Question

Find equivalent resistance between A and B

Answer 1

Answer:-

$\Large\boxed{\quad\sf{15\ \Omega}\quad}$

Solution:-

The three resistors of $\sf{2\ \Omega}$ at right are connected in series. Let them be replaced by another resistor having resistance,

$\longrightarrow\sf{R=(2+2+2)\ \Omega}$

$\longrightarrow\sf{R=6\ \Omega}$

Now the two resistors of $\sf{6\ \Omega}$ are connected in parallel. Let them be replaced by another resistor of resistance,

$\longrightarrow\sf{R=\left(\dfrac{1}{6}+\dfrac{1}{6}\right)^{-1}\ \Omega}$

$\longrightarrow\sf{R=\left(\dfrac{2}{6}\right)^{-1}\ \Omega}$

$\longrightarrow\sf{R=\left(\dfrac{1}{3}\right)^{-1}}$

$\longrightarrow\sf{R=3\ \Omega}$

Now the two resistors of $\sf{3\ \Omega}$ and the $\sf{4\ \Omega}$ resistor are connected in series. Let them be replaced by a resistor of resistance,

$\longrightarrow\sf{R=(3+3+4)\ \Omega}$

$\longrightarrow\sf{R=10\ \Omega}$

Now the two resistors of $\sf{10\ \Omega}$ are connected in parallel. Let them be replaced by another resistor of resistance,

$\longrightarrow\sf{R=\left(\dfrac{1}{10}+\dfrac{1}{10}\right)^{-1}\ \Omega}$

$\longrightarrow\sf{R=\left(\dfrac{2}{10}\right)^{-1}\ \Omega}$

$\longrightarrow\sf{R=\left(\dfrac{1}{5}\right)^{-1}\ \Omega}$

$\longrightarrow\sf{R=5\ \Omega}$

Now there exists only three resistors of resistance $\sf{5\ \Omega}$ each, connected in series. Hence the equivalent resistance between A and B is,

$\longrightarrow\sf{R=(5+5+5)\ \Omega}$

$\longrightarrow\sf{\underline{\underline{R=15\ \Omega}}}$