Question

the potential difference across the 3 ohm resistance in the following diagram is

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Answer 1

3Ω , 6Ω are connected in parallel connection ,

Let it be Rₚ ,

$:\implies \sf \dfrac{1}{R_{p}}=\dfrac{1}{3}+\dfrac{1}{6}$

$:\implies \sf \dfrac{1}{R_p}=\dfrac{2+1}{6}$

$:\implies \sf \dfrac{1}{R_p}=\dfrac{3}{6}$

$:\implies \sf R_p=\dfrac{6}{3}$

$:\implies \sf R_p=2\ \Omega$

Now Rₚ (3Ω , 6Ω) , 4Ω are connected in series ,

Let it be Rₛ ,

$:\implies \sf R_s=2+4$

$:\implies \sf R_s=6\ \Omega$

So , $\sf R_{eq}=R_s=6\ \Omega$

Potential difference , V = 3 V

Current , I = ? A

Apply Ohm's law ,

⇒ $\sf V=IR_{eq}$

⇒ 3 = I (6)

⇒ I = ³/₆

⇒ I = 0.5 A  $\orange{\bigstar}$

Potential difference across the 3Ω resistor :

➠ V = I (3)

➠ V = (0.5) (3)

➠ V = 1.5 V  $\green{\bigstar}$