Question

The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. (I) if a wire with aLength of 20cm and a diameter of 0.1cm has a relationship of 40ohms, determine the proportionality equation for this relationship. (I)what is the diameter of the wire if the length is 25cm and resistance is 64ohms

Answer 1

Answer:

$\begin{gathered} i)R=\frac{0.02l}{d^2} \\ ii)0.0884cm \end{gathered}$

Explanations:

According to the information given;

• Let the ,electrical resistance, be "R"

,

• Let the ,length ,of the wire be "l"

,

• Let the ,diameter ,of the wire be "d"

I) If the electrical resistance (R) of a wire varies directly as its length (l) and inversely as the square of its diameter (d²), this is mathematically expressed as:

$\begin{gathered} R\alpha\frac{l}{d^2} \\ R=\frac{kl}{d^2} \end{gathered}$

where k is the variation constant.

If a wire with a length of 20cm and a diameter of 0.1cm has a relationship of 40ohms, the proportionality equation for this relationship will be expressed as:

$\begin{gathered} R=\frac{kl}{d^2} \\ 40=\frac{20k}{0.1^2} \end{gathered}$

Simplify to determine the value of "k"

$\begin{gathered} 20k=40\times0.1^2 \\ 20k=40\times0.01 \\ 20k=0.4 \\ k=\frac{0.4}{20} \\ k=0.02 \end{gathered}$

Determine the required proportionality equation for this relationship

$\begin{gathered} R=\frac{kl}{d^2} \\ R=\frac{0.02l}{d^2} \end{gathered}$

II) If the length is 25cm (l) and the resistance (R) is 64ohms, the diameter of the wire will be calculated as:

$\begin{gathered} R=\frac{kl}{d^2} \\ d^2=\frac{kl}{R} \\ d=\sqrt[]{\frac{kl}{R}} \end{gathered}$

Substitute the given parameters:

$\begin{gathered} d=\sqrt[]{\frac{0.02(25)}{64}} \\ d=\sqrt[]{\frac{0.5}{64}} \\ d=\sqrt[]{0.0078125} \\ d\approx0.0884cm \end{gathered}$

Hence the diameter of the wire if the length is 25cm and resistance is 64ohms is approximately 0.0884cm