Electric Circuits
Fall 2015
Pingqiang Zhou
Solution #1
1. Determine the current flowing through an element if the charge flow is given by (a)
and (b), and find the charge q(t) flowing through a device if the current is (c) and (d).
(a) 𝑞(𝑡) = (8𝑡 2 + 4𝑡 − 2) C
(b) 𝑞(𝑡) = 20𝑒 −4𝑡 cos 50𝑡 𝜇C
(c) 𝑖(𝑡) = 20 cos(10𝑡 + 𝜋/6) 𝜇A, 𝑞(0) = 2𝜇C
(d) 𝑖(𝑡) = 10𝑒 −30𝑡 sin 40𝑡 A, 𝑞(0) = 0
Solution:
(a) 𝑖(𝑡) =
𝒅𝑞(𝑡)
(b) 𝑖(𝑡) =
𝒅𝑞(𝑡)
𝒅𝑡
𝒅𝑡
= (16𝑡 + 4) A
= −𝑒 −4𝑡 (80 cos 50𝑡 + 1000 sin 50𝑡) 𝜇A
𝑡
𝜋
(c) 𝑞(𝑡) = ∫0 20 cos(10𝜏 + 𝜋/6) 𝑑𝜏 + 𝑞(0) = (2 sin (10𝑡 + 6 ) + 1) 𝜇C
𝒕
1
𝑡
(d) 𝑞(𝑡) = ∫𝟎 𝟏𝟎𝒆−𝟑𝟎𝝉 𝐬𝐢𝐧 𝟒𝟎𝝉 𝒅𝝉 + 𝑞(0) = − 3 ∫0 sin 40𝜏 𝑑(𝑒 −30𝜏 ) =
1
− 3 𝑒 −30𝑡 sin 40𝑡 +
4
40
3
4
𝑒 −30𝑡 cos 40𝑡 + 9 −
9
𝑡
1
∫0 𝑒 −30𝜏 cos 40𝜏 𝑑𝜏 = (− 3 𝑒 −30𝑡 sin 40𝑡 −
16
9
𝒕
∫𝟎 𝟏𝟎𝒆−𝟑𝟎𝝉 𝐬𝐢𝐧 𝟒𝟎𝝉 𝒅𝝉) C
Then,
𝑡
9
1
4
𝑞(𝑡) = ∫0 10𝑒 −30𝜏 sin 40𝜏 𝑑𝜏 + 0 = 25 (− 3 𝑒 −30𝑡 sin 40𝑡 − 9 𝑒 −30𝑡 cos 40𝑡 +
4
9
) = (−𝑒 −30𝑡 (0.16 cos 40𝑡 + 0.12 sin 40𝑡) + 0.16) C
2. The charge entering a certain element is shown in Fig. 1.
Find the current at:
(a) t = 1 ms
(b) t = 6 ms
(c) t = 10 ms
Fig. 1. For Prob. 2.
Solution:
(a) At t = 1 ms, 𝑖(1) = (
𝒅𝑞(𝑡)
(b) At t = 6 ms, 𝑖(6) = (
𝒅𝑞(𝑡)
𝒅𝑡
𝒅𝑡
(c) At t = 10 ms, 𝑖(10) = (
)𝑡=1 =
30−0 mC
)𝑡=6 =
0 mC
𝒅𝑞(𝑡)
𝒅𝑡
2−0 ms
6 ms
)𝑡=10 =
= 15 A
=0A
0−30 mC
12−8 ms
= −7.5 A
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Electric Circuits
Fall 2015
Pingqiang Zhou
3. The current through an element is shown in Fig. 2. Find
the total charge that passed through the element at:
(a) t = 1 s
(b) t = 3 s
(c) t = 5 s
Fig. 2. For Prob. 3.
Solution:
1
1
(a) t = 0 s through t = 1 s, 𝑞 = ∫𝑡=0 𝑖(𝑡) 𝑑𝑡 = ∫𝑡=0 10 𝑑𝑡 = 10 C
3
2
3
(b) t = 0 s through t = 3 s, 𝑞 = ∫𝑡=0 𝑖(𝑡) 𝑑𝑡 = 10 + ∫𝑡=1 −5𝑡 + 15 𝑑𝑡 + ∫𝑡=2 5 𝑑𝑡 =
22.5 C
5
4
5
(c) t = 0 s through t = 5 s, 𝑞 = ∫𝑡=0 𝑖(𝑡) 𝑑𝑡 = 22.5 + ∫𝑡=3 5 𝑑𝑡 + ∫𝑡=4 −5𝑡 +
25 𝑑𝑡 = 30 C
In fact, you can also calculate the areas bounded by the curve and the axis.
