Answer to Question #306156 in Differential Equations for Rex

Question #306156

An inductance of L Henrys and a resistance of 10 ohms are connected in series with e.m.f. of 100 volts. If the current is initially zero, and is equal to 9 amperes after 1 second , find L and find the current after 0.5 second

1
Expert's answer
2022-03-07T15:51:02-0500

The instantenous emf equation is given as;

Ldidt+Ri=VL\frac{di}{dt}+Ri=V

didt=VRiL\frac{di}{dt}=\frac{V-Ri}{L}

didt=RL(VRi)\frac{di}{dt}=\frac{R}{L}(\frac{V}{R}-i)

di(VRi)=RLdt\frac{di}{(\frac{V}{R}-i)}=\frac{R}{L}dt

Integrating on both sides

0idi(VRi)=0tRLdt\int_0^i \frac{di}{(\frac{V}{R}-i)}=\int_0^t \frac{R}{L}dt

0idi(iVR)=RL0tdt\int_0^i \frac{di}{(\frac{i-V}{R})}=-\frac{R}{L}\int_0^tdt

ln(iVR)VR=RLtln\frac{(i-\frac{V}{R})}{-\frac{V}{R}}=-\frac{R}{L}t

(iVR)VR=eRLt\frac{(i-\frac{V}{R})}{-\frac{V}{R}}=e^{-\frac{R}{L}t}

i=VR(1eRLt)i={\frac{V}{R}}(1-e^{-\frac{R}{L}t})

When t=1, i(1)=9

9=10010(1e10L1)9={\frac{100}{10}}(1-e^{-\frac{10}{L}1})

910=1e10L\frac{9}{10}=1-e^{\frac{-10}{L}}

e10L=0.1e^\frac{-10}{L}=0.1

10L=ln(0.1)\frac{-10}{L}=ln(0.1)

ln(0.1)=-2.302

L=102.302L=\frac{10}{2.302}

L=4.443H

Current after 0.5s is;

i(0.5)=10010(1e104.3440.5)i(0.5)={\frac{100}{10}}(1-e^{-\frac{10}{4.344}*0.5})

=6.837A


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