Inductance

Assessment Problems

AP 6.1 [a] ig = 8e−300t − 8e−1200t A v=L dig = −9.6e−300t + 38.4e−1200t V, dt 38.4e−1200t = 9.6e−300t t > 0+

v(0+ ) = −9.6 + 38.4 = 28.8 V [b] v = 0 when or t = (ln 4)/900 = 1.54 ms [c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W dp = 0 when e1800t − 12.5e900t + 16 = 0 [d] dt Let x = e900t x = 1.45, x = 11.05, and solve the quadratic x2 − 12.5x + 16 = 0 t= ln 1.45 = 411.05 µs 900 ln 11.05 = 2.67 ms 900

t=

p is maximum at t = 411.05 µs [e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W [f] imax = 8[e−0.3(1.54) − e−1.2(1.54) ] = 3.78 A wmax = (1/2)(4 × 10−3 )(3.78)2 = 28.6 mJ [g] W is max when i is max, i is max when di/dt is zero. When di/dt = 0, v = 0, therefore t = 1.54 ms.

6–1

6–2

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance d dv = 24 × 10−6 [e−15,000t sin 30,000t] dt dt i(0+ ) = 0.72 A

AP 6.2 [a] i = C

= [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A, [b] i π ms = −31.66 mA, 80 p = vi = −649.23 mW [c] w = AP 6.3 [a] v = 1 Cv 2 = 126.13 µJ 2 1 C = t 0−

v

π ms = 20.505 V, 80

i dx + v(0− )

t 0−

1 0.6 × 10−6

3 cos 50,000x dx = 100 sin 50,000t V

[b] p(t) = vi = [300 cos 50,000t] sin 50,000t = 150 sin 100,000t W, [c] w(max) = AP 6.4 [a] Leq = p(max) = 150 W

1 2 Cvmax = 0.30(100)2 = 3000 µJ = 3 mJ 2

60(240) = 48 mH 300 [b] i(0+ ) = 3 + −5 = −2 A 125 t [c] i = (−0.03e−5x ) dx − 2 = 0.125e−5t − 2.125 A 6 0+ 50 t (−0.03e−5x ) dx + 3 = 0.1e−5t + 2.9 A [d] i1 = + 3 0 i2 = 25 6 t 0+

(−0.03e−5x ) dx − 5 = 0.025e−5t − 5.025 A

i1 + i2 = i AP 6.5 v1 = 0.5 × 106

t 0+

240 × 10−6 e−10x dx − 10 = −12e−10t + 2 V

t 0+

v2 = 0.125 × 106 v1 (∞) = 2 V, W =

240 × 10−6 e−10x dx − 5 = −3e−10t − 2 V

v2 (∞) = −2 V

1 1 (2)(4) + (8)(4) × 10−6 = 20 µJ 2 2

Problems AP 6.6 [a] Summing the voltages around mesh 1 yields di1 d(i2 + ig ) +8 + 20(i1 − i2 ) + 5(i1 + ig ) = 0 dt dt or 4 4 di1 di2 dig + 25i1 + 8 − 20i2 = − 5ig + 8 dt dt dt di1 d(i2 + ig ) +8 + 20(i2 − i1 ) + 780i2 = 0 dt dt

6–3

Summing the voltages around mesh 2 yields 16

or di1 di2 dig − 20i1 + 16 + 800i2 = −16 8 dt dt dt [b] From the solutions given in part (b) i1 (0) = −0.4 − 11.6 + 12 = 0; i1 (∞) = −0.4A; i2 (0) = −0.01 − 0.99 + 1 = 0

These values agree with zero initial energy in the circuit. At inﬁnity, i2 (∞) = −0.01A When t = ∞ the circuit reduces to

.·. i1 (∞) = −

7.8 7.8 + = −0.4A; 20 780

i2 (∞) = −

7.8 = −0.01A 780

From the solutions for i1 and i2 we have di1 = 46.40e−4t − 60e−5t dt di2 = 3.96e−4t − 5e−5t dt Also, dig = 7.84e−4t dt

Thus di1 = 185.60e−4t − 240e−5t 4 dt

6–4

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 25i1 = −10 − 290e−4t + 300e−5t 8 di2 = 31.68e−4t − 40e−5t dt

20i2 = −0.20 − 19.80e−4t + 20e−5t 5ig = 9.8 − 9.8e−4t 8 dig = 62.72e−4t dt

Test: 185.60e−4t − 240e−5t − 10 − 290e−4t + 300e−5t + 31.68e−4t − 40e−5t +0.20 + 19.80e−4t − 20e−5t = −[9.8 − 9.8e−4t + 62.72e−4t ] ?

−9.8 + (300 − 240 − 40 − 20)e−5t +(185.60 − 290 + 31.68 + 19.80)e−4t = −(9.8 + 52.92e−4t ) ?

−9.8 + 0e−5t + (237.08 − 290)e−4t = −9.8 − 52.92e−4t ?

−9.8 − 52.92e−4t = −9.8 − 52.92e−4t Also, 8 di1 = 371.20e−4t − 480e−5t dt di2 = 63.36e−4t − 80e−5t dt dig = 125.44e−4t dt

(OK)

20i1 = −8 − 232e−4t + 240e−5t 16

800i2 = −8 − 792e−4t + 800e−5t 16

Test: 371.20e−4t − 480e−5t + 8 + 232e−4t − 240e−5t + 63.36e−4t − 80e−5t −8 − 792e−4t + 800e−5t = −125.44e−4t ?

(8 − 8) + (800 − 480 − 240 − 80)e−5t +(371.20 + 232 + 63.36 − 792)e−4t = −125.44e−4t ?

(800 − 800)e−5t + (666.56 − 792)e−4t = −125.44e−4t ?

−125.44e−4t = −125.44e−4t

(OK)

Problems

6–5

Problems

P 6.1 [a] i = 0 50t A = 0.5 − 50t A t0

= 1.6e−20t mA, [e] i2 = −2 × 10−6 =

d [8e−20t + 17] dt

−2 × 10−6 (−20)8e−20t

= 0.32e−20t mA, t>0 −20t CHECK: i1 + i2 = 1.92e mA = io P 6.29 [a] w(0) = [ 1 (8 × 10−6 )(−5)2 + 1 (10 × 10−6 )(25)2 + 1 (2 × 10−6 )(25)2 ] 2 2 2 = 3850 µJ...