Question

Two resistors of resistance 12 Ώ and 4 Ώ are connected in parallel to a 12 V battery. Find out the effective resistance of the circuit ​

Answer 1

$\LARGE{\bf{\dag}\:{\underline{\underline{\frak{AnswEr\::}}}}}$

Given that: Two resistors of 12Ω and 4Ω are connected in parallel with the 12 V battery.

Need to find: The effective resistance of the circuit?

Let's do it!!

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We know that effective resistance is given by;

$\star\:\large{\boxed{\sf{\pink{\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}}}}}$

We have;

• R₁ = 12Ω
• R₂ = 4Ω
• Req = ?

Putting all values,

$\\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1}{12} + \dfrac{1}{4}$

$\\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1}{12} + \dfrac{1\:\times\:3}{4\:\times\:3}$

$\\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1}{12} + \dfrac{3}{12}$

$\\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1 + 3}{12}$

$\\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{4}{12}$

By doing cross multiplication we get;

$\\ :\implies\:\sf \dfrac{12}{4} = R_{eq}$

$\\ :\implies\:\sf R_{eq} = {\cancel{\dfrac{12}{4}}}$

$\\ :\implies\:\underline{\boxed{\purple{\frak{R_{eq} = 3\Omega}}}}\:\bigstar$

Hence, effective resistance is 3Ω.

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