EE 221 Final Exam Name: ____________________________________ 12/09/2002 _____ / 100 pts You need to show all of your steps in detail to get full credit! 1. Determine the current I using current division. Find the voltage V and the power P supplied by the source. (8 pts) + IS 12A V I 6Ω 18Ω 18Ω 3Ω − GND I = (1/18) / (1/18 + 1/18 + 1/9) * (−12A) = 1/4 (−12A) = −3 A V = I R = −3A 18Ω = −54 V Psupplied = −V I = −(−54V) (12A) = 648 W I = _−3 A__ V = _−54 V__ P = _648 W__ Example 1 of 10 2. Given the following circuit, determine the voltage V and current I. (8 pts) R1=20Ω R3=45Ω VS=24V R4=90Ω + − R2=50Ω − V I + R’ = R3 || R4 = R3 R4 / (R3 + R4) = 45 90 / 135 = 30 Ω V’ = VS R’ / (R1 + R’ + R2) = 24V 30/(20 + 30 + 50) = 24V 3/10 = 7.2 V I = V’ / R4 = 7.2V / 90Ω = 80 mA V = VS R2 / (R1 + R’ + R2) = 24V 50/100 = 12 V I = __80 mA__ V = _12 V__ Example 2 of 10 3. For the circuit shown below, use nodal analysis to find a consistent set of equations using the assigned node voltages V1, V2, V3, and V4. Do not substitute, simplify, or solve. V2 V1 3IR3 R2 V3 + − R1 R3 IR3 − + Vs (12 pts) Is V4 R4 GND A) (KVL) V2 = −VS B) (Supernode KCL) (V1 − V2) / R1 + V4 / R4 + (V4 − V3) / R3 = 0 C) (Supernode KVL) V1 − V4 = 3 IR3 = 3 (V3 − V4) / R3 D) (KCL) (V3 − V2) / R2 + (V3 − V4) / R3 = IS Example 3 of 10 4. Use mesh analysis in the phasor domain to find a consistent set of equations using the assigned mesh currents I1, I2, I3, and I4. Do not substitute, simplify, or solve. Use ω = 100 rad/s. (12 pts) + − 5I I4 20Ω 4Ω 20mH − + I2 I1 5mF I3 I GND 5 cos(ω t+50º)A 4 cos(ω t+25º)V Circuit into phasor-domain: L: 20mH ⇒ jωL = j 100 0.02 = j2 Ω C: 5mF ⇒ 1/jωC = −j 1/(100 0.005) = −j2 Ω Signals into phasor-domain: I: 5 cos(ωt + 50°)A ⇒ 5∠50° A V: 4 cos(ωt + 25°)V ⇒ 4∠25° V A) (KCL) I1 = −5∠50° A B) (KVL) −4∠25° V + 20Ω (I2 − I1) + j2Ω (I2 − I4) = 0 C) (KVL) 4∠25° V − j2Ω I3 = 0 D) (KVL) 5 I + 4Ω I4 + j2Ω (I4 − I2) = 0 where I = I3 Example 4 of 10 5. Find the output voltage VOUT of the following ideal OpAmp-circuit. (10 pts) R3=3kΩ R1=6kΩ VOUT Vs=8V + − R2=5kΩ R4=5kΩ V + = V- = V A) KCL at (-) node: (V − VS) / 6k + (V − VOUT) / 3k = 0 B) KCL at (+) node: V / 5k + (V-VOUT) / 5k = 0 ------------ A) V/2 − VS/2 + V − VOUT = 0 B) 2V − VOUT = 0 -----------B) ⇒ A) 3/2 VOUT/2 = VOUT + VS/2 VOUT (3/4 - 1) = VS/2 VOUT = −2 VS = −16 V VOUT = ___−16 V___ Example 5 of 10 6. The current through an inductor (4H) is shown below. Sketch the voltage across the inductor’s terminals and determine the energy stored in the inductor at t = 2.5s. (10 pts) i(t) in A 1.5 1 1 3 5 t (s) 1 3 5 t (s) v(t) in V 4 2 -2 -4 uL = L di / dt = 4H ∆I / ∆t time (s) 0-1 1-2 2-3 3-4 4-5 ∆I / ∆t (A/s) 1 ½ 0 -1/2 -1 uL (V) 4 2 0 -2 -4 w(2.5s) = ½ L I(2.5)2 = ½ 4H (1.5A)2 = 2 9/4 J = 4.5 J w(2.5s) = __4.5 J__ Example 6 of 10 7. For