# A Resistance of 100 Ohms and Inductance of 0.5 Henries Are Connected in Series with a Battery of 20 Volts. Find the Current at Any Instant If the Relation Between L,R,E is L D I D T + R I = E - Applied Mathematics 2

Sum

A resistance of 100 ohms and inductance of 0.5 henries are connected in series With a battery of 20 volts. Find the current at any instant if the relation between L,R,E is L (di)/(dt)+Ri=E.

#### Solution

L(di)/(dt)+Ri=E.

therefore (di)/(dt)+(Ri)/L=E/L

Solution is given by ,

i,e^(int(R/L)dt=inte^(int(R/L)dt)E/Ldt+c

therefore"i".e^((Rt)/L)=(Ee^((Rt)/L))/R+c

At t = 0, i = 0 thereforec=E/R

therefore "i".e^((Rt)/L)=(Ee^((Rt)/L))/R+(-E)/R

therefore "i" = E/R(1-e^(-(Rt)/L))

For given condition R = 100, L = 0.5, E = 20

therefore i=0.2(1-e^(-200t))

Concept: Linear Differential Equation with Constant Coefficient‐ Complementary Function
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