Question

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) The curent in the circuit, (b) the power delivered by the battery, (c) the power dissipated in heating the resistor and (d) the rate at which energy is being stored in magnetic field.

Answer 1

$\text{Here}L=2H,R=200W,E=2V,t=10ms\left(a\right)i=i_0(1-e^{-\frac{t}{\tau }})=\frac{2}{200}(1-e^{-\frac{10\times {10}^{-3\times 200}}{2}})=0.01(1-e^{-1})=0.01(1-03678)=0.01\times 0.632=6.3mA$
(b) Power dilivered by the battery
$P=ViP=E{i}_0(1-e^{-\frac{t}{\tau }})=\frac{E^2}{R}(1-e^{-\frac{t}{\tau }})P=\frac{2\times 2}{200}(1-e^{\frac{10\times {10}^{-3}\times 200}{2}})P=0.02(1-e^{-1})=0.01264=12mW.$
(c) Power dissipated in the resistor $=i^{2}R$
$=\left[i_0\right(1-e^{-\frac{t}{\tau }}){]}^{2}R=(6.3mA{)}^2\times 200=6.3\times 6.3\times 200\times {10}^{-5}=79.38\times {10}^{-4}=7.938\times {10}^{-3}=8mW.$
(d) Rate at which energy is stored in the
Magnetic field $=\frac{1}{2}L{i}^2=\frac{L}{2}{i}_0^2(1-e^{-\frac{t}{\tau }}{)}^2=2\times {10}^{-2}(0.225)=0.455\times {10}^{-2}=4.6\times {10}^{-3}=4.6mW.$