Question

The resistance between points A and B in the circuit shown in figure is:

• A. $4\Omega$
• B. $6\Omega$
• C. $10\Omega$
• D. $8\Omega$

Answer 1

Answer: (C)

$10\Omega$

In this case, the 8 ohms and 4 ohms resistances at the extreme right are connected in series. So their combined resistance is 8+4 = 12 ohms.
The 12 ohms and adjacent 6 ohms resistance are connected in parallel. So the combined resistance is $\frac{1}{\frac{1}{6}+\frac{1}{12}}=\frac{1}{0.25}=4ohm$
The two 8 ohms resistances at the bottom are connected in parallel. so, the combined resistance is $\frac{1}{\frac{1}{8}+\frac{1}{8}}=\frac{1}{0.25}=4ohm$
This combined 4 ohms resistance is again connected in parallel with another 4 ohms resistance at the bottom end. So the total resistance at this point is $\frac{1}{\frac{1}{4}+\frac{1}{4}}=\frac{1}{0.5}=2ohm$
Now, all the combined resistances are connected in series.
Therefore, the total resistance in the circuit is $4+4+2=10$ ohms.