v L di 50:0e 5000t V dt in Software

Print QR Code in Software v L di 50:0e 5000t V dt

In Fig. 2-11 the plots of i and v are given. Since the maximum current is 5.0 A, the maximum stored energy is Wmax 1 2 25:0 mJ LI 2 max

CHAP. 2]

Fig. 2-10

Fig. 2-11

An inductance of 3.0 mH has a voltage that is described as follows: for 0 > t > 2 ms, V 15:0 V and, for 2 > t > 4 ms, V 30:0 V. Obtain the corresponding current and sketch vL and i for the given intervals.

For 0 > t > 2 ms, i For t 2 ms, i 10:0 A For 2 > t > 4 ms, i 1 L t v dt 10:0

2 10 3

1 3 10 3

1 3 10 3

2 10 3

10:0

See Fig. 2-12.

A capacitance of 60.0 mF has a voltage described as follows: 0 > t > 2 ms, v 25:0 103 t (V). Sketch i, p, and w for the given interval and nd Wmax .

For 0 > t > 2 ms,

Fig. 2-12 dv d 60 10 6 25:0 103 t 1:5 A dt dt p vi 37:5 103 t W t wC p dt 1:875 104 t2 mJ i C

See Fig. 2-13. Wmax 1:875 104 2 10 3 2 75:0 mJ 1 1 Wmax CV2 60:0 10 6 50:0 2 75:0 mJ max 2 2

Fig. 2-13

A 20.0-mF capacitance is linearly charged from 0 to 400 mC in 5.0 ms. Find the voltage function and Wmax .

! 400 10 6 q t 8:0 10 2 t C 5:0 10 3 v q=C 4:0 103 t V Vmax 4:0 103 5:0 10 3 20:0 V Wmax 1 CV2 4:0 mJ max 2

A series circuit with R 2, L 2 mH, and C 500 mF has a current which increases linearly from zero to 10 A in the interval 0 t 1 ms, remains at 10 A for 1 ms t 2 ms, and decreases linearly from 10 A at t 2 ms to zero at t 3 ms. Sketch vR , vL , and vC .

vR must be a time function identical to i, with Vmax 2 10 20 V. For 0 < t < 1 ms, di 10 103 A=s dt When di=dt 0, for 1 ms < t < 2 ms, vL 0. and vL L di 20 V dt

CHAP. 2]

104 t dt 107 t2 V For 1 ms < t < 2 ms,

vC 20 10 t 10 3 10 V See Fig. 2-14.

Fig. 2-14

A single circuit element has the current and voltage functions graphed in Fig. 2-15. Determine the element.

Fig. 2-15

The element cannot be a resistor since v and i are not proportional. v is an integral of i. For 2 ms < t < 4 ms, i 6 0 but v is constant (zero); hence the element cannot be a capacitor. For 0 < t < 2 ms, di 5 103 A=s dt Consequently, 0 L v di 3 mH dt and v 15 V

(Examine the interval 4 ms < t < 6 ms; L must be the same.)

(a) i2 1 A, (b) i2 2 A,

Voltage v is the sum of the current-independent 10-V source and the current-dependent voltage source vx . Note that the factor 15 multiplying the control current carries the units . a b c v 10 vx 10 15 1 25 V v 10 vx 10 15 2 20 V v 10 15 0 10 V

Fig. 2-16

(a) v 50 V,

Fig. 2-17 Since the current enters the element at the negative terminal, a b p vi 50 8:5 425 W p vi 50 8:5 425 W

Find the power delivered by the sources in the circuit of Fig. 2-18.