Fundamentals of Electric Circuits
Tutorial 1 -Voltage, Current, Resistance, and Ohm’s law
1. Find the current in amperes if 650 C of charge passes through a wire in 50 s.
Solution: Q = 650 C
T = 50 s
I = Q/ T = 650/50 = 13 A.
2. Will a fuse rated at 1 A “blow” if 86 C pass through it in 1.2 minute?
Solution: Q = 86 C
T = 1.2 min = 1.2 *60 second = 72 seconds
I = Q/ T = 86/ 72 = 1.194 A
Yes the fuse with 1A rating will blow.
3. If 21.847x1018 electrons pass through a wire in 12 s, find the current.
Solution: Q = 1.6*10-19*21.847*1018 = 3.49 C
I = Q/T = 3.49C/12s = 0.29 A
4. How many electrons pass through a conductor in 1 min and 30 s if the current is 4 mA?
Solution: Q = I*T = 4*10-3*90s = 0.36 C
Number of electrons Ne = Q/charge of electron = 0.36/(1.6*10-19)
= 2.25*1018 electrons
5. Determine the current flowing through an element if the charge flow is given by
(a) q(t) = (8t2 + 4t -2) C
(b) q(t) = 10 sin120πt pC
(c) q(t) = 20e-4t µC
Solution: (a) 𝑖 (𝑡 ) =
(b) 𝑖(𝑡 ) =
(c) 𝑖 (𝑡 ) =
𝑑
𝑑𝑡
𝑑
𝑑𝑡
𝑑 𝑞(𝑡)
𝑑𝑡
=
𝑑
𝑑𝑡
(8t2 + 4t -2) = (16t + 4) A
(10 sin120πt pC) = 10cos120 πt *120π = 1200πcos120πt pA
20e-4t µC = 20e-4t (-4) = -80e-4t µA
6. Find the charge q(t) flowing through a device if the current is:
(a) i(t) = 3 A, q(0) = 1 C
(b) i(t) = (2t + 5) mA, q(0) = 0
(c) i(t) = 20 cos(10t + π/6) µA, q(0) = 2 µC
Solution: (a) q(t) = ʃ i(t) dt + q(0) = ʃ 3 A dt + 1 = 3t + 1 C
(b) q(t) = ʃ (2t + 5) mA + 0 = t2 + 5t = (t2 + 5t) mC
(c) q(t) =ʃ 20 cos(10t + π/6) µA + 2 = 20 sin(10t + π/6)/10 + 2 µC
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7. The charge entering a certain element is shown in figure. Find the current at:
(a) t = 1 ms
(b) t = 6 ms
(c) t = 10 ms
Solution: First find the charge equations using the formulae for the equation of line with coordinates as
(0, 0), and (2ms, 80mC)
𝑦 −𝑦
𝑦 − 𝑦1 = 1 2 (𝑥 − 𝑥1 )
q(t) – 0 =
𝑥1 −𝑥2
0−80𝑚𝐶
(t – 0)
0−2𝑚𝑠
q(t) = 40t C, 0 ≤ t ≥ 2 ms
q(t) = 80 mC, 2 ms ≤ t ≥ 8 ms
8 ms ≤ t ≥ 12 ms; (8 ms, 80 mC), (12 ms, 0)
q(t) – 80 mC =
80𝑚𝐶−0
8𝑚𝑠−12𝑚𝑠
(t – 8 ms)
q(t) = 80 m – 20(t – 8m)
q(t) = - 20 t + 80 m + 160 m
q (t) = - 20 t + 0.24 C, 8 ms ≤ t ≥ 12 ms
The charge equations are,
q(t) = 40t C, 0 ≤ t ≥ 2 ms
q(t) = 80 mC, 2 ms ≤ t ≥ 8 ms
q (t) = - 20 t + 0.24 C, 8 ms ≤ t ≥ 12 ms
(a) t = 1 ms; i =
(b) t = 6 ms; i =
𝑑
𝑑𝑡
𝑑
𝑑𝑡
(c) t = 10 ms; i =
𝑞 (𝑡 ) =
𝑞 (𝑡 ) =
𝑑
𝑑𝑡
𝑞 (𝑡 ) =
𝑑
𝑑𝑡
𝑑
𝑑𝑡
40 𝑡 𝐶 = 40 𝐴
80 𝑚𝐶 = 0 𝐴
𝑑
𝑑𝑡
(−20 𝑡 + 0.24)𝐶 = −20 𝐴
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8. The current flowing past a point in a device is shown in figure. Calculate the total charge through
the point.
