When a potential difference of 2 V is applied across the ends of a wire of 5 m length, a current of 1A is found to flow through it. Calculate: (i) the resistance per unit length of the wire (ii) the resistance of 2 m length of this wire (iii) the resistance across the ends of the wire if it is doubled on itself.
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(i) V = I R ----- Ohm's law
(ii) Resistance of 2 m length of the wire = 0.4 x 2
(iii) When the wire is doubled on itself:
(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.
(2) The length becomes half i.e.
(i) V = I R ----- Ohm's law
(ii) Resistance of 2 m length of the wire = 0.4 x 2
(iii) When the wire is doubled on itself:
(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.
(2) The length becomes half i.e.
(i) V = I R ----- Ohm's law
(ii) Resistance of 2 m length of the wire = 0.4 x 2
(iii) When the wire is doubled on itself:
(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.
(ii) Resistance of 2 m length of the wire = 0.4 x 2
(iii) When the wire is doubled on itself:
(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.
(2) The length becomes half i.e.
(i) V = I R ----- Ohm's law
(ii) Resistance of 2 m length of the wire = 0.4 x 2
(iii) When the wire is doubled on itself:
(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.
(2) The length becomes half i.e.
(i) V = I R ----- Ohm's law
(ii) Resistance of 2 m length of the wire = 0.4 x 2
(iii) When the wire is doubled on itself:
(1) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A.
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