Question

How can three resistors of resistance 2Ω, 3Ω and 6Ω be connected to give a total resistance of 4Ω and 1Ω?

Answer 1

Step 1: Given

$$\begin{gathered} R1=2\Omega \\ R2=3\Omega \\ R3=6\Omega \end{gathered}$$

Step 2: Combining resistors

Considering the second and third resistors connected in parallel, the effective resistance becomes

$$\begin{gathered} \frac{1}{R}=\frac{1}{R2}+\frac{1}{R3}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \\ R=2\Omega \end{gathered}$$

Now connecting $R$and $R1$in series total resistance becomes

$R+R1=4\Omega$

Step 3: Equivalent Resistance

Now considering all three resistance in parallel, the total resistance becomes

$\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$

$R=1\Omega$

Thus, By connecting the $2\Omega$ resistances in series with the parallel combination of $3\Omega$ and $6\Omega$, we can obtain a total resistance of $4\Omega$ and By connecting resistors $2\Omega$, $3\Omega$, and $6\Omega$ in parallel, we can produce a total resistance of $1\Omega$.