ECE 231 -11

# ECE 231 -11 - ECE-231 Circuits and Systems I Spring 2011...

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ECE-231 Circuits and Systems I Spring 2011 Session 11 Professor Stewart Personick Office: ECEC Room 321 Stewart.Personick@NJIT.edu

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Solving for the response of an RLC circuit to a time-varying source
From: HW#8 Problem #1 (RL Circuit) is= 1,000A R1 = 0.01 Both switches are closed for all times: t<0 seconds. At time t= -1 second, the current i(t) = 0. Switch A opens at time: t=0 seconds, and remains open thereafter. At time: t=2 seconds, the switch B opens ; and remains open thereafter. Calculate and plot (on graphs that are neatly drawn) the current i(t), and the voltage Vab(t), as a function of time, for values of time between: t=-1 second and t= 5 seconds. How much energy (Joules) is stored in the inductor at each of the following times: t=-0.000001 second, t=+.000001 second, t= 2 seconds, t= 5 L= .005H R2 =0.01 a b t< 0s i(t)= 0A i(t) Switch B Switch A 1,000A 1,000A Vab(t)= 0V

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Initial current at t=0s is 0A Target current for 0s<t<2s is1000A Target current for 2s<t is 0A 1000A Initial current at t=2s is 865A
General Example (RLC Circuit) C1= 0.01 μ F R2= 2000 Node B Node C: VC= 0 volts i1(t) = ? L1 = 100 μ H Node A VB (t) = ? is(t) R1= 2000 Possible solution approaches: A) Use analysis for special cases. E.g. if is(t) vs. t is: a step function, or a sine wave, or a complex exponential function; or if is(t) is a constant, Is , and there is a switch that opens or closes at some time: t=t0 B) Use computer-based numerical [e.g. f (t + t) = f (t) + f ’(t) t] integration methods

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Time Domain Analysis of the response of an RLC Circuit to a current source whose value is(t) vs. t is a step function 0.01A t is(t) t=0 s is(t) is(t) = .01A for t<0s is(t) =0A for t>0s
Example 1 C1= 0.01 μ F R2= 2000 Node B Node C: VC= 0 volts i1(t) = ? L1 = 100 μ H R1= 2000 Node A VB (t) = ? is(t) = .01A for t<0s is(t) =0A for t>0s 0.01A t is(t)

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Example 1 C1= 0.01 μ F R2= 2000 Node B Node C: VC= 0 volts i1(t) = ? In the steady state (t<0s), the current flowing through the inductor L1 is constant. I.e. d[i1(t)]/dt = 0 . Therefore the voltage across the inductor is 0 volts In the steady state (t<0s), the voltage across the capacitor C1 is constant. I.e. d[VB (t)]/dt = 0 Therefore the current flowing through the L1 = 100 μ H R1= 2000 Node A VB (t) = ? is(t) = .01A for t<0s
Example 1 R2= 2000 R1= 2000 Node C: VC= 0 volts Node A Node B VB (t) = 10 Volts i1(t) = 0.005A In the steady state (t<0s), the current flowing through the inductor L1 is constant. I.e. d[i1(t)]/dt = 0 . Therefore the voltage across the inductor is 0 volts In the steady state (t<0s), the voltage across the capacitor C1 is constant. I.e. d[VB (t)]/dt = 0 Therefore the current flowing through the is(t) = .01A for t<0s

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Example 1 C1= 0.01 μ F R2= 2000 Node B Node C: VC= 0 volts i1(t) = 0.005A In the steady state
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## ECE 231 -11 - ECE-231 Circuits and Systems I Spring 2011...

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