an inductor of inductance 2.00 H is joined in series with a resistor of resistance `200 Omega` and a battery of emf 2.00 V. At t = 10 ms, find (a) th current in the circuit, (b) the power delivered by the batter, (c ) the power dissipated in heating the resistor and (d) the rate at which energy is being stored in magnetic field.

Explanation

Here `L=2 H,R=200W,`
`E=2V,t=10ms`
`(a) i=i_(0)(1-e^(-t//(tau)))`
`=(2)/(200)(1-e^(-10xx10^(-8)xx200/2))`
`=0.01(1-e^(-1))`
`0.01xx0.632=6.3 mA`
(b) Power delivered by the battery
`P=V_(i)`
`P=Ei_(0)(1-e^(-t//(tau)))`
`P=(2xx2)/(200)(1-e^(10xx10^(-3)xx(200)/(2)))`
`P=0.02(1-e^(-1))`
`=0.01264=12mW.`
`(c) Power dissipated in the resistor =i^(2)R`
=[i_(0)(1-e^(-t//(tau)))]^(2) R`
`=(6.3xx6.3xx200xx10^(-5))`
`=79.38xx10^(-4)`
`=7.938xx10^(-3)=8mW.`
(d) Rate at which energy is stored in the magnetic field=(1/2)(Li^2)`
`=(L/2)(i_0)^2(1-e^(-t//(tau)))^2`
=2xx10^(-2)(0.225)`
`=0.455xx10^(-2)`
=4.6xx10^(-3)`
=4.6mW.