A 10 m long nichrome wire having 80Ω 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).
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answer is 2.
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Detailed Solution
∵I=VR=20080=2.5A&P=V2R=(200)280=500 watt When the wire is divided into two equal parts and connected in parallel we get maximum power∴Resistance of each part =40Ω∴lmax=5A⇒Pmax=V2Req=(200)2(20)=2000W=2kw.