Question

The equivalent resistance across the points A and B in the given circuit is:

• A. 8$\Omega$
• B. 12$\Omega$
• C. 16$\Omega$
• D. 32$\Omega$

Answer 1

The correct option is A 8$\Omega$

Starting from inner loop the between point F and point E resistances $10\Omega$ and $2.5\Omega$ are in parallel connection, their equivalent resistance will be $R_{eq1}=\frac{R_1{R}_2}{R_1+R_2}=\frac{10\times 2.5}{10+2.5}\Omega =2\Omega$ resistance between F and E can be replaced with $2\Omega$
Now resistance between point F and point G and $R_{eq1}$ are in series their equivalent resistance will be $R_{eq2}=10+2\Omega =12\Omega$
In next step resistance between point F and point G and $R_{eq2}$ are in series their equivalent resistance will be $R_{eq3}=10+2\Omega =12\Omega$ resistance between G and D can be replaced with $12\Omega$
The resistance between point G and point E and $R_{eq3}$ are in parallel, their equivalent resistance will be $R_{eq4}=\frac{R_1{R}_2}{R_1+R_2}=\frac{12\times 12}{12+12}\Omega =6\Omega$ resistance between G and E can be replaced with $6\Omega$
for resistances between point C and point G and $R_{eq4}$ are in series their equivalent resistance will be $R_{eq5}=10+6\Omega =16\Omega$ resistance between C and D can be replaced with $16\Omega$
Finally resistance between point B and point E and $R_{eq5}$ are in parallel, their equivalent resistance will be $R_{eq6}=\frac{R_1{R}_2}{R_1+R_2}=\frac{16\times 16}{16+16}\Omega =8\Omega$ resistance between A and B can be replaced with $8\Omega$
Thus eqvivalent resistance between A and B is $8\Omega$ .