math4b-14 - Math 4B Lecture 14 Doug Moore Our next goal is...

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Math 4B Lecture 14 May 15, 2013 Doug Moore
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. Our next goal is to consider the circuit given by a resistor of resistance R , an inductor of inductance L , and a capacitor of capacitance C , connected in series to a source of electromotive force E ( t ).
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. Let x ( t ) denote the current at time t . As the current flows through the circuit elements, the voltage drops in accordance with the following rules, which are discussed in more detail in courses on electrical circuit theory: The voltage drop across the resistor is E R ( t ) = Rx ( t ). The voltage drop across the inductor is E L ( t ) = L ( dx/dt )( t ). The voltage drop across the capacitor is E C ( t ) = (1 /C ) q ( t ), where q ( t ) is the charge on the capacitor. (By definition of current, dq/dt = x .)
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According to Kirchhoff’s law , the sum of voltage drops around the circuit is equal to the voltage applied to the circuit, in other words E R ( t ) + E L ( t ) + E C ( t ) = E ( t ) , or equivalently, Rx + L dx dt + 1 C q = E ( t ) . If R , L , and C are constant, we can differentiate this equation obtaining L d 2 x dt 2 + R dx dt + 1 C x = E 0 ( t ) , a nonhomogeneous linear equation, for which the associated ho- mogeneous equation having constant coefficients.
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A particular choice of the differential equation, L d 2 x dt 2 + R dx dt + 1 C x = E 0 ( t ) , corresponding to the appropriate choice of initial conditions, might be regarded as the response of the circuit to the electro- motive force E ( t ).
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  • Winter '14
  • Math, Equations, Elementary algebra, dt, Electrical network, Voltage drop

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