Question

A 10 m long nichrome wire having $80\Omega$ resistance has current carrying capacity of 5 A. This wire can be cut into equal parts and equal parts can be connected in series or parallel. What is the maximum power which can be obtained as heat by the wire from a 200 V mains supply? (in KW)

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$\begin{array}{c}I=\frac{V}{R}=\frac{200}{80}=2.5A\\ andP=\frac{V^2}{R}=\frac{{\left(200\right)}^2}{80}=500watt\\ whenthewireisdivededintotwoequalpartsandconnectedinparallelwegetmaximumpower,\\ \therefore Resis\tan ceofeachpart=40\Omega \\ \therefore I_{max}=5A\\ \Rightarrow P_{max}=\frac{V^2}{R_{eq}}=\frac{{\left(200\right)}^2}{20}=2000W=2kw\end{array}$

Hence,

option $\left(B\right)$ is correct answer.