Solution: equation of line for the points (0,0) and (1,10) is
𝑦1 − 𝑦2
𝑦 − 𝑦1 =
(𝑥 − 𝑥1 )
𝑥1 − 𝑥2
i(t) – 0 =
10𝑚𝐴 −0
1𝑚𝑠−0
(t – 0)
i(t) = 10t A , 0≤ t ≥1ms
i(t) = 10 mA , 1≤ t ≥2ms
2𝑚𝑠
q(t) = ∫0
1𝑚𝑠
𝑖(𝑡 )𝑑𝑡 =∫0
1𝑚𝑠
= 5 𝑡 2 ∫0
2𝑚𝑠
10t A 𝑑𝑡 + ∫1𝑚𝑠 10 mA 𝑑𝑡
2𝑚𝑠
+ 10 𝑚 𝑡 ∫1𝑚𝑠 = 5 [(1 × 10−3 )2 − (0)2 ] + 10 × 10−3 [ 2 × 10−3 − 1 × 10−3 ]
= 5 × 10−6 + 10 × 10−3 (1 × 10−3 ) = 15 × 10−6 𝐶 = 15 𝜇𝐶
9. What is the voltage between two points if 96 mJ of energy are required to move 50×1018 electrons
between the two points?
Solution: Charge of electron = 1.6 × 10-19 coulombs
Total charge Q = 50*1018 * 1.6 × 10-19 = 8.01 C
Energy required W = 96 mJ
Voltage V = W/Q = 96*10-3 J/8.01 C = 12 mV
10. The potential difference between two points in an electric circuit is 24 V. If 0.4 J of energy were
dissipated in a period of 5 ms, what would the current be between the two points?
Solution: V= 24 V , W = 0.4 J , T = 5 ms = 5*10-3 s
V = W/ Q
Q = W/ V = 0.4/ 24 = 0.01667 C
I = Q/ T = 0.01667/ 5*10-3 = 3.333 A
11. How much charge passes through a radio battery of 9 V if the energy expended is 72 J?
Solution: V= 9 V, W = 72 J
V = W/ Q
Q = W/ V = 72/ 9 = 8 C
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12.
A coil consists of 2000 turn of copper wire having a cross-sectional area of 0.08mm2. The mean
length pre turn is 80 cm and the ρ = 0.02µΩ.m. Find the resistance of the coil.
Solution: N = 2000, ρ = 0.02µΩ.m = ρ = 0.02 x10-6 Ω.m
L = N x80 x10-2 m = 2000 x80 x10-2 m = 1600 m
A = 0.08mm2 =0.08 x10-6 m2
𝑅=
𝜌𝑙
𝐴
=
0.02 x10−6 x1600
0.08 x10−6
= 400 Ω
13. What is the potential drop across a 6Ω resistor if the current through it is 2.5 A?
Solution: V = IR , from ohm's law
V = 6Ω*2.5A = 15 V
14. What is the current through a 72 Ω resistor if the voltage drop across it is 12 V?
Solution: I = V/R , from ohm's law
I = 12V/72Ω = 0.167 A
15. The current through a 4 Ω resistor is 7 mA. What is the power delivered to the resistor?
Solution: P = I2R = (7*10-3)2 *4 = 196*10-6 W = 196 µW
16. Refer to the following figure, calculate:
a) Current Is
b) Voltage VR.
c) Power in resistance R.
Solution: Is = E/R = 12V/4kΩ = 3 mA
VR = E = 12 V
P = I*V = 3*10-3 *12V = 36 mW
17. A small, portable black-and-white television draws 0.455 A at 9V.
a. What is the power rating of the television?
b. What is the internal resistance of the television?
c. What is the energy converted in 6 h of typical battery life?
Solution: (a) P = I*V = 0.455A*9V = 4.095 W
(b) Rin = V/I = 9V/0.455A = 19.78 Ω
(c) W = P*t = 4.095W*6*60*60s = 88.452 kJ
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18 a. If a house is supplied with 120 V, 100 A service, find the maximum power capability.
b. Can the homeowner safely operate the following loads at the same time?
5 hp motor
3000 W clothes dryer
2400 W electric range
1000 W steam iron
c. If all the appliances are used for 2 hours, how much energy is converted in kWh?
Solution: (a) maximum power capability P = V*I = 120V*100A = 12000 W
(b) Total power used = 5*746 + 3000 + 2400 + 1000 = 10130 W (1 hp = 746W)
Owner can use all the mentioned appliances, because total power used is less than maximum power
capability.
(c) W = P*t = 10130W*2h = 20260 Wh = 20.26 kWh.
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