Question

A wire of resistence R is bent in form of closed circle,what is the resistence of a diameter of a circle​

Answer 1

Key Points :

When resistors are connected in series ,

$\bigstar\ \; \sf R_{eq}=R_1+R_2+R_3+...$

When resistors are connected in parallel ,

$\bigstar\ \; \sf \blue{\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...}$

Solution :

R/2 , R/2 are connected in parallel ,

$\sf \to \dfrac{1}{R_{eq}}=\dfrac{1}{R/2}+\dfrac{1}{R/2}\\\\\to \sf \dfrac{1}{R_{eq}}=\dfrac{2}{R}+\dfrac{2}{R}\\\\\to \sf \dfrac{1}{R_{eq}}=\dfrac{4}{R}\\\\\to \sf \green{R_{eq}=\dfrac{R}{4}\ \Omega }\; \bigstar$

Resistance of a diameter of a circle​ is R/4 Ω .

More Info :

Ohm's Law :

Voltage is directly proportional to current passing through circuit / conductor .

$\to \sf V \propto I\\\\\to \sf V=IR\ \; \bigstar$

Answer 2

GIVEN :-

• A ᴡɪʀᴇ ᴏғ ʀᴇsɪsᴛᴀɴᴄᴇ R ɪs ʙᴇɴᴛ ɪɴ ғᴏʀᴍ ᴏғ ᴀ ᴄʟᴏsᴇᴅ ᴄɪʀᴄʟᴇ .

TO FIND :-

• Tʜᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴀᴄʀᴏss ᴀ ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴛʜᴇ ᴄɪʀᴄʟᴇ .

SOLUTION :-

➪ sᴇᴇ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ ғɪɢᴜʀᴇ .

✯ As ᴡᴇ ᴄᴀɴ sᴇᴇ ᴛʜᴀᴛ ᴛʜᴇ ᴘᴏɪɴᴛ A ᴀɴᴅ B ᴅɪᴠɪᴅᴇs ᴛʜᴇ ᴄɪʀᴄʟᴇ ɪɴ ᴛᴡᴏ ᴇǫᴜᴀʟ ᴘᴀʀᴛs .

✯ Hᴇɴᴄᴇ, ɪғ ᴡᴇ ғʟᴏᴡ ᴀ ᴄᴜʀʀᴇɴᴛ ғʀᴏᴍ ᴘᴏɪɴᴛ A ᴛᴏ B ɪᴛ ᴡɪʟʟ ᴘᴀssᴇs ғʀᴏᴍ AOB ᴀɴᴅ ACB, ᴡʜᴇʀᴇ O ᴀɴᴅ B ᴀʀᴇ ᴘᴏɪɴᴛs ᴏɴ ᴛᴡᴏ ᴀʀᴄs, ᴛʜᴇʀᴇғᴏʀᴇ ᴛʜᴇ ᴄᴏɴɴᴇᴄᴛɪᴏɴ ᴡɪʟʟ ʟᴏᴏᴋ ʟɪᴋᴇ ᴛʜɪs, ɪ.ᴇ. Pᴀʀᴀʟʟᴇʟ-ᴄᴏɴɴᴇᴄᴛɪᴏɴ .

✯ Hᴇɴᴄᴇ, ᴇғғᴇᴄᴛɪᴠᴇ ᴏʀ ɴᴇᴛ ʀᴇsɪsᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴀɴʏ ᴘᴏɪɴᴛ ᴡɪʟʟ ʙᴇ,

$\huge\red\star$$\bf\green{\dfrac{1}{R_{eq}}\:=\:\dfrac{1}{R_1}\:+\:\dfrac{1}{R_2}\:}$

$\rm{:\implies\:\dfrac{1}{R_{eq}}\:=\:\dfrac{1}{R/2}\:+\:\dfrac{1}{R/2}\:}$

$\rm{:\implies\:\dfrac{1}{R_{eq}}\:=\:\dfrac{2}{R}\:+\:\dfrac{2}{R}\:}$

$\rm{:\implies\:\dfrac{1}{R_{eq}}\:=\:2\times{\dfrac{2}{R}}\:}$

$\rm{:\implies\:\dfrac{1}{R_{eq}}\:=\:\dfrac{4}{R}\:}$

$\bf\purple{:\implies\:R_{eq}\:=\:\dfrac{R}{4}\:\Omega\:}$

$\huge\red\therefore$ Tʜᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴀᴄʀᴏss ᴀ ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴛʜᴇ ᴄɪʀᴄʟᴇ ɪs $\bf\pink{\dfrac{R}{4}\:\Omega\:}$ .