Question

ᴀ ᴘɪᴇᴄᴇ ᴏғ ᴡɪʀᴇ ʜᴀᴠɪɴɢ ᴀ ʀᴇsɪsᴛᴀɴᴄᴇ ʀ ɪs ᴄᴜᴛ ɪɴᴛᴏ ғɪᴠᴇ ᴇϙᴜᴀʟ ᴘᴀʀᴛs.

(ɪ) ʜᴏᴡ ᴡɪʟʟ ᴛʜᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғ ᴇᴀᴄʜ ᴘᴀʀᴛ ᴏғ ᴛʜᴇ ᴡɪʀᴇ ᴄᴏᴍᴘᴀʀᴇ ᴡɪᴛʜ ᴛʜᴇ ᴏʀɪɢɪɴᴀʟ ʀᴇsɪsᴛᴀɴᴄᴇ?

(ɪɪ) ɪғ ᴛʜᴇ ғɪᴠᴇ ᴘᴀʀᴛs ᴏғ ᴛʜᴇ ᴡɪʀᴇ ᴀʀᴇ ᴘʟᴀᴄᴇᴅ ɪɴ ᴘᴀʀᴀʟʟᴇʟ, ʜᴏᴡ ᴡɪʟʟ ᴛʜᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴄᴏᴍʙɪɴᴀᴛɪᴏɴ ᴄᴏᴍᴘᴀʀᴇ ᴡɪᴛʜ ᴛʜᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴏʀɪɢɪɴᴀʟ ᴡɪʀᴇ? ᴡʜᴀᴛ ᴡɪʟʟ ʙᴇ ʀᴀᴛɪᴏ ᴏғ ʀᴇsɪsᴛᴀɴᴄᴇ ɪɴ sᴇʀɪᴇs ᴛᴏ ᴛʜᴀᴛ ᴏғ ᴘᴀʀᴀʟʟᴇʟ?
$\rule{150pt}{}$
ɴᴏ ᴘᴀʟɢᴀʀɪᴢᴇᴅ ᴄᴏɴᴛᴇɴᴛ, ᴘʟᴇᴀsᴇ !
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Answer 1

(a)

We know the equation:

$\boxed{\bold{\sf{R=\dfrac{l}{A}\implies R \propto l}}}$

So here after the cutting of 5 pieces,

$\sf{\implies r=\dfrac{l}{5}}$

Therefore, new resistance:

$\boxed{\bold{\sf{R'=\dfrac{r}{A}\implies R' \propto l'}}}$

So the ratio is :

$\sf{\implies Ratio=\dfrac{R}{R'}}$

$\sf{\implies Ratio=\dfrac{l}{\dfrac{l'}{5}}}$

$\sf{\implies Ratio=\dfrac{5}{1}}$

$\huge{\boxed{\bold{\sf{Ratio=5:1}}}}$

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(b)

When the five parts of the wire are placed in parallel,

$\boxed{\bold{\sf{Effective \ Resistance \ (\dfrac{1}{R_p})=\dfrac{1}{\frac{R}{5}}+\dfrac{1}{\frac{R}{5}}+\dfrac{1}{\frac{R}{5}}+\dfrac{1}{\frac{R}{5}}+\dfrac{1}{\frac{R}{5}}}}}$

$\sf{\implies \dfrac{1}{R_p}=\dfrac{1+1+1+1+1}{\frac{R}{5}}}$

$\sf{\implies \dfrac{1}{R_p}=\dfrac{5}{\frac{R}{5}}}$

$\sf{\implies \dfrac{1}{R_p}=\dfrac{5\times 5}{R}}$

$\sf{\implies \dfrac{1}{R_p}=\dfrac{25}{R}}$

$\sf{\implies R_p=\dfrac{R}{25}}$

$\boxed{\bold{\sf{ Resistance \ of \ the \ combinations=\dfrac{R}{25}}}}$

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(c)

When the five parts of the wire are placed in series,

$\boxed{\bold{\sf{Effective \ Resistance \ (R_s)=\dfrac{R}{5}+\dfrac{R}{5}+\dfrac{R}{5}+\dfrac{R}{5}+\dfrac{R}{5}}}}$

$\sf{\implies R_s=\dfrac{R+R+R+R+R}{5}}$

$\sf{\implies R_s=\dfrac{5R}{5}}$

$\sf{\implies R_s=R}$

Therefore ratio in series to parallel is:

$\sf{\implies R_s:R_p=R:\dfrac{R}{25}}$

$\sf{\implies R_s:R_p=25:1}$

$\huge{\boxed{\bold{\sf{R_s:R_p=25:1}}}}$

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