Question

find the equivalent resistance between AB​

Answer 1

Answer:

70/ 17

Explanation:

hope the answer will help you

Answer 2

Answer:-

$\Large\boxed{\sf{\dfrac{70}{17}\ \Omega}}$

Solution:-

• $\sf{R_{AD}=2\Omega}$

• $\sf{R_{DC}=4\Omega}$

• $\sf{R_{AC}=6\Omega}$

• $\sf{R_{CB}=4\Omega}$

• $\sf{R_{AB}=10\Omega}$

The equivalent resistance through the path ADC is (series),

$\longrightarrow\sf{R_{ADC}=R_{AD}+R_{DC}}$

$\longrightarrow\sf{R_{ADC}=2\Omega+4\Omega}$

$\longrightarrow\sf{R_{ADC}=6\Omega}$

So the equivalent resistance through the path AC with ADC is (parallel),

$\longrightarrow\sf{R'_{AC}=\left(\dfrac{1}{R_{ADC}}+\dfrac{1}{R_{AC}}\right)^{-1}}$

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

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

$\longrightarrow\sf{R'_{AC}=3\Omega}$

The equivalent resistance through the path ACB is (series),

$\longrightarrow\sf{R_{ACB}=R'_{AC}+R_{CB}}$

$\longrightarrow\sf{R_{ACB}=3\Omega+4\Omega}$

$\longrightarrow\sf{R_{ACB}=7\Omega}$

So the equivalent resistance through AB is,

$\longrightarrow\sf{R'_{AB}=\left(\dfrac{1}{R_{ACB}}+\dfrac{1}{R_{AB}}\right)^{-1}}$

$\longrightarrow\sf{R'_{AB}=\left(\dfrac{1}{7\Omega}+\dfrac{1}{10\Omega}\right)^{-1}}$

$\longrightarrow\sf{R'_{AC}=\left(\dfrac{17\Omega}{70\Omega^2}\right)^{-1}}$

$\longrightarrow\sf{\underline{\underline{R'_{AC}=\dfrac{70}{17}\ \Omega}}}$