A uniform wire of resistance 9 Ω is cut into equal parts. They are connected in the form of equilateral triangle ABC. A cell of emf 2 V and negligible internal resistance is connected across B and C. Potential difference across AB is:

Question

A uniform wire of resistance 9 Ω is cut into equal parts. They are connected in the form of equilateral triangle ABC. A cell of emf 2 V and negligible internal resistance is connected across B and C. Potential difference across AB is: (A) 1 V (B) 2 V (C) 3 V (D) 0.5 V

Tardigrade Admin · Accepted Answer

R=(9/3)=3 Ω Rnet=2 Ω PD across AB V=i× R=(V/Rnet)× R =(2/2)× 3=3V

Q. A uniform wire of resistance $9 Ω$ is cut into equal parts. They are connected in the form of equilateral triangle ABC. A cell of emf 2 V and negligible internal resistance is connected across B and C. Potential difference across AB is:

1 V

2 V

3 V

0.5 V

0.25 V

$R = \frac{9}{3} = 3 Ω$
$R_{n e t} = 2 Ω$
PD across AB
$V = i \times R = \frac{V}{R _{n e t}} \times R$
$= \frac{2}{2} \times 3 = 3 V$

Q. A uniform wire of resistance 9Ω is cut into equal parts. They are connected in the form of equilateral triangle ABC. A cell of emf 2 V and negligible internal resistance is connected across B and C. Potential difference across AB is:

1 V

2 V

3 V

0.5 V

0.25 V

R=39​=3Ω Rnet​=2Ω PD across AB V=i×R=Rnet​V​×R =22​×3=3V

Q. A uniform wire of resistance $9 Ω$ is cut into equal parts. They are connected in the form of equilateral triangle ABC. A cell of emf 2 V and negligible internal resistance is connected across B and C. Potential difference across AB is:

$R = \frac{9}{3} = 3 Ω$
$R_{n e t} = 2 Ω$
PD across AB
$V = i \times R = \frac{V}{R _{n e t}} \times R$
$= \frac{2}{2} \times 3 = 3 V$