Five resistances of resistance R Ω are there, 3 are connected in parallel and are joined to them in series. Find resultant resistance:

Question

Five resistances of resistance R Ω are there, 3 are connected in parallel and are joined to them in series. Find resultant resistance: (A)  ( (3/7) )R Ω  (B)  ( (7/3) )R Ω  (C)  ( (7/8) )R Ω  (D)  ( (8/7) )R Ω

Tardigrade Admin · Accepted Answer

(1/RP) =(1/R)+(1/R)+(1/R)=(3/R) ⇒ RP =(R/3) Ω <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0d0235ad4639c17251885d9d918ecd37-.png" /> R5 =R+R=2 R Ω R text net =RP+RS =2 R+(R/3)=(7 R/3) Ω

Q. Five resistances of resistance $R Ω$ are there, $3$ are connected in parallel and are joined to them in series. Find resultant resistance:

$(\frac{3}{7}) R Ω$

$(\frac{7}{3}) R Ω$

$(\frac{7}{8}) R Ω$

$(\frac{8}{7}) R Ω$

$\frac{1}{R _{P}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}$
$\Rightarrow R_{P} = \frac{R}{3} Ω$

$R_{5} = R + R = 2 R Ω$
$R_{net} = R_{P} + R_{S}$
$= 2 R + \frac{R}{3} = \frac{7 R}{3} Ω$

Q. Five resistances of resistance RΩ are there, 3 are connected in parallel and are joined to them in series. Find resultant resistance:

(73​)RΩ

(37​)RΩ

(87​)RΩ

(78​)RΩ