Question

An inductor - coil of inductance 20 mH having resistance $10\Omega$ is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at t = 0, (b) t = 10 ms and (c) t = 1.0 s.

Answer 1

$L=20mH,e=5.0V,R=10\Omega \tau =\frac{L}{R}=\frac{20\times {10}^{-3}}{10}{i}_0=\frac{5}{10}\Rightarrow i=i_0(1-e^{-\frac{t}{\tau }})\Rightarrow i=i_0-i_0{e}^{-\frac{t}{\tau }}\Rightarrow iR=i_{0}R-i_{0}R{e}^{-\frac{t}{\tau }}\left(b\right)\frac{Rdi}{dt}=R{i}_0=\frac{1}{\tau }\times e^{-\frac{t}{\tau }}Fort=10ms=10\times {10}^{-3}s\frac{dE}{dt}=10\times \frac{5}{10}\times \frac{101}{20\times {10}^{-3}}\times e^{-\frac{0.01}{2}}\times {10}^{-2}=16.844=17V/s\left(c\right)Fort=1s\frac{dE}{dt}=\frac{Rdi}{dt}=\frac{5}{2}\times {10}^{-3}{e}^{-\frac{1}{2}\times 10-2}=0.00V/s$