when a potential difference osf 2v is applied across the ends of a wire of 5cm length,a current of 1 a is found to flow through it.calculate(i:the resistance per unit length of the wire(ii)the resistance of 2m length of that wire(iii)the resistance across the ends of the wire if it is doubled on itself.
Asked by snigdhaa138_461 | 1st Sep, 2010, 02:30: AM
R = V/I = 2 Ω
R/L = 2 Ω/ 0.05 m = 40 Ω/m ..... Ans (i)
Since R α L
R' = 40 Ω/m x 2 m = 80 Ω .... Ans (ii)
Since it's doubled on itself, the L is now L/2,
Therefore the resistance of wire of 2.5 cm is 1 Ω, since originally it's 2 Ω.
Now these two 1 Ω's are in parallel with each other.
Hence the resistance is 1/2 = 0.5 Ω ... Ans (iii)
Answered by | 1st Sep, 2010, 11:45: AM
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