Tutorial 1 Basic DC Theory Part 1 Solutions PDF

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ENGINEERING PRINCIPLES (16 – 4019) Electrical Engineering Tutorial 1: Basic DC Theory Part 1 Solutions 1. Suggest four different types of electric power sources (supply). Indicate which one is classified as ‘clean’ power that is environmentally friendly, and why? These are, but not limited to: Fossil Fuel (oil & gas, coal, and others burning fuels). Fossil fuel-fired electric power plants emit carbon dioxide, a greenhouse gas, contributing to global warming. Combustion of fossil fuels also produces other air pollutants, such as nitrogen oxides, sulphur dioxide, volatile organic compounds and heavy metals. Renewable Energy (wind farms/turbines, solar, hydroelectric, tidal), environmentally friendly, doesn’t cause air pollution. Nuclear power (nuclear power plants produce electricity by nuclear fission). Proponents argue that nuclear power is a sustainable energy sources that reduces carbon emissions. Opponents believe that nuclear power poses many threats to people and the environment. Where do you stand?

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2. Which one of the following circuits is a ‘closed circuit’, ‘short circuit’, and an ‘open circuit’? The electric current I, is flowing from the battery (power source) through the two series resistors (R1 and R2), and back in a ‘closed loop circuit’.

This is an example of a ‘short circuit’ because almost all the current flows through the wire (≅0Ω) and practically none through the two series resistors. This results in an excessive current limited only by the internal resistance of the source (not shown) and potentially cause circuit damage. This is an example of an ‘open circuit’ because the wire is ‘open’ before the two series resistors and therefore, no current will flow in the circuit.

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3. Calculate the electrical resistance (in Ohm) for the following different lengths of different conductors (Resistivity for each material is given in the following Table): Material

Resistivity (ρ) (in Ωm at °C)

Aluminium

2.7 X 10-8

Brass

7.2 X 10-8

Copper

1.59 X 10-8

Carbon

6500 X 10-8

Tungsten

5.35 X 10-8

Zinc

5.37 X 10-8

a. A brass rod of cross-sectional area of 0.005 m2, and length of 1 m. 𝑅=𝜌

1 𝑙 = (7.2 × 10−8 ) � � = 1.44 × 10−5 = 14.4 × 10−6 = 14.4 µΩ 𝐴 0.005

b. A copper wire of cross-sectional area of 0.82 mm2, and length of 50 m. 𝑅=𝜌

𝑙 50 = (1.59 × 10−8 ) � � = 0.97Ω 𝐴 0.82 × 10−6

c. A tungsten wire of cross-sectional area of 0.52 mm2, and length of 5 cm. 𝑅=𝜌

5 × 10−2 𝑙 = (5.35 × 10−8 ) � � = 5.14 × 10 −3 = 5.14 mΩ 𝐴 0.52 × 10−6

d. A zinc rod of cross-sectional area of 2 cm2, and length of 0.5 m connected in series with a carbon rod of cross-sectional area of 2 cm2, and length of 0.25 m.

The two rods would be equivalent to two resistors connected in series, hence The Total Resistance of this arrangement is 𝑅𝑡𝑡𝑡𝑡𝑡 = 𝑅1 + 𝑅2 . 𝑅𝑧𝑧𝑧𝑧 = 𝑅1 = 𝜌

𝑙 0.5 = (5.37 × 10−8 ) � � = 0.13425 × 10−3 Ω 𝐴 2 × 10−4

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0.25 𝑙 −8 −3 𝑅𝑧𝑡𝑐𝑐𝑡𝑧 = 𝑅2 = 𝜌 𝐴 = (6500 × 10 ) �2 × 10−4 � = 81.25 × 10 Ω

𝑅𝑡𝑡𝑡𝑡𝑡 = 𝑅1 + 𝑅2 = 0.13425 × 10−3 + 81.25 × 10−3 = 81.4 × 10−3 = 81.4 mΩ e. Two copper rods connected in parallel, both of identical cross sectional area of 1 cm2, and length of 20 cm.

