How can three resistors of resistances
Question.
How can three resistors of resistances $2 \Omega, 3 \Omega$, and $6 \Omega$ be connected to give a total resistance of

(a) $4 \Omega$

(b) $1 \Omega$ ?

solution:

(a) To get a total resistance of $4 \Omega$ from resistors of resistances $2 \Omega, 3 \Omega$ and $6 \Omega$, the resistors are joined as shown below.

The resistors having resistances $3 \Omega$ and $6 \Omega$ are connected in parallel. This combination is connected in series with the resistor of resistance $2 \Omega$. Let us check it mathematically, equivalent resistance of $3 \Omega$ and $6 \Omega$ resistors is,

$R_{1}=\frac{3 \times 6}{3+6}=\frac{3 \times 6}{9}=2 \Omega$

Now, $R_{1}$ and $2 \Omega$ resistors are in series, their equivalent resistance is $R_{e}=R_{1}+2=2+2=4 \Omega$.

(b) To get a resistance of $1 \Omega$ from three given resistors of resistances $2 \Omega, 3 \Omega, 6 \Omega$, are joined as shown below.

They all are connected in parallel. Their equivalent resistance is given by,

$\frac{1}{\mathrm{R}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1 \therefore \mathrm{R}=1 \Omega$
Editor