Question

A 16 W resistance wire is bent to form a square. A source of emf 9 V is connected across one of its sides as shown. Calculate the current drawn from the source. Find the potential difference between the ends C and D. If now the wire is stretched uniformly to double the length and once again the same cell is connected in the same way, across one side of the square formed, what will now be the potential difference across one of its diagonals?

Answer 1

Given:

A 16 Ohm resistance wire is bent to form a square. A source of emf 9 V is connected across one of its sides as shown.

To find:

• Current drawn from the source

• Potential difference between C and D

• Potential difference across the diagonals after wire is stretched to double length.

Calculation:

Net resistance be r ;

$\therefore \: r = \dfrac{3R \times R }{3R + R }$

$= > \: r = \dfrac{3R }{4 }$

$= > \: r = \dfrac{3 \times 4 }{4 }$

$= > \: r = 3 \: ohm$

So, net Current drawn from battery:

$\therefore \: i = \dfrac{V}{r}$

$= > \: i = \dfrac{9}{3}$

$\boxed{ = > \: i = 3 \: amp}$

Since AD , DC , CB resistors are in series, they will equally divide the potential difference from battery ;

Potential difference between C and D be e :

$\boxed{ = > \: e = \dfrac{9}{3} = 3 \: volt}$

After stretching the wire to twice length, the resistance will become 4 times ;

So, each resistance becomes 4 × 4 = 16 Ohm.

Net current be I ;

$\therefore \: I = \dfrac{9}{ (\frac{3 \times 16}{4}) } = \dfrac{3}{4} \: amp$

Potential difference across its diagonal will be E:

Potential difference will remain equal to potential drop of DC and CB resistors :

$\boxed{E = 9 \times \dfrac{2}{3} = 6 \: volt}$

Hope It Helps.