Question

$\footnotesize\sf\red{Calculate \: the \: equivalent \: resistance \: \: between \: \: the} \\ \footnotesize\sf\red{points \: \: A \: \: and \: \: B \: for \: \: the \: following \: \: combination} \\ \footnotesize\sf \red {of \: \: resistors : \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$
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Answer 1

GiveN :-

• A electrical circuit is given to us with some resistances of 5Ω , 4Ω , 2Ω , and 6Ω .

To FinD :-

• The equivalent resistance .

CalculationS :-

We would be simplifing the circuit step by step to obtain the equivalent resistance . Firstly we see that the three 4Ω resistances are connected in series . We know that when the resistors are connected in series then the equivalent resistance is equal to the sum of individual resistors .

• Hence the net resistance would be ,

$\sf\to R_{net}= R_1+R_2+R_3 = 4\Omega \times 3 = \red{12 \Omega }$

$\rule{200}2$

Similarly if we look at the below then three 2Ω resistance are connected in series , then the equivalent resistance will be ,

$\sf\to R_{net}= R_1+R_2+R_3 = 2\Omega \times 3 = \red{6 \Omega }$

Refer to the attachment for easy understanding.

$\rule{200}2$

Now we see that the resistances of 12Ω , 4Ω and 6Ω are connected in parallel .

• So the net resistance will be ,

$\sf\to \dfrac{1}{R_{net}}= \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3} \\\\\sf\to \dfrac{1}{R_{net}}= \dfrac{1}{12\Omega }+\dfrac{1}{4\Omega}+\dfrac{1}{6\Omega}\\\\\sf\to \dfrac{1}{R_{net}}=\dfrac{1+3+2}{12\Omega }\\\\\sf\to \dfrac{1}{R_{net}}=\dfrac{6}{12\Omega } \\\\\sf\to \boxed{\red{\sf R_{net}= 2\Omega }}$

$\rule{200}2$

Finally we see that the resistances of 5Ω , 2Ω and 6Ω are in series . So the final net resistance will be ,

$\sf\to R_{(net)}=R_1+R_2+R_3 \\\\\sf\to R_{(net)}= (5 + 2 + 6 )\Omega \\\\\sf\to \underset{\blue{\sf Required \ Resistance }}{\underbrace{\boxed{\pink{\frak{ Resistance_{(net)}= 13 \Omega }}}}}$

Answer 2

Explanation:

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