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Unformatted text preview: The RLC Circuit The RLC circuit is the electrical circuit consisting of a resistor of resistance R , a coil of inductance L , a capacitor of capacitance C and a voltage source arranged in series. We’re going to think of the voltage x ( t ) +- x ( t ) R L C i ( t ) +- y ( t ) as an input signal and the voltage y ( t ) as an output signal. The goal is to determine the output voltage for a given input voltage. If i ( t ) is the current flowing at time t in the loop as shown and q ( t ) is the charge on the capacitor, then the voltage across R , L and C , respectively, at time t are Ri ( t ), L di dt ( t ) and y ( t ) = q ( t ) C . By the Kirchhoff’s law that says that the voltage between any two points has to be independent of the path used to travel between the two points, these three voltages must add up to x ( t ) so that Ri ( t ) + L di dt ( t ) + q ( t ) C = x ( t ) (1) Assuming that R, L, C and x ( t ) are known, this is still one differential equation in two unknowns, i ( t ) and q ( t ). Fortunately, there is a relationship between the two. Namely i ( t ) = dq dt ( t ) = Cy ( t ) (2) This just says that the capacitor cannot create or destroy charge on its own. All charging of the capacitor must come from the current. Subbing (2) into (1) gives LCy 00 ( t ) + RCy ( t ) + y ( t ) = x ( t ) (3) For an ac voltage source, choosing the origin of time so that x (0) = 0, x ( t ) = E sin( ωt ) and the differential equation becomes LCy 00 ( t ) + RCy ( t ) + y ( t ) = E sin( ωt ) (4) One Solution We first guess one solution of (4) by trying y p ( t ) = A sin( ωt- ϕ ) with the amplitude A and phase ϕ to be determined. That is, we are guessing that the circuit responds to an oscillating applied voltage with a current that oscillates at the same frequency. For y p ( t ) to be a solution, we need LCy 00 p ( t ) + RCy p ( t ) + y p ( t ) = E...
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- Spring '10
- RLC, RC circuit, general solution, e-t c1 cos