Question

An inductor-coil inductance $17mH$ is constructed from a copper wire of length $100m$ and cross-sectional area $1m{m}^2$. Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper $=1.7\times {10}^{-8}\Omega$ $m$

Answer 1

Given,

Inductance, $L=17mH$

Length of the wire, $l=100m$

Cross-sectional area of the wire, $A=1m{m}^2=1\times {10}^{-6}{m}^2$

Resistivity of copper, $\rho =1.7\times {10}^{-8}\Omega -m$

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

$\tau =\frac{L}{R}=\frac{17\times {10}^{-3}}{1.7}={10}^{-2}s=10ms$

L=17mH,l=100m,A=1mm2=1×10−6m2,fcu=1.7×10−8Ω−m