Answer to Question #116210 in Electric Circuits for please

Question

[img]https://upload.cc/i1/2020/05/09/dwlmCx.jpg[/img]



R=4

Answer 1

Explanations & Calculations

• For the ease of calculations start from the right & workout to left.
• R+1 = 4+1 =5$\small \Omega$
• Equivalent resistance of a set of parallel connected resistors is given by,

$\qquad\qquad \frac{1}{R_{eq}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots$ , while it is reduced to

$\qquad\qquad R_{eq} = \frac{R_1\times R_2}{R_1+R_2}$ for 2 resistors.

• When those (n number of resistors) are identical, $R_{eq} = \frac{R}{n}$
• Equivalent resistance of a set of series connected resistors is given by,

$\qquad\qquad R_{eq} = R_1+R_2+R_3+ \cdots$

1) Consider LMCD section,

• 7$\small \Omega$ & 2$\small \Omega$ resistors are in series which gives (7+2=) 9$\small \Omega$ . Now all those 9 ;this one ,one in between BC & the one between MC are in parallel which gives the equivalent of 9/3 = 3$\small \Omega$ .
• Now this 3, other 3 & the 2 at the top are in series which gives an equivalent of 2+3+3 = 8$\small \Omega$ .

2) Now leave it there & consider 6 & 2 in between nodes EL.

• They are in series which gives 6+2 = 8$\small \Omega$ .

3) Now consider the arrangement in between nodes GH.

• Vertical 5 & horizontal 3 are in series, so equals to (5+3 =) 8 $\small \Omega$ which gives $\big(\frac{8\times5}{8+5} =\big) \frac{40}{13}\Omega$ with the adjacent 5 as the are parallel.
• Now that is in series with those—1$\small \Omega$ & 5$\small \Omega$ —above (between HI) and gives an equivalent of $\frac{40}{13}\Omega$ +1 + 5 = $\bold{\frac{118}{13}\Omega}$ .

4) Now consider the arrangement between JK.

• Vertical & horizontal 5 are in series equals to (5+5=) 10$\small \Omega$ which gives an equivalent of $\big(\frac{10\times 5}{10+5} =\big) \frac{10}{3}\Omega$ with the adjacent 5 as they are parallel.
• Now this and the 5 in between KL are in parallel which gives $\big(\frac{\frac{10}{3}\times 5}{\frac{10}{3}+5}\big) = 2\Omega$
• Now that 2 & the 1 in between KG are in series which gives (2+1 =) 3$\small \Omega$ .

Now the simplified arrangement is as follows.

5) Now I& L nodes are equivi-potential and G,F,E,D points are also equivi-potential which set all the resistors within I,L,D,G in a parallel combination. Now the equivalent of that is

$\qquad\qquad \begin{aligned} \small \frac{1}{R_{eq1}} &= \small \frac{1}{8}+\frac{1}{8}+\frac{1}{3}+\frac{1}{3}+\frac{13}{118}\\ \small R_{eq} &= \small \bold{0.9739\Omega} \end{aligned}$

6) Now required equivalent resistance of the system can be calculated to be 7$\small \Omega$ + 0.9739$\small \Omega$ = 7.9739$\small \Omega$ as they are in series.