seventh

5.The current through a 47F capacitor is shown in Fig.7.40.Calculate the voltage across the device at (a)t=2ms;(b) t=4ms;(c) t=5ms.

iCvdv,so dt1tidt 0C1000000110333.42mV 4721000000110333.42mV when t=4ms,v472when t=2ms,vwhen t=5ms,v10000001110350.13mV 474211.With reference to Fig.7.43:(a)sketch vL as a function of time,0(b)find the value of time at which the inductor is absorbing a maximum power;

(c) find the value of time at which it is supplying a maximum power ;and (d) find the energy stored in the inductor at t=40ms.

(a)We use \" vLdi \" to obtain the voltage waveform dt•if t<0ms, i=0 then v=0

di5250then v=0.2×250＝50V 3dt2010di5500then v=0.2×（－500）＝－100V •if 20≤t<40，

dt10103di5500then v=0.2×500＝100V •if 40≤t<50,

dt10103•if 0≤t<20,

•if t≥50， i=0 then v=0

(b)(c)Now,we use pLidi to obtain p20 ,p20,p40 and p40dtp200.25250250W p200.25(500)500W p400.2(5)(500)500W p400.2(5)500500W

•So that ,the value of time at which the inductor is absorbing a maximum power is 40-ms. So that ,the values of time at which the inductor is absorbing a maximum power are 20+ ms and 40+ ms. (d)

12Li21 0.2(5)2

2 2.5JwLSo that ,the energy stored in the inductor at t=40ms is 2.5J.

13.(a)If is0.4t2A for t>0 in the circuit of Fig.7.44a,find and sketch vin(t) for t>0. (b) If is40tV for t>0 and iL(0)5A , find and sketch iin(t) for t>0 in the circuit of Fig.7.44b.

(a)First,applying KVL,we obtain that

vin10isvL0

•Next,we use the ralationgship of voltage-current to obtain vL

did(0.4t2)vL5 dtdt 50.42t4tso that

vin10isvL 100.4t4t4t4t22

The circuit is shown in Fig 7.44c.

(b)First,applying KCL and Ohm's law,we obtain that

iiniLi10i10 vs4t10•Next, using the voltage-current ralationgship of inductor ,we obtain that

iL1tvsdtiL(0)0L1t 40tdt5

50 4t25•so that

iin4t24t5

The circuit is shown in Fig 7.44d

15.The voltage vLacross a 0.2-H inductor is 100V for 0t10ms; decreases linearly to zero in the interval 10iL02A.(b)Determine the stored energy at t=22ms if iL00.(c)If the circuit shown in

Fig.7.45 has been connected for a very long time ,find ix.

iL02A

(a)

vLi(b)

di,so dt1t1vdt2810010322AL00.2

i11010051001037.5A 0.2when t=22ms, W(c)

12Li0.50.27.525.63J 2

find R

8020168020R210 8020168020using voltage division

V100880V 10so, using Ohm’s law

i

801A 8017.A long time after all connections have been made in the circuit shown in Fig.7.46,find vxif (a)a capacitor is present between x and y and (b)an inductor is present between x and y.

(a) Using superposition principle,we first open-circuit the 5-A source .The equation is

Vx11204066.67V

402012Next,we short-circuit the 120-V source .The equation is

Vx2so

1254033.33V

402012VxVx1Vx2100V

(b)

Using superposition principle,we first open-circuit the 5-A source .The equation is

60152Vx1120601540V

60153126015Next,we short-circuit the 120-V source .The equation is

Vx2so

1215151254020V 1215601512VxVx1Vx260V

19. Let vs400t2V for t>0 and iL(0)0.5A in the circuit of Fig.7.48,At t>0.4s.finf the values of energy ;(a)stored in the capacitor;(b) stored in the inductor ;and (c)dissipated by the resistor since t=0.

(a)we know

vs0.44000.464V, vCvs

2so,WC(b)

122CvC0.5101066420.5J 21t80iLvdtt30.5

L03so

iL0.4WL122LiL0.5101.359.16J 2800.430.51.35A3

23.For the circuit of Fig.7.52 (a)reduce the circuit to the fewest possible components using series/parallel combinations;(b)determine vx if all resistors are 10k,all capacitor are

50F,and all inductors are 1mH.

(a) The cicuit id shown in Fig 7.52a.

(b) Applying the principle of voltage division,we obtain that

vx1093.6V

101525.Reduce the network of Fig.7.54 to a single equivalent capacitance as seen looking into terminals a and b.

1004012212C278.71F

212so

C110040127586.05nF

CeqC1C285.21nF

C1C229.For the network of Fig.7.57,L11H,L2L32H,L4L5L63H.(a)Find the equivalent inductance .(b)Derive an expression for a general network of this type having N stage, assuming stage N is composed of N inductors, each having inductance N henrys.

(a)

Leq12213H 11122333(b)

Leq12211...11...1NH 12211111N...333NNN80t37. Let vs0100eV and v1020V in the circuit of Fig.7.62.(a)Find i(t)for all

t.(b)Find v1(t)for t0.(c)Find v2(t)for t0.

(a)

We know

v1020V,vs0100V

so, using KVL,

v2080V

We know

Ceqso

140.8F 14dv0.8100e80t801066.4e80tmA dtitCeq(b)

iC1so

dv1 dtdv11idt C1t0v16.4e80t1000dtv1080e80t60V

(c)

iC2so

dv2 dtdv2v21idt C21t80t80t6.4e1000dtv020e60V 204