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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.
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Solution
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
\[\tau = \frac{L}{R} = \frac{17 \times {10}^{- 3}}{1 . 7}\]
\[= {10}^{- 2} s = 10 ms\]
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