Inductance, Capacitance, and Mutual Inductance Assessment Problems
6 Inductance, Capacitance, and Mutual
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 +
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 +