A particular resistance wire has a resistance of 3.0 ohm per meter. Find —
(a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel.
(b) The potential difference of the battery which gives a current of 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that resistance of the battery is negligible).
(c) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.

Question

A particular resistance wire has a resistance of 3.0 ohm per meter. Find —
(a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel.
(b) The potential difference of the battery which gives a current of 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that resistance of the battery is negligible).
(c) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.

KnowledgeBoat · Accepted Answer

(a) Resistance of 1 m of wire = 3 Ω. Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 Ω. As three such wires are joined in parallel and if the equivalent resistance of this part is R_p then Hence, total resistance of circuit = 1.5 Ω (b) I = 2 A From Ohm's Law V = IR Substituting the values in the formula above we get, V = 2 x 4.5 = 9 V Hence, potential difference = 9 V (c) R = 3 Ω for 1 meter wire Therefore, for 5 m R = 3 x 5 = 15 Ω Here the area is twice and resistance is inversely proportional to area. Thus, resistance becomes half R = 15 / 2 = 7.5 Ω Hence, resistance = 7.5 Ω

Physics

Current Electricity

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