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iren [92.7K]
3 years ago
13

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

s (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) =
Mathematics
1 answer:
ser-zykov [4K]3 years ago
5 0

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})

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