Question

Show how you would connect three resistors, each of resistance $6\mathrm{\Omega }$, so that the combination has a resistance of $4\mathrm{\Omega }$.

Answer 1

Step 1: Given data

We have three resistors who have a resistance of $6\mathrm{\Omega }$ each.

Step 2: Formulas used

When resistors are in series, the equivalent resistance is given as,

$R_{eq}=R_1+R_2+...$

When the resistors are in parallel, the equivalent resistance is given as,

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...$

Step 3: Determine the combination of resistors

We have three resistors of $6\mathrm{\Omega }$. Connecting two of them in series results in $6+6=12\mathrm{\Omega }$.

If we connect the above combination in parallel to the other resistor, we will have an equivalent resistance of,

$\begin{array}{cc}& \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{12}+\frac{1}{6}\\ \Rightarrow & \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1+2}{12}\\ \Rightarrow & {\mathrm{R}}_{\mathrm{eq}}=\frac{12}{3}\\ \Rightarrow & {\mathrm{R}}_{\mathrm{eq}}=4\mathrm{\Omega }\end{array}$

The overall combination will look like the following,

Therefore, when two of the given resistors are in series and that combination is in parallel to the remaining one, the overall resistance of the combination will be $4\mathrm{\Omega }$.