the circuit below, find the equivalent input impedance of the network at ω = 20 rad/s. (10 pts) 25mF 2Ω 50mF 0.2H Circuit into phasor-domain: L: 200mH ⇒ jωL = j 20 0.2 = j4 Ω C: 25mF ⇒ 1/jωC = −j 1/(20 0.025) = −j2 Ω C: 50mF ⇒ 1/jωC = −j 1/(20 0.05) = −j Ω Z1 = j4Ω || -jΩ = j4 (−j) / (j4 − j) = −j4/3 Ω ZEQU = 2Ω − j2Ω − j4/3Ω = 2Ω − j10/3Ω = 2Ω − j3.33Ω ZEQU = ___2Ω − j10/3Ω___ Example 7 of 10 8. For the circuit shown below, find the Thevenin- and Norton equivalent circuit (VTH, IN, and ZTH). Find the load impedance ZL that maximizes it’s own power consumption. Compute the value of the maximum power transfer (P). (10 pts) IS -j20Ω 20Ω + − Vs=100∠0ºV j10Ω IS 10Ω VTH: (open terminals) (KCL, 1 equation in 1 unknown) 2 (100∠0 − VTH) / (20 − j20Ω) = VTH / (10 + j10Ω) ⇒ 2 100∠0 / (20 - j20Ω) = VTH ( 1/(10 − j10Ω) + 2/(20 − j20Ω) ) ⇒ VTH = (20 + j20Ω)/40 100∠0 = (1/2 + j/2Ω) 100∠0 = 100/√2∠45 ISC: (terminals shortened) (KCL, 1 equation in 1 unknown) IN = 2 IS = 100∠0 / (20 - j20Ω) = 100∠0 / (√200∠-45) = 100/√200 ∠45 A use ZTH = VTH / ISC = (100/√2∠45) / (100/√200 ∠45 A) = 10 Ω ZL = ZTH* = 10 Ω P = ½ (VTH/2)2 / RL= 1/8 1002/2 1/10 = 62.5 W VTH = ___100/√2∠45__, IN = __100/√200 ∠45 A__, ZTH = _____10 Ω____ ZL = ____10 Ω___, P = _____62.5 W___ Example 8 of 10 9. From the figure below, determine a) the admittance Y b) the functions v(t) and i(t) in the time domain (ω = 120π rad/s) c) the value of the power factor d) is the PF leading or lagging? e) is the load more inductive or capacitive? f) the average and reactive power absorbed by the impedance. I + (10 pts) Im I = 45ARMS V Z V = 60VRMS 110° 60° − Re ω = 2πf = 377 rad/s V: V = 60∠60 VRMS ⇒ v(t) = 60 √2 cos(377t + 60º) V I: I = 45∠110 ARMS ⇒ i(t) = 45 √2 cos(377t + 110º) A Y = I / V = 45∠110º / 60∠60º = 0.75∠50º S ϕLOAD = −50º (voltage lags current by 50º ⇒ capacitive load) PF = cos(ϕLOAD) = cos(−50º) = 0.643 leading S = V I* = 60∠60º 45∠-110º = 2700∠-50º = 1735.5 W − j2068.3 VAR Y = __0.75∠50º S_____ v(t) = _60 √2 cos(377t + 60º) V__ PF = _0.643 leading__ ......capacitive.... load P = ____1735.5 W__ i(t) = _45 √2 cos(377t + 110º) A_ Q = _−2068.3 VAR__ Example 9 of 10 10. Find the total average and reactive power consumed by the two loads (impedances). How much reactive power must a capacitor absorb to achieve a PF = 1. Determine the capacitor’s value assuming ω = 120π rad/s. Z1: absorbs 2kW and 1kVAR (10 pts) Z2: absorbs 1kVA at 0.8 lagging + 100VRMS Z1 Z2 C − S1 = 2 kW + j1kVAR S2 = 1000 * 0.8 W + j 1000 sin(cos-1(0.8)) VAR = 800 W + j 600 VAR S = S1 + S2 = 2800 W + j 1600 VAR capacitor must absorb -1600VAR to achieve PF = 1 SC = V I* = V V* / Z* = V2 / (-j/ωC)* = -jωCV2 -1600 = -120π C 1002 ⇒ C = 1600 / (120π 1002) = 424.4 µF P = ___2800 W__ Q = _1600 VAR__ QC = _-1600 VAR__ C = ___424.4 µF__ Example 10 of 10