The two copper rods would be equivalent to two resistors connected in parallel, hence the total resistance of this arrangement is 𝑅𝑧𝑡𝑐𝑐𝑐𝑐

1

𝑅𝑡𝑡𝑡𝑡𝑡

1

1

=𝑅 + 𝑅 . 1

2

2 × 10−1 𝑙 −8 ) (1.59 = × 10 � � = 3.18 × 10 −5 Ω = 𝑅1 = 𝑅2 = 𝜌 𝐴 1 × 10−4 𝑅

As 𝑅1 = 𝑅2 , then 𝑅𝑡𝑡𝑡𝑡𝑡 = 1= 1.59 × 10−5 = 15.9 𝜇Ω 2 This follows from:

1

𝑅𝑡𝑡𝑡𝑡𝑡 ∴ 𝑅𝑡𝑡𝑡𝑡𝑡

2 2𝑅1 1 1 1 1 = = + = = + 𝑅1 𝑅2 𝑅1 𝑅1 (𝑅1 )2 𝑅1

𝑅1 3.18 × 10−5 = = = 1.59 × 10−3 = 15.9 µΩ 2 2

We observe that the electrical resistance is dependent on the length and the cross-sectional area of the conductor (wire), which leads us to two important conclusions: First, for any given length and thickness of an electrical cable, we can easily calculate the total resistance of that cable and therefore, we can calculate the power loss in the cable Power = 𝑃 = 𝑉𝑉 = 𝑉2 𝑅 =

𝑉2 Watt. 𝑅

Second, what will happen if the length and thickness of a ‘wire’ change due to temperature change or strain (due to an external mechanical force)? We expect that the electrical resistance would change as well. We use this physical property to sense and measure temperature (Resistance Tutorial 1: Basic DC Theory Part 1 Solutions

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Temperature Detectors, RTD), and to sense and measure mechanical forces, deformation and the growth of a crack in a structure (the strain gauge), and many other applications. 4. Calculate the resistance that would be measured between the terminals of the following resistive networks:

𝑅 = 𝑅1 + 𝑅2 = 100 + 50 = 150 Ω 1 1 1 1 1 = 0.01 + 0.02 = 0.03 + + = = 𝑅 𝑅1 𝑅 2 100 50 ∴𝑅=

1

1

𝑅 𝑐𝑡𝑐𝑡𝑡𝑡𝑐𝑡 𝑅2 =

+

1

𝑅3

=

1 = 33.3 Ω 0.03

1 1 + = 0.01 + 0.02 = 0.03 100 50

1 𝑅𝑐𝑡𝑐𝑡𝑡𝑡𝑐𝑡 = = 33.3 Ω 0.03 The total resistance (adding the above with the other resistor in series, R 1 is: 𝑅 = 𝑅1 + 33.3 = 100 + 33.3 = 133.3 Ω

5. By means of Ohm’s law, explain why a ‘short circuit’ might seriously damage electrical equipment and result in people being injured. What type of electrical device is often installed to provide overcurrent protection? 𝑉

According to Ohms law, we have the current 𝑉 = 𝑅, where R is the denominator of

the equation. A ‘short circuit’ means that the electrical resistance R is almost ZERO which will result in a very high current (increasing towards infinity), flowing through the short. However, it wouldn’t reach infinity as the very high current will melt the conductor. This could cause serious damage to equipment, unless there is a FUSE in the circuit. A fuse is a type of low resistance resistor that acts as a sacrificial device which melts when too much current flows. The melted fuse ‘opens’ (breaks) the circuit, disconnecting the faulty circuit and keeping you safe.

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6. Calculate the current and power in each resistor in the following circuit:

The voltage across the terminals is 200 V. To find the total current I (the current flowing through the 10 Ω resistor) in the circuit, we must first calculate the total resistance. Once the total resistance has been calculated, it is an easy task to calculate the voltage across each resistor, and hence the current flowing through each resistor and its power. Let’s call the series resistor R1 and the parallel combination R2. Now we have

The resistance of the parallel combination, R2, is given by: 1 1 1 1 7 = + + = R2 5 20 10 20 ∴ R2 =

20 = 2.857 Ω 7

The total resistance is calculated as follows: 𝑅𝑡𝑡𝑡𝑡𝑡 = R1 + R2 = 10 + 2.857 = 12.857 Ω Hence, the total current (flowing through the 10 Ω resistor) is: 𝑉=

𝑉

𝑅𝑡𝑡𝑡𝑡𝑡

Tutorial 1: Basic DC Theory Part 1 Solutions

200 = 15.56 A = 12.857

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The circuit now looks like:

