A uniform wire of resistance 9 Ω is cut into 3 equal parts. They are connected in the form of equilateral triangle A B C. A cell of e.mf 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 3 equal parts. They are connected in the form of equilateral triangle A B C. A cell of e.mf 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

The circuit can be drawn as follows Equivalent resistance R=(3 ×(3+3)/3+(3+3))=2 Ω Current i=(2/2)=1 A, So i1=1 ×((3/3+6))=(1/3) A Potential difference between A and B=(1/3) × 3=1 volt

Q. A uniform wire of resistance $9Ω$ is cut into $3$ equal parts. They are connected in the form of equilateral triangle $A BC$ . A cell of e.mf $2 V$ and negligible internal resistance is connected across $B$ and $C$ . Potential difference across $A B$ is

1 V

2 V

3 V

0.5 V

The circuit can be drawn as follows Equivalent resistance
$R = \frac{3 \times ( 3 + 3 )}{3 + ( 3 + 3 )} = 2 Ω$
Current $i = \frac{2}{2} = 1 A,$
So $i_{1} = 1 \times (\frac{3}{3 + 6}) = \frac{1}{3} A$
Potential difference between $A$ and
$B = \frac{1}{3} \times 3 = 1$ volt

Q. A uniform wire of resistance 9Ω is cut into 3 equal parts. They are connected in the form of equilateral triangle ABC. A cell of e.mf 2V and negligible internal resistance is connected across B and C. Potential difference across AB is

1 V

2 V

3 V

0.5 V

The circuit can be drawn as follows Equivalent resistance R=3+(3+3)3×(3+3)​=2Ω Current i=22​=1A, So i1​=1×(3+63​)=31​A Potential difference between A and B=31​×3=1 volt

Q. A uniform wire of resistance $9Ω$ is cut into $3$ equal parts. They are connected in the form of equilateral triangle $A BC$ . A cell of e.mf $2 V$ and negligible internal resistance is connected across $B$ and $C$ . Potential difference across $A B$ is

The circuit can be drawn as follows Equivalent resistance
$R = \frac{3 \times ( 3 + 3 )}{3 + ( 3 + 3 )} = 2 Ω$
Current $i = \frac{2}{2} = 1 A,$
So $i_{1} = 1 \times (\frac{3}{3 + 6}) = \frac{1}{3} A$
Potential difference between $A$ and
$B = \frac{1}{3} \times 3 = 1$ volt