Question

$\normalsize\sf\red{Find \: \: the \: \: equivalent \: resistance \: between } \\ \normalsize\sf\red{points \: \: A \: \: and \: \: B. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$
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Answer 1

GiveN :-

• A circuit is given to us with some rahi resistances of 6Ω , 4Ω , 12Ω , 5Ω and 2Ω .

To FinD :-

• The equivalent resistance .

CalculationS :-

We will simply the circuit step by step to obtain the equivalent resistance . The given circuit looks a bit complex so let's transform it. Refer to the attachment.

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 2Ω , 2Ω and 4Ω are in series . So the final net resistance will be ,

$\sf\to R_{(net)}=R_1+R_2+R_3 \\\\\sf\to R_{(net)}= (2 + 2 + 4 )\Omega \\\\\sf\to \underset{\blue{\sf Required \ Resistance }}{\underbrace{\boxed{\pink{\frak{ Resistance_{(net)}= 8 \Omega }}}}}$

Answer 2

Answer:

tum ne to bola tha ki wo meri id waha se points le lo chahiye to yad aaya