Question

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volts (time t in seconds). The current through the circuit I(t) (in amperes) satisfies the differential equation:dI/dt+(R/L)I=V(t)/L
Find the solution to this equation with the initial condition I(0) = 0, assuming that R = 150 Ω, L= 5 H, and V(t) is constant with V(t) = 10 V.
I(t) =

Answer 1

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$