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ICSE Class 10 Physics Question 18 of 31

Current Electricity — Question 25

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Question 25

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.

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 Rp then

1Rp=14.5+14.5+14.51Rp=34.5⇒Rp=4.53⇒Rp=1.5Ω\dfrac{1}{R_p} = \dfrac{1}{4.5} + \dfrac{1}{4.5} + \dfrac{1}{4.5} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3}{4.5} \\[0.5em] \Rightarrow R_p = \dfrac{4.5}{3} \\[0.5em] \Rightarrow R_p = 1.5 Ω

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