First–Order CircuitsFirst–Order Circuits Example 1: Determine the voltage . ()ovt Solution: This is a first order circuit containing an inductor. First, determine . ()Lit Consider the circuit for time t < 0. Step 1: Determine the initial inductor current. The circuit will be at steady state before the source voltage changes abruptly at time 0t=. The source voltage will be 2 V, a constant. The inductor will act like a short circuit. ()()L220010|| 25 15 8i ===+.25 A 0t<, at steady state: Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time . Determine the Norton equivalent circuit for the part of the circuit connected to the inductor. 0t = Replacing the resistors by an equivalent resistor, we recognize oc6 Vv =− and t8 R =ΩConsequently sc60.75 A8i−==−Step 3. The time constant of a first order circuit containing an inductor is given by tLRτ= Consequently t40.5 s8LRτ=== and 112 saτ== Step 4. The inductor current is given by: () ( )()()()22scLsc0 0.75 0.25 0.75 0.75at t tit i i ie e e−−=+ − =− + −− =− +− for 0t ≥ Step 5. Express the output voltage as a function of the source voltage and the inductor current. Using current division: ()RL100.210 25 15ii==++LiLi Then Ohm’s law gives oR15 3vi== Step 6. The output voltage is given by ()2o2.25 3tvt e−=− + for 0t ≥Example 2: Determine the current . ()oit Solution: This is a first order circuit containing a capacitor. First, determine . ()Cvt Consider the circuit for time t < 0. Step 1: Determine the initial capacitor voltage. The circuit will be at steady state before the source voltage changes abruptly at time 0t=. The source voltage will be 5 V, a constant. The capacitor will act like an open circuit. Apply KVL to the mesh to get: ()xx110 2 3 5 0 A3ii++ −= ⇒ = Then ()Cx03 1 Vvi== 0t<, at steady state: Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time . Determine the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. First, determine the open circuit voltage, : 0t =ocv Apply KVL to the mesh to get: ()xx10 2 3 15 0 1 Aii++ − = ⇒ = Then oc x33vi V== Next, determine the short circuit current, : sciExpress the controlling current of the CCVS in terms of the mesh currents: x1siiic=− The mesh equations are ()()1 1 sc 1 sc 1 sc10 2 3 15 0 15 5 15iii ii ii+−+−−=⇒ −= And ()sc 1 sc 1 sc4303iii i−−=⇒=i so sc sc sc415 5 15 1 A3ii i⎛⎞−=⇒=⎜⎟⎝⎠ The Thevenin resistance is t33 1R==Ω Step 3. The time constant of a first order circuit containing an capacitor is given by tRCτ= Consequently t130.212RCτ⎛⎞== =⎜⎟⎝⎠5 sand 114 saτ== Step 4. The capacitor voltage is given by: () ( )()()44CCoc oc03133at t tvt v v v e e e2−−−=+ − =+− =− for 0t ≥ Step 5. Express the output current as a function of the source voltage and the capacitor voltage. () () ()oCC112ddit C vt vtdt dt== Step 6. The output current is given by ()()()()444to11232 2 412 12 3ttdit e e edt−−−=−=−−=0≥ for