First slide

combination of resistance

Question

An electrical cable of copper has just one wire of radius 9 mm. Its resistance is 5 ohm. This single copper wire of the cable is replaced by 6 different well insulated copper wires each of radius 3 mm. The total resistance of the cable will now be equal to

Difficult

A

7.5 ohm
B

45 ohm
C

90 ohm
D

270 ohm

Solution

Let the resistance of single copper wire be .R₁. If p is the specific resistance of copper wire, then
$R_{1} = \frac{ρ \times l}{A_{1}} = \frac{ρ \times l}{{πr}_{1}^{2}}$ ………………(1)
When the wire is replaced by six wires, let the resistance of each wire be R₂. Then
$R_{2} = \frac{ρ \times l}{A_{2}} = \frac{ρ \times l}{{πr}_{2}^{2}}$ ………………(2)
From eqs. (1) and (2), we get
$\begin{array}{l} \frac{R_{1}}{R_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}} or \frac{5}{R_{2}} = \frac{{(3 \times 10^{- 3})}^{2}}{{(9 \times 10^{- 3})}^{2}} \\ R_{2} = 45 Ω \end{array}$
These six wires are in parallel. Hence the resistance of the combination would be
$R = \frac{45}{6} = 7 \cdot 5 Ω$

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