Question

please give the correct answer with explanation.....
Need help...​

Answer 1

Answer:

(4) 3Ω

Explanation:

all the 2Ω resistor are connected in series

and there resultant is in parallel to 4Ω resistor

therefore, resultant of 2Ω resistor = (2+2+2+2+2+2) = 12Ω

now 12Ω & 4Ω are in parallel

let R be there equivalent resistance

(1/R) = (1/4)+(1/12)

taking LCM on RHS

(1/R) = { (1×3) + (1×1) } ÷ (12)

(1/R) = ( 3+1)÷12

(1/R) = 4/12 = 1/3

=> R = 3Ω

Answer 2

☆ We are given a circuit have resistors having values 4Ω , 2Ω , 2Ω , 2Ω , 2Ω , 2Ω , 2Ω .

Find the attachment for entire scenario .

Stretch the wire having resistances 2Ω .

Circuit will be modified as 2nd figure in the attachment .

Make sure to know about potential method , which denotes , ' When there is no resistance in the wire , potential difference passing through the wire will be zero (0) '

2Ω , 2Ω , 2Ω , 2Ω , 2Ω , 2Ω are connected in series ,

Let it be R$\sf _s$

➠ $\sf R_s=2+2+2+2+2+2$

➠ $\textsf{\textbf{\pink{R _{\text{s}} \ =\ 12\ }}}\Omega\ \; \bigstar$

R$\sf _s$ , 4Ω are connected in parallel ,

$:\implies \sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_s}+\dfrac{1}{4}$

$:\implies \sf \dfrac{1}{R_{eq}}=\dfrac{1}{12}+\dfrac{1}{4}$

$:\implies \sf \dfrac{1}{R_{eq}}=\dfrac{1+3}{12}$

$:\implies \sf \dfrac{1}{R_{eq}}=\dfrac{4}{12}$

$:\implies \sf \dfrac{1}{R_{eq}}=\dfrac{1}{3}$

$:\implies \textsf{\textbf{\green{R _{\text{eq}} \ =\ 3\ }}}\Omega\ \; \bigstar$

Option 4