EECE 290 Problems For Quiz 2, Spring 2012

EECE 290 - Selected Problems for Quiz 2 Spring 2011-2012 Excited First-Order Circuits 5. A voltage impulse Kt) Vs is applied at t= 0 with zero + Vo - initial conditions in the circuit. If K = 10, determine Vc | | at t=0". + 1F . . Kat) Vs C) S10 Q Solution: The voltage impulse appears across R and = causes a current impulse of 10.4(f)/R through C. The voltage across C at f = 0° is 104{¢)/RC. Alternatively, the voltage source and 10 © resistor are transformed to a current source of 104(f)/R in parallel with R and a 1 F capacitor. The current impulse causes a jump in voltage of 104f)/RC = 1V across the capacitor at f = 0°. 1F a ' oo 4H 47. v,(t) = d(t) is applied to the circuit shown, 108) | L with zero initial conditions. Determine /, (t) . ( P ZA, etu(t)A vid C) 12 ix 2e*u(t) A 4eu(t)A yi Be tu(t)A None of the above moo o Switched First-Order Circuits + Sf. 7. Cyis initially uncharged and the switch has al been in position “a” for a long time. The switch @) io is moved to position “b” at t= 0. Assuming 15pF iF} _ Vsre = 10 V dc, determine: (a) Vc; at t = 0°; (b) | Voy as t > 00; (C) Vo,(f) for t= 0°. Solution: (a) After the switch has been in position “a” for a long time, v,, at t=O" is Vspc¢ = 10 V, and remains at this voltage at t = 0°. 1x15 25 WN 5kQ Cc 1 Ver (Db) Cogs = =0.6 uF; Vigs = Vere V, SO that gage = 6 V. The initial charge on C; is 1x10 0.410 -4Vv. uC. The final charge on C; is (1 — 0.6)10= 4 wC. Hence, vc; as [> wis (c) r= 5 kOx0.6 uF = 3 ms. Hence, for t> 0°, Voi(t) = Vege0.4+(1-0.4)e*?] = 10/0.4 +0.6e~*|, where tis in ms. 1. Given an inductor of 2 H and 5 © resistance carrying an initial current of 1 A and another inductor of 3 H and 5 resistance carrying an initial current of 2 A. At f= 0, the two inductors are connected in a loop, the directions of initial currents being in the same sense around the loop. Determine the current in the two inductors for f > 0°. Solution: Flux linkage is conserved at the instant of switching. The initial flux linkage is 2x1 + 3x2 = 8 Wb-turns. The initial current through both inductors is 8/(2 + 3) = 1.6 A; r= 5/(5 + 5) s and 1/r=2"'; hence, i(t) = 1.6e** Problem 2 f= isa RQ 30 AN AWN AW + x0 VeRQ 02H», 2600 2200 WA 22 The switch has been closed a long time. At t=0, it is opened. Find v(t) for t20 given R=100. Vo(t) = —30e FV € V_(t) = —30e- ty vy(t) = —15e7 Vv vy (t) = —15¢e76- ty None of the above ganoro 3. The switch has been in position “a” for a long time. It is moved to position “b” 2 0 1A 4A R 2H3“% at t= 0, adding 1 A to the 4 A current source. If R= 5, determine y, at f= = 0. Solution: At f = 0°, the current in the inductor does not change. The additional current must therefore pass though R, making v,(0*) = 1xRV =5V. Both switches have been closed for a long time. At t=0, both switches open simultanously. Find i,{t) for t20 given R=8. ig(t) = 10.72e-*' ma ig(t) = 10.72¢-5°° ma i,(t) = 10e7** mA i,(t) = 10e5° ma € None of the above 2. Consider the circuit below. Switches K] and K2 open at t= 0, after being closed for a long time. Find final values of voltages V1 and V2 respectively across C1 and C2.°C1= 1p F, C2= 2uF panes >a) VI=+4V, V2= -4V b) V1=+6V, V2= -6V c) VI=+8V, V2=-8V d) V1= 0V, V2= 0V e) None of the above 13. An inductor L2 = 2H, an inductor L1 = 1H and a resistor R = 6Q are connected in series to a voltage source v(t) = 2 t e“u(t). The initial energy stored in L2 was 25J. No initial energy was stored in L1. Find the current i(t) in this circuit at t = 0. i(t) ere Ee pein 12 eo i(t) = [4 5(t) + 4e7 + 2/3 te™] ult) i(t) =[4e7 -2/3 e*-2/3 te“) u(t) it) =[4e7+2/3 e+ 2/3 te*] u(t) i(@t) =[4e7 - 2/3 ej u(t) None of the above epa0cp The switch in the circuit has been in position x for a long time. At t= 0, the switch moves instantaneously to position y. Find: a) v(t) fort >=0 b) volt) for t>= 0" c) (ft) fort >=0° d) The total energy dissipated in the 60 KQ resistor 10kN x 100V + 8, $240 kN $ 60 kD 05 uF - ve a) Because the switch has been in position x for a long time, the 0.5 uF capacitor will charge to 100 V and be positive at the upper terminal. We can replace the capacitor at t= 0° with an equivalent resistance of 80 KQ. Hence the time constant of the circuit is: (0.5*10°)*(80*10° )=40ms then: v.(t)= 100 e?