\dfrac{1}{R_1}+\dfrac{1}{R_2}$

Now take the resistor of resistance 12 Ω and 24 Ω in parallel, we get,

$\implies$ $\dfrac{1}{R_t}$ = $\dfrac{1}{12}+\dfrac{1}{24}$

Taking LCM,

$\implies$ $\dfrac{1}{R_t}$ = $\dfrac{2+1}{24}$

$\implies$ $\dfrac{1}{R_t}$ = $\cancel{\dfrac{3}{24}}$

$\dfrac{1}{R_t}$ = $\dfrac{1}{8}$

Hence, we get the equivalent resistance is of 8 Ω.