Question

please solve this guys

a wire of 15 ohm resistance is gradually stretched to double its original length it is then cut into two equal parts these parts are then connected in parallel across 3 volt battery find the current drawn from the battery ​

Answer 1

Answer:

I = 0.2 A

Explanation:

Given:

Resistance (R ) = 15 ohm

V = 3 volt

By using the fact:

whenever any register is stretched to double its original length then the new resistance becomes four times of its original resistance.

•°• New resistance (R) = 4R

= 4 × 15

= 60 ohm

Now individual resistance that is of each part

= 60/2 = 30 ohm

•°• $\implies{\mathsf{R_1 = 15ohm \: and\: R_2 = 30ohm}}$

Now

calculate effective resistance in parallel combination

As we know that

$\boxed{\bold{\mathsf{\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}}}}$

Put in the values

$\implies{\mathsf{\frac{1}{R_p} = \frac{1}{30} + \frac{1}{30} }}\\ \\ = \frac{2}{30} \\ \\ = \frac{1}{15}$

•°• Rp = 15ohm

Now , To find Current

$\boxed{\bold{\mathsf{I = \frac{V}{R_p} }}}$

Put in the values

$\implies{\mathsf{=\frac{3}{15}}}$

= 0.2A