Question

Show how you would connect three resistors, each of resistance $6\Omega$, so that the combination has a resistance of
(i) $9\Omega$,
(ii) $4\Omega$.

Answer 1

If resistor are connected in series, the equivalent resistance is given by,

$6\Omega +6\Omega +6\Omega =18\Omega$. So this is not correct.

When they are connected in parallel, the equivalent resistance is given by,

$\frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=2\Omega$. This is also not correct.

(i) When two $6\Omega$ are connected in parallel, the equivalent resistance is given by,

$\frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{6}$

$R_{eq}=\frac{6\times 6}{6+6}=3\Omega$

If the $3^{rd}$ resistor of $6\Omega$ and $R_{eq}$ are connected in series, it becomes $6\Omega +3\Omega =9\Omega$.

(ii) When two $6\Omega$ are connected in series, the equivalent resistance is given by,

$R=6\Omega +6\Omega =12\Omega$

If $3^{rd}$ resistor of $6\Omega$ and R is connected in parallel, then

$\frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{12}$

$R_{eq}=\frac{6\times 12}{6+12}=4\Omega$.