A uniform wire of resistance R and length L is cut into four equal parts, each of length L/4 which are then connected in parallel combination. The effective resistance of the combination will be

R

4R

$\frac{R}{4}$

$\frac{R}{16}$

Solution

$\frac{R}{16}$

Resistivity of materials is the resistance to the flow of an electric current with some materials resisting the current flow more than others

$R = \frac{ρ L}{A}$

where R is resistance in ( Ω )

ρ is resistivity

L is length in meters (m)

A is the area in square meters (m²)

⇒ $R \propto \frac{L}{A}$

$∴ \frac{R_{1}}{R_{2}} = \frac{L_{1}}{L_{2}}$

= $\frac{L}{L / 4}$

$\frac{R_{1}}{R_{2}} = 4$

⇒ R₂ = $\frac{R}{4}$ .....( since R₁ = R )

In parallel combination of such four resistance

$\frac{1}{R'} = \frac{1}{R_{2}} + \frac{1}{R_{2}} + \frac{1}{R_{2}} + \frac{1}{R_{2}}$

⇒ $\frac{1}{R'} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$

⇒ $\frac{1}{R'} = \frac{16}{R}$

⇒ R' = $\frac{R}{16}$

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