4. The charge entering the positive terminal of an element is
𝑞 = 5 sin 4𝜋𝑡 mC
while the voltage across the element (plus to minus) is
𝑣 = 3 cos 4𝜋𝑡 V
(a) Find the power delivered to the element at t = 0.3 s.
(b) Calculate the energy delivered to the element between 0 and 0.6 s.
Solution:
dq
(a) 𝑃(𝑡) = 𝑣𝑖 = 𝑣 ( dt ) = 3 cos 4𝜋𝑡 V ∗ 20𝜋 cos 4𝜋𝑡 mA = 60πcos 2 (4𝜋𝑡) mW
At t = 0.3 s, 𝑃(0.3) = 60πcos2 (4𝜋 ∗ 0.3) = 123.37 mW
0.6
0.6
(b) W = ∫𝑡=0 60πcos2 (4𝜋𝑡) 𝑑𝑡 = 30𝜋 ∫𝑡=0[1 + cos 8𝜋𝑡] 𝑑𝑡 = 30𝜋 [0.6 +
1
8𝜋
[sin 4.8𝜋]] = 58.76 mJ
5. Find the power absorbed by each of the elements in Fig. 3.
Fig. 3. For Prob. 5.
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Electric Circuits
Fall 2015
Pingqiang Zhou
Solution:
𝑝1 = 30 V ∗ (−10 A) = −300 W
𝑝2 = 10 V ∗ 10 A = 100 W
𝑝3 = 20 V ∗ 14 A = 280 W
𝑝4 = 8 V ∗ (−4 A) = −32 W
𝑝5 = 12 V ∗ (−0.4 ∗ 10 A) = −48 W
6. Find 𝑉𝑜 and the power absorbed by each element in the circuit of Fig. 4.
Fig. 4. For Prob. 6.
Solution:
KVL: 12 V + 𝑉𝑜 − 30 V = 0 V, or 28 V − 5𝐼𝑜 − 𝑉𝑜 = 0 V, 𝑉𝑜 = 18 V
𝑝30 𝑣𝑜𝑙𝑡 𝑠𝑜𝑢𝑟𝑐𝑒 = 30 V ∗ (−6 A) = −180 W
𝑝12 𝑣𝑜𝑙𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 12 V ∗ 6 A = 72 W
𝑝28 𝑣𝑜𝑙𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑤𝑖𝑡ℎ 2 𝑎𝑚𝑝𝑠 𝑓𝑙𝑜𝑤𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑖𝑡 = 28 V ∗ 2 A = 56 W
𝑝28 𝑣𝑜𝑙𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑤𝑖𝑡ℎ 1 𝑎𝑚𝑝𝑠 𝑓𝑙𝑜𝑤𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑖𝑡 = 28 V ∗ 1 A = 28 W
𝑝𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑠𝑜𝑢𝑟𝑐𝑒 = 5 ∗ 2 V ∗ (−3 A) = −30 W
𝑝18 𝑣𝑜𝑙𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 18 V ∗ 3 A = 54 W
Or 𝑝18 𝑣𝑜𝑙𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 72 − 56 − 28 + 30 = 54 W
7. Fig. 5 shows the power consumption of a
certain household in 1 day. Calculate:
(a) the total energy consumed in kWh,
(b) the average power per hour over the
total 24 hour period.
Fig. 5. For Prob. 7.
Solution:
(a) Energy = Energy = ∑ 𝑝𝑡 = 200 ∗ 6 + 800 ∗ 2 + 200 ∗ 10 + 1200 ∗ 4 +
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Electric Circuits
Fall 2015
Pingqiang Zhou
200 ∗ 2 = 10 kWh
(b) Average power = 10 kWh/24 h = 416.7 W
8. In the network graph shown in Fig. 6, determine the number of branches, nodes and
loops.
Fig. 6. For Prob. 8.
Solution:
n = 14; l = 8; b = n + l – 1 = 21
9. Find 𝑖1 , 𝑖2 and 𝑖3 in Fig. 7.
Fig. 7. For Prob. 9.
Solution:
KCL:
𝑖1 = 1 A + 6 A = 7 A, or 𝑖1 = 5 A + 2 A = 7 A
𝑖2 = 6 A − 7 A = −1 A
𝑖3 = 4 A − (−1 A) = 5 A, or 𝑖3 = 7 A − 2 A = 5 A
10. In the circuit in Fig. 8, obtain 𝑣1 , 𝑣2 and 𝑣3 .
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Fig. 8. For Prob. 10.
