Question

Question 11 (ii)
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of
$\left(ii\right)4\Omega$?

Answer 1

(b) Two resistors in series

Two $6\Omega$ resistors are in series. Their equivalent resistance will be the sum $6\Omega +6\Omega =12\Omega$.
The third $6\Omega$ resistor is in parallel with $12\Omega$. Hence, equivalent resistance will be
$R=\frac{1}{\frac{1}{12}+\frac{1}{6}}=\frac{12\times 6}{12+6}=4\Omega$
Therefore, the total resistance is $4\Omega$.