Question

Calculate the equivalent resistance between points A and B

CLASS 10th​

Answer 1

ANSWER:

Given:

• $\sf R_1=2\Omega$
• $\sf R_2=4\Omega$
• $\sf R_3=6\Omega$
• $\sf R_4=7\Omega$
• $\sf R_5=10\Omega$

To Find:

• Equivalent Resistance between A and B

Procedure:

• First we'll find Equivalent resistance of $\sf R_1$ and $\sf R_2$.
• Then, we'll find the equivalent resistance of $\sf R_3$ and that of $\sf R_1$ and $\sf R_2$.
• After that, we'll find the equivalent resistance of the above equivalent resistance and $\sf R_4$
• Finally, we will find the Equivalent resistance of the above one and $\sf R_5$

Solution:

(Refer Attachment 1)

We are given that,

$:\implies\sf R_1=2\Omega,\:R_2=4\Omega,\:R_3=6\Omega,\:R_4=7\Omega,\:R_5=10\Omega$

(Refer Attachment 1)

We can see that,

$:\implies\sf R_1\:\&\:R_2\xrightarrow{Connected\:in} Series\:Combination$

We know that, for a Series Combination:

$:\hookrightarrow\sf R_{eq}=R_1+R_2+. . . .+R_n$

So,

$:\implies\sf R_{eq}^1=R_1+R_2$

Here, $\sf R_{eq}^1$, is the equivalent resistance of $\sf R_1$ and $\sf R_2$. (1 in $\sf R_{eq}^1$ is just for notation)

As, $\sf R_1=2\Omega\:\&\:R_2=4\Omega$,

Hence,

$:\implies\sf R_{eq}^1=2+4$

$:\implies\sf R_{eq}^1=6\Omega - - - -(1)$

(Refer Attachment 2)

Now, we can see that,

$:\implies\sf R_{eq}^1\:\&\:R_3\xrightarrow{Connected\:in} Parallel\:Combination$

We know that, for a Parallel Combination:

$:\hookrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+. . . .+\dfrac{1}{R_n}$

So,

$:\implies\sf \dfrac{1}{R_{eq}^2}=\dfrac{1}{R_{eq}^1}+\dfrac{1}{R_3}$

Here, $\sf R_{eq}^2$, is the equivalent resistance of $\sf R_{eq}^1$ and $\sf R_3$. (2 in $\sf R_{eq}^2$ is just for notation)

Using (1),

As, $\sf R_{eq}^1=6\Omega\:\&\:R_3=6\Omega$,

Hence,

$:\implies\sf \dfrac{1}{R_{eq}^2}=\dfrac{1}{6}+\dfrac{1}{6}$

$:\implies\sf \dfrac{1}{R_{eq}^2}=\dfrac{2\!\!\!/}{6\!\!\!/_{\:3}}$

$:\implies\sf \dfrac{1}{R_{eq}^2}=\dfrac{1}{3}$

$:\implies\sf R_{eq}^2=3\Omega - - - -(2)$

(Refer Attachment 3)

We can see that,

$:\implies\sf R_{eq}^2\:\&\:R_4\xrightarrow{Connected\:in} Series\:Combination$

We know that, for a Series Combination:

$:\hookrightarrow\sf R_{eq}=R_1+R_2+. . . .+R_n$

So,

$:\implies\sf R_{eq}^3=R_{eq}^2+R_4$

Here, $\sf R_{eq}^3$, is the equivalent resistance of $\sf R_{eq}^2$ and $\sf R_4$. (3 in $\sf R_{eq}^3$ is just for notation)

Using (2),

As, $\sf R_{eq}^2=3\Omega\:\&\:R_4=7\Omega$,

Hence,

$:\implies\sf R_{eq}^3=3+7$

$:\implies\sf R_{eq}^3=10\Omega - - - -(3)$

(Refer Attachment 4)

Now, we can see that,

$:\implies\sf R_{eq}^3\:\&\:R_5\xrightarrow{Connected\:in} Parallel\:Combination$

We know that, for a Parallel Combination:

$:\hookrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+. . . .+\dfrac{1}{R_n}$

So,

$:\implies\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_{eq}^3}+\dfrac{1}{R_5}$

Here, $\sf R_{eq}$, is the equivalent resistance of $\sf R_{eq}^3$ and $\sf R_5$.

Using (3),

As, $\sf R_{eq}^3=10\Omega\:\&\:R_5=10\Omega$,

Hence,

$:\implies\sf \dfrac{1}{R_{eq}}=\dfrac{1}{10}+\dfrac{1}{10}$

$:\implies\sf \dfrac{1}{R_{eq}}=\dfrac{2\!\!\!/}{10\!\!\!\!\!/_{\:\:5}}$

$:\implies\sf \dfrac{1}{R_{eq}}=\dfrac{1}{5}$

$:\implies\bf{R_{eq}=5\Omega}$

Hence, the Total Equivalent Resistance between A and B is 5 $\bf{\Omega}$.