A wire of given material having length $l$ and area of cross-sectional area $A$ has a resistance of 4 ohms. What would be the resistance of another wire of the same material having length $\frac{l}{2}$ and area of cross-section $2 A$ ?

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Solution

We know that, $R = \frac{ρ l}{A} = 4 Ω$ .
Now, $R^{'} = \frac{ρ l^{'}}{A^{'}} = \frac{\frac{ρ l}{2}}{2 A}$
$R^{'} = \frac{ρ l}{4 A} = \frac{R}{4} = 1 Ω$ .

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A wire of given material having length l and area of cross-sectional area A has a resistance of 4 ohms. What would be the resistance of another wire of the same material having length l2 and area of cross-section 2A?

We know that, R=ρlA=4 Ω. Now, R′=ρl′A′=ρl22AR′=ρl4A=R4=1 Ω.

A wire of given material having length $l$ and area of cross-sectional area $A$ has a resistance of 4 ohms. What would be the resistance of another wire of the same material having length $\frac{l}{2}$ and area of cross-section $2 A$ ?

We know that, $R = \frac{ρ l}{A} = 4 Ω$ .
Now, $R^{'} = \frac{ρ l^{'}}{A^{'}} = \frac{\frac{ρ l}{2}}{2 A}$
$R^{'} = \frac{ρ l}{4 A} = \frac{R}{4} = 1 Ω$ .