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 = 1.0s.

Explanation

Here L=20mH,
e=5.0V, R=10 Omega`
` tau=(L/R)=(20xx10^(-3))/10`
` implies i_0=5/10`
implies i=i_0(1-e^(-t//tau))`
` impliesi=i_0-i_0 e^(-t//tau))`,brgt`implies iR=i_0R=i_0Re^(-t//tau)`
`(b) (Rdi)/(dt)=Ri_0=(1/tau)xxe^(-t//tau)`
`For t=10ms=10xx10^(-3)`
`(dE)/(dt0=10xx(5/10)xx(10)/(20xx10^(-3))`
`xxe^(-0.01//2)xx10^(-2)`
`=16.844=17 V//s`.(c) For t=1s
(dE)/(dt)=(Rdi)/(dt)`
`=(5/2)xx10^(-3)xxe^(-1/2)xx10-2`
`=0.00V//s`.