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,

\[R = \frac{\rho l}{A}\]

\[ = \frac{1 . 7 \times {10}^{- 8} \times 100}{1 \times {10}^{- 6}} = 1 . 7 \Omega\]

The time constant of the L-R circuit is given by

\[\tau = \frac{L}{R} = \frac{17 \times {10}^{- 3}}{1 . 7}\]

\[=  {10}^{- 2}   s = 10  ms\]