Question

calculate the equivalent resistance in this figure?​

Answer 1

Given:-

• A circuit diagram is given to us.

To Find:-

• The equivalent resistance of the circuit.

Formulae Used:-

• If n resistances are connected in series :

$\large{\underline{\boxed{\red{\sf{\dag R_{eff}= R_1+R_2+R_3+......R_n}}}}}$

• If n resistances are in parallel :

$\large{\underline{\boxed{\red{\sf{\dag \dfrac{1}{R_{eff}}= \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+......\dfrac{1}{R_n}}}}}}$

Solution:-

[For labelled circuit diagram refer to attachment :]

Between points G & H is a resistance of 3Ω and between E & I is 6Ω . And they are in parallel .

So ,

⇒ 1/R = 1/3Ω + 1/6Ω.

⇒ 1/R = 2+1/6Ω.

⇒ 1/R = 3/6Ω.

⇒ 1/R = 1/2Ω.

⇒R = 2Ω.

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[For more simplified circuit refer to attachment:]

Now we can see that between points A and B , two 2Ω resistances are present .

So , net resistance would be :-

⇒ R = 2Ω + 2Ω .

⇒ R = 4Ω.

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[For more simplified circuit refer to attachment:]

We can see here that Between points A and B is a resistance of 4Ω and between points C and D is a resistance of 4Ω . Also these are in parallel .

⇒ 1/R = 1/4Ω + 1/4Ω .

⇒ 1/R = 1+1/4Ω.

⇒ 1/R = 2/4Ω.

⇒ 1/R = 1/2Ω.

⇒ R = 2Ω.

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Now this above 2Ω resistance and another 2Ω resistance are in series . So , net resistance will be :-

⇒ R = 2Ω + 2Ω .

⇒ R = 4Ω .

$\underline\red{\tt{\leadsto Hence\; the\; net \;resistance\; is \;4\Omega.}}$