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Two wires of copper have the same length but cross-sectional area in the ratio of 3:1. The resistance of the thicker wire is 12 ohm. (a) Calculate the resistance of the thinner wire. (b) Find the equivalent resistance of both these wires, when they are connected (1) in series (ii) parallel (c) If both the wires are connected in parallel to a battery of 24 volt, then find out the ratio of heat produced in the thicker wire to the thinner wire identify the wire if both of them are connected to the same source for the same time duration.
Asked by swetalinasamantaray022 | 20 Jan, 2023, 10:54: AM Expert Answer

Given that,

Ratio of area of cross-section = 3:1

Resistance of thick wire, R1 = 12 Ω

(a)

Now,

The resistance of a wire is inversely proportional to its cross-sectional area.

i.e., Thus,

Resistance of the thinner wire, R2 = 3 × R1 = 3× 12 = 36 Ω

(b)

(i) When two wires are connected in series, their equivalent resistance will be

Req = 12 Ω + 36 Ω = 48 Ω

(ii) When two wires are connected in parallel, their equivalent resistance will be

1/Req = 1/R1 + 1/R2

i.e., 1/(1/12 + 1/36) = 9 Ω.

(c)

The heat produced in a wire is inversely proportional to the resistance of the wire. Answered by Jayesh Sah | 20 Jan, 2023, 12:36: PM
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