First Order Circuit Capacitors and inductors RC and RL circuits

RC and RL circuits (first order circuits) Circuits containing no independent sources Circuits containing independent sources Complete response = Natural response + forced response ‘source-free’ circuits Excitation from stored energy Natural response DC source (voltage or current source) Sources are modeled by step functions Step response Forced response

RC circuit – natural response Assume that capacitor is initially charged at t = 0  v c (0) = V o  Taking KCL,  Objective of analysis: to find expression for v c (t) for t >0 i.e. to get the voltage response of the circuit  +vc+vc icic iRiR R C OR

RC circuit – natural response Can be written as ,  = RC  time constant This response is known as the natural response  Voltage decays to zero exponentially  At t= , v c (t) decays to 37.68% of its initial value  The smaller the time constant the faster the decay VoVo t=  t v C (t) V o

RC circuit – natural response The capacitor current is given by:  And the current through the resistor is given by The power absorbed by the resistor can be calculated as: The energy loss (as heat) in the resistor from 0 to t:

RC circuit – natural response As t  , E R  As t  , energy initially stored in capacitor will be dissipated in the resistor in the form of heat

RC circuit – natural response PSpice simulation +vc+vc icic iRiR R C 0 1 RC circuit c e-6 IC=100 r tran 7e-6 7e-3 0 7e-6 UIC.probe.end

RC circuit – natural response PSpice simulation +vc+vc icic iRiR R C 0 1 RC circuit.param c=1 c1 1 0 {c} IC=100 r step param c list 0.5e-6 1e-6 3e-6.tran 7e-6 7e-3 0 7e-6.probe.end c1 = 3e-6 c1 = 1e-6 c1 = 0.5e-6

RL circuit – natural response Assume initial magnetic energy stored in L at t = 0  i L (0) = I o  Taking KVL,  Objective of analysis: to find expression for i L (t) for t >0 i.e. to get the current response of the circuit  iLiL vL+vL+ R L +vR+vR OR

RL circuit – natural response Can be written as ,  = L/R  time constant This response is known as the natural response  Current exponentially decays to zero  At t= , i L (t) decays to 37.68% of its initial value  The smaller the time constant the faster the decay IoIo t=  t i L (t) I o

RL circuit – natural response The inductor voltage is given by:  And the voltage across the resistor is given by The power absorbed by the resistor can be calculated as: The energy loss (as heat) in the resistor from 0 to t:

RL circuit – natural response As t  , E R  As t  , energy initially stored in inductor will be dissipated in the resistor in the form of heat

RL circuit – natural response PSpice simulation 0 1 RL circuit L IC=10 r tran 7e-6 7e-3 0 7e-6 UIC.probe.end vL+vL+ R L +vR+vR

RL circuit – natural response PSpice simulation L1 = 3H L1 = 1H L1 = 0.5H RL circuit.param L=1H L1 0 1 {L} IC=10 r step param L list tran 7e-6 7e-3 0 7e-6 UIC.probe.end 0 1 vL+vL+ R L +vR+vR