A piece of wire of resistance $6\, \Omega$ is connected to $a$ battery of $12 \,V$. Find the amount of current flowing through it. Now, the same wire is redrawn by stretching it to double its length. Find the resistance of the new (redrawn) wire.
A piece of wire of resistance $6\, \Omega$ is connected to $a$ battery of $12 \,V$. Find the amount of current flowing through it. Now, the same wire is redrawn by stretching it to double its length. Find the resistance of the new (redrawn) wire.
Here, resistance $( R )=6 \Omega$, potential difference $( V )=12 V$
$V = IR$
or $I=\frac{V}{R}=\frac{12 V}{6 \Omega}=2 A$
Original resistance,
$R =\rho \frac{l}{ A }$
If the wire is doubled, then its new length $l^{\prime}=2 l$ and new area of cross-section $A ^{\prime}=\frac{ A }{2}$
Then new resistance,
$R ^{\prime}=\rho \frac{l^{\prime}}{ A ^{\prime}}=\rho \frac{2 l}{\frac{ A }{2}}=\frac{4 \rho l}{ A }=4 R$
or $\quad R^{\prime}=4 R,$ i.e., the resistance increases to four time.
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