1. Electric Circuits Assessment Problems
Chapter 6
Circuit Elements

2. Ans: i=Leq ∫𝑡𝑜 v dτ+i(t0) (a) t Leq = 60(240) 60+240 = 48mH #
6.4 The initial values of i1 and i2 in the circuit shown are +3 A and -5 A, respectively. The voltage at the terminals of the parallel inductors for t > 0 is -30e-5t mV.
a) If the parallel inductors are replaced by a single inductor, what is its inductance?
b) What is the initial current and its reference direction in the equivalent inductor?
c) Use the equivalent inductor to find i(t).
d) Find i1(t) and i2(t). Verify that the solutions for i1(t), i2(t), and i(t) satisfy Kirchhoff’s current law.
Ans:
i=Leq ∫𝑡𝑜 v dτ+i(t0)
(a)
Leq = 60(240) = 48mH # t

3. t t t 1 48𝑚 ∫𝑜 -0.03e-5τdτ-2 I1(t) = 1 60𝑚 ∫𝑜 -0.03e-5τdτ+3
(b)
i1(0)=+3A(向下)
i2(0)=-5A(向上)
i(0)=-2(向上) #
(c)
1 48𝑚 ∫𝑜 -0.03e-5τdτ-2
=0.125e-5t , t≥0 #
(d)
I1(t) = 1 60𝑚 ∫𝑜 -0.03e-5τdτ+3
=0.1e-5t+2.9 , t≥0 #
I2(t) = 𝑚 ∫𝑜 -0.03e-5τdτ-5
=0.025e-5t ,t≥0 # t t t

4. 6.5 The current at the terminals of the two capacitors shown is 240e-10tA for t ≥ 0. The initial values of v1 and v2 are - 10 V and -5V,respectively. Calculate the total energy trapped in the capacitors as t ∞. (Hint: Don't combine the capacitors in series — find the energy trapped in each, and then add.)
Ans: V(t)= 1 𝐶 ∫𝑡0 i dt+v(t0) V1(t)= 1 2μ ∫0 240μe-10t dt-10 = 2V V2(t)= 1 8μ ∫0 240μe-10t dt-5 = -2V W1 = 1 2 CV2 =0.5(1.6 μ) 22 = 4μJ W2= 1 2 CV2 =0.5(1.6 μ) -22 = 16μJ , Wt=16μ+4μ =20μJ # t ∞ ∞

5. Summing the voltages around the 𝑖1 mesh yields
a) Write a set of mesh-current equations for the circuit in Example 6.6 if the dot on the 4 H inductor is at the right-hand terminal,the reference direction of ig is reversed, and the 60Ω resistor is increased to 780 Ω.
b) Verify that if there is no energy stored in the circuit at t = 0,and if ig = e-4t A, the solutions to the differential equations derived in (a) of this Assessment Problem are
𝑖1 = e-4t + 12e-5tA,
𝑖2 = e-4t + e-5tA.
Ans:
(a)
Summing the voltages around the 𝑖1 mesh yields
25𝑖1+4 𝑑𝑖1 𝑑𝑡 -20𝑖2+5𝑖g+8 𝑑𝑖𝑔 𝑑𝑡 +8 𝑑𝑖2 𝑑𝑡 =0 #
780 Ω

6. Summing the voltages around the 𝑖2 mesh yields
-20 𝑖 𝑖2 +16 𝑑𝑖2 𝑑𝑡 +16 𝑑𝑖𝑔 𝑑𝑡 + 8 𝑑𝑖1 𝑑𝑡 =0 #
(b)
25𝑖1+4 𝑑𝑖1 𝑑𝑡 -20𝑖2+5𝑖g+8 𝑑𝑖𝑔 𝑑𝑡 +8 𝑑𝑖2 𝑑𝑡 = ①
𝑑𝑖1 𝑑𝑡 = 46.40e−4t − 60e−5t ; 4 𝑑𝑖1 𝑑𝑡 = e−4t − 240e−5t
𝑑𝑖2 𝑑𝑡 = 3.96e−4t − 5e−5t ; 8 𝑑𝑖2 𝑑𝑡 = 31.68e−4t − 40e−5t
𝑑𝑖𝑔 𝑑𝑡 = 7.84e−4t ; 8 𝑑𝑖𝑔 𝑑𝑡 = 62.72e−4t
25𝑖1 = −10 − 290e−4t + 300e−5t
20𝑖2 = −0.20 − 19.80e−4t + 20e−5t
5𝑖g = 9.8 − 9.8e−4t

7. -10-290e-4t+300e-5t+185. 6e-4t-240e-5t+0. 2+19. 8e-4t-20e-5t+9. 8-9
e-4t+300e-5t+185.6e-4t-240e-5t e-4t-20e-5t e-4t+31.68e-4t- 40e-5t+62.72e-4t= ②
①=② OK
-20 𝑖 𝑖2 +16 𝑑𝑖2 𝑑𝑡 +16 𝑑𝑖𝑔 𝑑𝑡 + 8 𝑑𝑖1 𝑑𝑡 = ③
8 𝑑𝑖1 𝑑𝑡 =371.20e−4t − 480e−5t
16 𝑑𝑖2 𝑑𝑡 =63.36e−4t − 80e−5t
16 𝑑𝑖𝑔 𝑑𝑡 =125.44e−4t
20 𝑖1 = −8 − 232e−4t + 240e−5t
800𝑖2 = −8 − 792e−4t + 800e−5t
8+232e-4t-240e-5t-8-792e-4t+800e-5t+63.36e-4t-80e-5t e-4t e-4t
-480e-5t= ④
③=④ OK