What if a resistor is inserted into an LC circuit?

# What if a resistor is inserted into an LC circuit? - o-iR =...

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Physics 202 Tuesday, April 20, 1999 Announcements: The final exam will be 110 minutes long. It will consist of 25 multiple choice questions. 15 of the 25 will be on new material with the remaining 10 on material covered on the first and second midterm. Lecture notes: Continued from last lecture... Solution for d 2 q/ dt 2 = -q /LC q = Q cos( ϖ t + δ ) where δ is the initial condition ϖ = (1/ LC) 1/2 Since we said the capacitor is charged to a maximum value of Q at t= 0, δ = 0 i = dq/dt = -Q ϖ Sin ( ϖ t) = -i max Sin ( ϖ t) U = U E + U B = q 2 /2C + (1/2)Li 2 = Q 2 /2C cos 2 ( ϖ t) + (1/2)L (i max ) 2 sin 2 ( ϖ t) = Q 2 /2C cos 2 ( ϖ t) + (1/2)L( Q 2 /LC)sin 2 ( ϖ t) = (1/2) L I 2 max See problem 33-25 What if a resistor is inserted into an LC circuit? As current flows through the resistor, electrical energy is dissipated as heat Without R; U = U E + U B = q 2 /2C + (1/2)Li 2 = a constant With R: dU/dt = -i 2 R = d/dt [Li 2 /2 + q 2 /2C] o or... -i 2 R = Li di/dt + q/c dq/dt

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Unformatted text preview: o-iR = L di/dt + q/c o L di/dt + q/c + iR = 0 o L d 2 q/dt 2 + q/c + R dq/dt = 0 • Solution? o q = Qe-Rt/2L cos ( ϖ l t + δ ) Charge on a capacitor o ϖ l = ( ϖ 2- (R/2L) 2 ) 1/2 where ϖ = (1/ LC) 1/2 o U = Q 2 /2C e-Rt/L AC Circuits • We learned that by rotating a coil in a magnetic field, a sinusoidal emf can be generated. What if such a generator (instead of a battery) is placed in a circuit? • Resistive load • ξ = ξ max sin ϖ t = v R o Let's assume that instantaneously the current i R is clockwise, i R = v R / R o By the loop rule, ξ- v R = 0, therefore v R =V R sin ϖ t = ξ max sin ϖ t and i R = I R sin ϖ t = (v R / R) sin ϖ t o v R and i R are said to be phase, T = one period o Phasor Diagram • i R = I R sin ϖ t and v R =V R sin ϖ t ( or the projection of the phasors on y axis)...
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