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) $1 \Omega$ ?
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,
$\frac{1}{\mathrm{R}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1 \therefore \mathrm{R}=1 \Omega$
(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,
$\frac{1}{\mathrm{R}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1 \therefore \mathrm{R}=1 \Omega$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.