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30 September, 23:34

When a wire that has a large diameter and a length L is connected across the terminals of an automobile battery, the current is 40 A. If we cut the wire to half of its original length and connect one piece that has a length L 2 across the terminals of the same battery, the current will be

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  1. 30 September, 23:48
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    If we cut the wire to half of its original length and connect one piece that has a length L/2 across the terminals of the same battery, the current will be * *twice* * of the initial value when the length of wire was L.

    Explanation:

    V = IR

    I = V/R

    V = voltage or potential difference across the wire

    I = current flowing in the wire

    R = Resistance across the length of the wire

    But the resistance is a function of the wire's physical attributes like its resistivity (ρ), cross sectional Area (A) and the length of wire (L)

    R = ρL/A

    If all other parameters are constant, R varies directly as the length of wire if all other parameters are constant

    Let all the other parameters be k

    R = kL

    Initially, the length of the wire is L₁

    R₁ = kL₁

    I₁ = V/R₁

    Then the wire is cut in half

    L₂ = L₁/2

    R₂ = kL₂

    R₂ = kL₁/2 = R₁/2

    I₁ = V/R₁

    I₂ = V/R₂

    Substituting R₂ = R₁/2 into the expression

    I₂ = V / (R₁/2) = 2V/R₁ = 2 I₁

    This means the current flowing in the wire of length L/2 is twice that flowing in the wire of length L.
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