Question

An inductor-coil of inductance 17 mH is constructed from a copper wire of length 100 m and cross-sectional area 1 mm². Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper = 1.7 × 10⁻⁸ Ω-m.

Answer 1

Given:
Inductance, L = 17 mH
Length of the wire, l = 100 m
Cross-sectional area of the wire, A = 1 mm² = 1 × 10⁻⁶ m²
Resistivity of copper, ρ = 1.7 × 10⁻⁸ Ω-m
Now,
$$\begin{gathered} R=\frac{\mathrm{\rho }l}{A} \\ =\frac{1.7\times {10}^{-8}\times 100}{1\times {10}^{-6}}=1.7\mathrm{\Omega } \end{gathered}$$

The time constant of the L-R circuit is given by
$$\begin{gathered} \tau =\frac{L}{R}=\frac{17\times {10}^{-3}}{1.7} \\ ={10}^{-2}\mathrm{s}=10\mathrm{ms} \end{gathered}$$