Question

A 60 Ω resistor is made by winding platinum wire into a coil at 20°C. If the platinum has resistivity

r=10.4x10-8 Wm at 20°C and the platinum wire has a diameter of 0.10 mm, what length of wire is

needed?

Answer 1

We use the formula

$R=\frac{\rho*L}{A}$

Where R= resistance in ohms

$\rho$ = resistivity in Wm

L = Length of wire

A = Cross sectional area

We have,

R=6$\Omega$, $\rho$ =10.4 x 10^-8Wm at 20°c

radius (r)= $\frac{diameter}{2}$ = $\frac{0.10mm}{2}$ = 0.05mm

We convert to meters

0.05mm = 5 x 10^-5m

Now, we find the Area

Area = $\pi$r²

Where $\pi$ = $\frac{22}{7}$, r = 5 x 10^-5

Area = $\frac{22}{7}$ * (5 x 10^-5)²

= $\frac{22}{7}$ * 2.5 x 10^-9

= 7.86 x 10^-9

Now, using the formula

$R=\frac{\rho*L}{A}$

Making L the subject of the formula

$L=\frac{R*A}{\rho}$

= $\frac{60 *7.86*10^{-9}}{10.4 * 10^{-8}}$

= $\frac{4.716*10^{-7}}{10.4*10^-{-8}}$

= $\frac{1179}{260} = 4.53m (2 d.p)$

Therefore, the length L= 4.53m