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EXAMPLE 4 Consider the electric circuit shown in the figure and modeled by the differential equation below. [ L frac{d I}{d t}+R I=E(t) ] Find an expression for the current in a circuit where the resistance is 8 ?, the inductance is 4 H, a battery gives a constant voltage of 8 V, and the switch is turned on when t = 0. What is the limiting value of the current? SOLUTION With L = 4, R = 8, and E(t) = 8, the equation becomes [ 4 frac{d I}{d t}+8 I=8 quad ext{or} quad frac{d I}{d t}=2-2 I ] and the initial value problem is [ frac{d I}{d t}= 2-2 I quad I(0)=0. ] We recognize this equation as being separable, and we solve it as follows: [ int frac{d I}{2-2 I} = int d t quad (2-2 I eq 0) ] [ -frac{1}{2} ln |2-2 I| = t+C ] [ |2-2 I| = e^{-2(t+c)} ] [ 2-2 I = pm e^{-2(t+c)} = A e^{-2 t} ] [ I = 1 - 1/2 A e^{-2 t}. ] Since I(0) = 0, we have 1 - A/2 = 0, so A = 2 and the solution is [ I(t) = 2-2e^{-2t} ] The limiting current, in amperes, is [ lim _{t ightarrow infty} I(t) = lim _{t ightarrow infty} (1 - 1 e^{-2t}) = 1 - 1 lim _{t ightarrow infty} e^{-2t} = 2 ]

          EXAMPLE 4 Consider the electric circuit shown in the figure and modeled by the differential equation below.
[ L frac{d I}{d t}+R I=E(t) ]
Find an expression for the current in a circuit where the resistance is 8 ?, the inductance is 4 H, a battery gives a constant voltage of 8 V, and the switch is turned on when t = 0. What is the limiting value of the current?
SOLUTION With L = 4, R = 8, and E(t) = 8, the equation becomes
[ 4 frac{d I}{d t}+8 I=8 quad 	ext{or} quad frac{d I}{d t}=2-2 I ]
and the initial value problem is
[ frac{d I}{d t}= 2-2 I quad I(0)=0. ]
We recognize this equation as being separable, and we solve it as follows:
[ int frac{d I}{2-2 I} = int d t quad (2-2 I 
eq 0) ]
[ -frac{1}{2} ln |2-2 I| = t+C ]
[ |2-2 I| = e^{-2(t+c)} ]
[ 2-2 I = pm e^{-2(t+c)} = A e^{-2 t} ]
[ I = 1 - 1/2 A e^{-2 t}. ]
Since I(0) = 0, we have 1 - A/2 = 0, so A = 2 and the solution is
[ I(t) = 2-2e^{-2t} ]
The limiting current, in amperes, is
[ lim _{t 
ightarrow infty} I(t) = lim _{t 
ightarrow infty} (1 - 1 e^{-2t}) = 1 - 1 lim _{t 
ightarrow infty} e^{-2t} = 2 ]

$EXAMPLE 4 Consider the electric circuit shown in the figure and modeled by the differential equation below. [ L fracd Id t+R I=E(t) ] Find an expression for the current in a circuit where the resistance is 8 ?, the inductance is 4 H, a battery gives a constant voltage of 8 V, and the switch is turned on when t = 0. What is the limiting value of the current? SOLUTION With L = 4, R = 8, and E(t) = 8, the equation becomes [ 4 fracd Id t+8 I=8 quad extor quad fracd Id t=2-2 I ] and the initial value problem is [ fracd Id t= 2-2 I quad I(0)=0. ] We recognize this equation as being separable, and we solve it as follows: [ int fracd I2-2 I = int d t quad (2-2 I eq 0) ] [ -frac12 ln |2-2 I| = t+C ] [ |2-2 I| = e^-2(t+c) ] [ 2-2 I = pm e^-2(t+c) = A e^-2 t ] [ I = 1 - 1/2 A e^-2 t. ] Since I(0) = 0, we have 1 - A/2 = 0, so A = 2 and the solution is [ I(t) = 2-2e^-2t ] The limiting current, in amperes, is [ lim t ightarrow infty I(t) = lim t ightarrow infty (1 - 1 e^-2t) = 1 - 1 lim t ightarrow infty e^-2t = 2 ]$

Added by Jay K.

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