A 8 volt battery is connected to a series circuit

percibaa8

percibaa8

Answered question

2021-12-31

A 8 volt battery is connected to a series circuit in which the inductance is 4 henry and the resistance is 5 ohms. Find the current if the initial current is zero. What will the current be after a very long time?

Answer & Explanation

levurdondishav4

levurdondishav4

Beginner2022-01-01Added 38 answers

Let the current flows in the circuit is (t)
According to KVL,
8+vL+vR=0
VL+VR=8
Ldidt+i×R=8
4didt+5i=8
4didt=85i
di85i=dt4
Integrate both side
di85i=dt4
Let 85i=x
di=dx5
dx5x=dt4
15lnx=t4+c
15ln(85i)=t4+c
At t=0;i=0
15ln8=c
15ln(85i)=t415ln8
15ln815ln(85i)=t4
15ln(885i)=t4(lnab=lnalnb)
ln(885i)=5t4
885i=e5t4
85i=8e5t4
5i=88e5t4
usaho4w

usaho4w

Beginner2022-01-02Added 39 answers

Let I be current at ant time t
EMF E=5 volts, resistance R=50 ohms and inductance L=1 henry
The basic equation of current for RL circuit is dIdt+RLI=EL
dIdt+50I=5
This is a linear equation and the solution of it is I=ce50t+110
At t=0,I=0. So we get c=110.
Therefore the equation for current will become I=110e50t+110
As t tends to infinite , the value of I tends to 110
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

L=20HR=10Ωi(t)=VR[1eRTL]di(t)dt|t=0=RLVR[1eRTL]12×510e0=140.25Amp/s

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