Problems 7–61
λ1(1)=λ2(1) = 40 Wb-turns;
1(1) and i2(1) are consistent with λ1(1) and λ2(1).
P 7.68 [a] From Example 7.10,
Leq =L1L2M2
L1+L2+2M=50 25
15 + 10 = 1 H;
[c] vo=5
di1
dt 5di2
dt = 80e−20tV;
io=i1+i2;
[d] i2=ioi1=44e−20t2.4+2.4e−20t
[e] io(0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy.
dio
7–62 CHAPTER 7. Response of First-Order RL and RC Circuits
vo= 10di2
dt 5di1
dt = 80e−20tV,t0+.(checks)
P 7.69 [a] From Example 7.10,
Leq =L1L2M2
L1+L22M=36 16
20 8=5
3H;
3io= 100 50
3(6 6e−10t) = 100e−10tVt0+.
[c] vo=2
di1
dt +4di2
dt ;
io=i1+i2;
Problems 7–63
[d] i2=ioi1
[e] vo=L2
di2
dt +Mdi1
dt
P 7.70 [a] Leq =5+102.5(2) = 10 H;
[d] i(0) = 2 2=0,which agrees with initial conditions.
7–64 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.71 [a] Leq =5+10+2.5(2) = 20 H;
[d] i(0) = 0,which agrees with initial conditions.
P 7.72 t<0:
0t5:
Problems 7–65
5t<1:
P 7.73 For t<0:
0t10 ms:
7–66 CHAPTER 7. Response of First-Order RL and RC Circuits
10 ms t20 ms:
20 ms t<1:
P 7.74 From the solution to Problem 7.73, the initial energy is
0.04w(0) = 0.1 J;
Problems 7–67
Again, from the solution to Problem 7.73, tmust be between 10 ms and 20 ms
since
P 7.75 [a] t<0:
Using Ohm’s law,
Using current division,
[b] 0t1 ms:
[c] i(1 ms) = 7.5e−1=2.7591 A.
1 ms t:
7–68 CHAPTER 7. Response of First-Order RL and RC Circuits
[d] 0t1 ms:
i=7.5e−1000t;
[e] 1 ms t1:
P 7.76 0 t10 µs:
τ=RC = (4 ⇥103)(20 ⇥10−9)=80µs; 1/τ= 12,500;
10 µst<1:
Problems 7–69
t!1:
i=50 V
20 kΩ=2.5 mA;
P 7.77 0 t200 µs:
7–70 CHAPTER 7. Response of First-Order RL and RC Circuits
200 µst<1:
vc= 110.36e−6000(t−200 µs)V;
P 7.78 t<0:
Problems 7–71
0t50 ms:
t50 ms:
Summary:
P 7.79 Note that for t>0,v
o= (4/6)vc, where vcis the voltage across the 0.5µF
capacitor. Thus we will find vcfirst.
t<0:
7–72 CHAPTER 7. Response of First-Order RL and RC Circuits
0t800 µs:
800 µst1.1 ms:
1.1 ms t<1:
Problems 7–73
vc(1.1ms) = 6.74e−333.33(1100−800)10−6=6.74e−0.1=6.1 V;
P 7.80 w(0) = 1
2(0.5⇥10−6)(15)2= 56.25 µJ.
0t800 µs:
1.1 ms t1:
vc=6.1e−1000(t−1.1×10−3)V; v2
c= 37.19e−2000(t−1.1×10−3);
The total energy dissipated by the 3 kΩresistor is
7–74 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.81 [a] 0t75 µs:
io(0) = 0; io(1) = 25 mA;
75 µst<1:
·
.. t<0: vo=0;
P 7.82 [a] 0t2.5 ms :
vo(0+) = 80 V; vo(1)=0;
Problems 7–75
[b]
[c] vo(5 ms) = 16.35 V;
P 7.83 [a] t<0; vo=0.
0t4 ms:
4 ms t8 ms:
[b]
7–76 CHAPTER 7. Response of First-Order RL and RC Circuits
[c] t0: vo= 0.
0t4 ms:
4 ms t8 ms:
P 7.84 [a] 0t1 ms:
vc(0+)=0; vc(1) = 50 V;
vc(1) = 0 V;