Electric Circuits
Fall 2015
Pingqiang Zhou
Solution:
KVL:
𝑣1 − 40 V − 50 V + 20 V = 0 V, 𝑣1 = 70 V
𝑣2 − 30 V + 20 V = 0 V, 𝑣2 = 10 V
𝑣3 − 𝑣1 + 𝑣2 = 0 V, 𝑣3 = 60 V
11. Identify all the nodes, branches, and independent loops in Fig. 9.
Fig. 9. For Prob. 11.
Solution:
n = 8; l = 7; b = n + l – 1 = 14
12. Find 𝑉𝑜 in the circuit in Fig. 10 and the power absorbed by the dependent source.
Fig. 10. For Prob. 12.
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Electric Circuits
Fall 2015
Pingqiang Zhou
Solution:
𝑉𝑜 = −(2𝑉𝑜 + 25) A ∗ 10 Ω, 𝑉𝑜 = −
250
21
= −11.905 V
P = (−11.905 V − 10 Ω ∗ (25 A − 2 ∗ 11.905 A)) ∗ (−2 ∗ 11.905 )A = 566.8 W
13. Find the equivalent resistance 𝑅𝑎𝑏 in the circuit of Fig. 11.
Fig. 11. For Prob. 13.
Solution:
5 Ω ∥ 20 Ω = 4 Ω
6Ω ∥3Ω= 2Ω
𝑅𝑎𝑏 = 10 + 4 + 2 + 8 = 24 Ω
14. Calculate 𝐼𝑜 in the circuit of Fig. 12.
Fig. 12. For Prob. 14.
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Electric Circuits
Fall 2015
Pingqiang Zhou
Solution:
We convert the T to Δ.
𝑅𝑎𝑏 =
𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅1 𝑅3 20 ∗ 40 + 40 ∗ 10 + 10 ∗ 20
=
= 35 Ω
𝑅3
40
Similarly, 𝑅𝑎𝑐 =
1400
10
= 140 Ω, 𝑅𝑏𝑐 =
1400
20
= 70 Ω
𝑅𝑒𝑞 = 35 ∥ (140 ∥ 60 + 70 ∥ 70) = 24.0625 Ω
𝐼𝑜 =
24𝑉
𝑅𝑒𝑞
= 997.4 mA
15. Find 𝑅𝑒𝑞 and I in the circuit of Fig.13.
Fig. 13. For Prob. 15.
Solution:
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Electric Circuits
Fall 2015
𝑅𝑎𝑏 =
6∗12+12∗8+8∗6
𝑅𝑑𝑒 =
4∗2+2∗8+8∗4
12
8
=
=
216
56
8
12
= 18 Ω, 𝑅𝑎𝑐 =
= 7 Ω, 𝑅𝑒𝑓 =
56
4
216
8
Pingqiang Zhou
= 27 Ω, 𝑅𝑏𝑐 = 36 Ω
= 14 Ω, 𝑅𝑑𝑓 =
56
2
= 28 Ω
Combining resistors in parallel,
10 ∥ 28 = 7.368 Ω, 36 ∥ 7 = 5.868 Ω, 27 ∥ 3 = 2.7 Ω
(Correction: the 7.568 𝛺 resistor in the upper figure should be 7.368 𝛺.)
18∗2.7
𝑅𝑎𝑛 = 18+2.7+5.867 = 1.829 Ω, 𝑅𝑏𝑛 =
𝑅𝑐𝑛 =
5.868∗2.7
26.567
18∗5.868
26.567
= 3.977 Ω
= 0.5904 Ω
𝑅𝑒𝑞 = 4 + 1.829 + (3.977 + 7.368) ∥ (0.5964 + 14) = 12.21 Ω
20
𝐼=𝑅
𝑒𝑞
= 1.64 A
16. Find 𝑅𝑎𝑏 in the eight-way divider shown in Fig. 14. Assume each element is 1 Ω.
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Electric Circuits
Fall 2015
Pingqiang Zhou
Solution:
𝑅𝑎𝑏 = 1 + 1 = 2 Ω
17. In a certain application, the circuit in Fig. 15 must be designed to meet these two
criteria:
(a) 𝑉𝑜 /𝑉𝑠 = 0.05 (b) 𝑅𝑒𝑞 = 40 kΩ
If the load resistor 5 kΩ is fixed, find 𝑅1 and 𝑅2 to meet the criteria.
Fig. 15. For Prob. 17.
Solution:
Let 𝑅1 and 𝑅2 be in kΩ.
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 ∥ 5 kΩ = 40 kΩ
𝑅2 ∥ 5 kΩ
0.05
=
𝑅1
1 − 0.05
⇒
𝑅1 = 38 kΩ, 𝑅2 = 10/3 kΩ
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