* V, 0 b) The easiest way to find v,(t) is to note that the resistive circuit forms a voltage divider across the terminals of the capacitor. Thus. vo(t) = (48 / 80 ) v(t) = 60 eV, t= 0" vo(t) valid for t >= 0” because vo(0-) is zero. Thus we have an instantaneous change in the voltage across the 240 KQ resistor. c) Using ohms law: io(t) = ( vo(t) / 60*10° y= eo 'mA. t= 0" d) The power dissipated in the 60 KQ resistor is: Peo xo(t) = io(t) * (60*10°) = 60e°" mW, t>= 07 The total energy dissipated is: Weoxo= Jo io(t)*( 60*10°) dt= 1.2 mJ 13. The switch closes at 0. Find v(t) in volts for t= 0, tis in seconds. 2002 0.067" PA. 2.25 + 0.75¢ 0.033 B. 2.75 + 0.5¢ O05 C.2.5+15¢ it D.3+2¢ E. None of the above 14. The switch opens again at t = 10sec. Find v(t) for t = 10sec. -0.067(t - 10) A.2-0.63e -0.038(t— 10) B.4-0.6e -0.067(t- 10) C.2.5-0.5e -0.05(t- 10) >D.3-0.37e E. None of the above. 21. At time t=0 the switch move Vi=10V. V2=20V, Ri=SKQ., : a) V=10V b) V.=20V c) V=-10V >d) V=-20V e) None of the above ny Open t=6 Close t=3 Vs 0.02uf aT Ov R1 Ve ~ 10.06uf R2 = 19. If V, = -9V, Ry=5KQ, R9=2KQ, R3=2KQ, then the voltage V, across the capacitors at R1: time t=0 is: vic a) V.=-9V — Sb) V=9V c) V.=20V d) V.= -20V e) None of the above 20. In the circuit shown above find in milliseconds the value of tau for the time interval 35 t<6. >a) tau= 0.08 msec b) tau= 0.48 msec c) tau= 0.96 msec d) tau= 0.64 msec e) None of the above The switch in the circuit has been im position | for a long time. At t =0. the switch moves to position 2.Find a) volt) fort >=0 b) igft) fort >= 0" iV a) The switch has been in position | for a long time. so the initial value of v0 is 40( 60/80 ) = 30 V. Using Norton equivalent: Vee = (( 160 * 10?) / (( 40+ 160) * 10°) ) *(-75) =-60V To find Rg. we short circuit the -75V source and making series and parallel combinations of the resistors: Re = 8000 + ( 40000 || 160000 ) = 40 KQ The time constant: RC = 10 ms So the solution for vo(t) become: volt) = -60 + (30 = (-60 ) jt e'™ = -60+90e™™ V, =0 b) ig =c (dv / dt) = (0.257 10-6 -9000e"™) —The switch in the circuit has been open for a long time. The initial charge on the capacitor is zero, At t =0, the switch is closed. Find the expression for: a) i(t) fort >=0” b) v(t) when t>= 07 a) The initial voltage on the capacitor is zero. after switching, i(0.) = ( (7.5*20) /50)=3 mA The final value of the capacitor current will be zero because the capacitor will become an open circuit. Thus ig= 0 The time constant: Ry,*C = (20 + 30)* 10°*(0.1)*10° = Sms i(t) =0 + 3 — Oe = 36° mA, P= 07 b) The initial voltage across the capacitor is zero The final voltage across the capacitor is: (7.5)*(20) = 150V The time constant: 5 ms V(t) = 150 + (0-150) eo? =(150—150e7") Vv, t>=0 Therefore v(t) become: v(t) = 150 — 15027 + (30)(3)e7™ =, Find the expression for: a) v(t) when t>= 0+ b) i(t) when = 0 it) — 4 20v{ ” ) ci) 330mH >. a) The initial current in the inductor is 5 A After the switch closes, the current still is 5 A, and the initial voltage across the inductor becomes ( 20 - 5(1) )=15 V The final value of the inductor voltage is 0 V The time constant is ($0*107/ 1) = 80 ms V(t) =0 + (15-0 eS) = 15e ty, t=0+ b) The initial value of the inductor current is 5 A. The switch has been closed for a long time, the inductor reaches ( 20/1)=20A The time constant is 80 ms i(t)=20+(5-20)e?* =(20-15eP"*) A p=0 = In the circuit shown, both switches operate together: that is, they either open or close at the same time. The switches are closed a long time before opening at t= 0. a) How many microjoules of energy have been dissipated in the 12 KQ resistor 12 ms after the switches open? b) How long does it take to dissipate 75% of the initially stored energy? t=0 t=0 1.8 kO 68 kQ a) t<0: fig. at left Req = 12K || 68 K= 10.2 KQ vo(0) = ( 10200 / ( 10200 + 1800 ) ) * (-120) = -102 V to fig. at right tT=( (10/3) *10-°)*(12,000) =40 ms: 1/t=25 vo= 102 "VY PO, p=Vo0/ 12000 = 867 *107°e"™" W Waiss = Jo"? aa At the time the switch is closed in the circuit. the capacitors are = 7.82 mJ charged as shown. w(0)=(1/2)( a) Find vo(t) fort > 0" 0.75w(0)=13 6) Find v,(t) fort 0 Jot (867 * 10° c) Find v,(t) fort >0 1-220 = 975: d) Find the energy trapped in the ideal capacitors 4.¢ ———— ~ iL ps0 | 3V Nb uF vi + a) Ce=((2+1 )*6)/(2+1+6) =2 uF vo(0) = -5 +30=25V T= (2 * 10°°)(250 *10*) =0.5 s: fig. down vo=25e"%V. t>0