By using the potential divider equation 𝑉 = 𝐸

the resistor R2, we have 𝑉=𝐸

R1

R1+R2

where V is the voltage across

2.857 𝑅2 = 44.443 V = 200 10 + 2.857 𝑅1 + 𝑅2

This is the voltage across each of the parallel resistors; hence, we can now calculate the current flowing through each of the parallel resistors and its power. For the 5 Ω resistor: 𝑉=

44.443 = 8.886 A, 𝑃 = 𝑉 × 𝑉 = 44.443 × 8.886 = 395 Watt 5

For the 10 Ω resistor: 𝑉=

44.443 = 4.444 A, 𝑃 = 𝑉 × 𝑉 = 44 .443 × 4.444 = 197 .5 Watt 10

For the 20 Ω resistor: 𝑉=

44.443 = 2.222 A, 𝑃 = 𝑉 × 𝑉 = 44 .443 × 2.222 = 98.76 Watt 20

Notice that the power dissipated by the 5 Ω resistor is twice that dissipated in the 10 Ω resistor and four times the power dissipated in the 20 Ω resistor. The power dissipated in each resistor may also be calculated using the other equations for power, namely 𝑃 =

𝑉2 𝑅

and 𝑃 = 𝑉2 𝑅.

Lastly, the power dissipated in resistor R1 is calculated as follows: 𝑃 = 𝑉2 𝑅 = (15.56)2 × 10 = 2.421 kW

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7. A starter-motor of winding resistance 0.1Ω is connected across the terminals of a 12V car battery of internal resistance 0.1Ω, as shown:

Calculate: (a) the current drawn by the starter motor; 𝑉=

12 𝑉 = 60 A = 𝑅 0.1 + 0.1

(b) the voltage at the motor terminals; Using the potential divider rule, and letting the internal resistance of the battery be R1 and the starter-motor winding resistance be R2. 𝑉=𝐸

0.1 𝑅2 =6V = 12 0.1 + 0.1 𝑅1 + 𝑅2

(c) the fault current that would be drawn from the battery if the startermotor winding became a short-circuit. 𝑉=

12 𝑉 = 120 A = 𝑅 0.1 + 0

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8. Each of the resistors below is of the same nominal value, 100Ω and tolerance ± 5%. Calculate the minimum and maximum voltages that would be measured at the output terminals of each network below on a large production run.

(a) 𝑉=𝐸

𝑅2

𝑅1 +𝑅2

= 10

100

100+100

= 5 V (if all resistors = 100Ω) 105

With tolerances, Maximum voltage, 𝑉𝑚𝑡𝑚 = 10 95+105 = 5.25 V With tolerances, Minimum voltage, 𝑉𝑚𝑧𝑧 = 10

95

105+95

= 4.75 V

(b) 1 1 1 1 , hence 𝑅1 = 50 Ω = + = 𝑅1 100 100 50 𝑉=𝐸

𝑅2

𝑅1 +𝑅2

= 10

100

50+100

= 6.67 V (if all resistors = 100Ω)

With tolerances, Maximum voltage, 𝑉𝑚𝑡𝑚 = 10 With tolerances, Minimum voltage, 𝑉𝑚𝑧𝑧 = 10

105

47.5+105 95

52.5+95

= 6.89 V

= 6.44 V

Note: parallel 5% tolerance resistors simply give 5% overall error. (c) Let R1 = 100 + 100 = 200 Ω. Hence, R1,min = 190 Ω and R1,max =210 Ω . 𝑉=𝐸

𝑅2

𝑅1 +𝑅2

= 10

100

200+100

= 3.33 V (if all resistors = 100Ω)

With tolerances, Maximum voltage, 𝑉𝑚𝑡𝑚 = 10 With tolerances, Minimum voltage, 𝑉𝑚𝑧𝑧 = 10

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105

190+105 95

210+95

= 3.56 V

= 3.22 V

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9. Use Thévenin’s theorem to determine the voltage across and the current flowing through the 8Ω resistor.

First, remove the 8 Ω resistor (load) and calculate the open-circuit voltage across terminals A and B.

The voltage across the series resistors (E2) is 15 V. The right-hand lead of the 4 Ω resistor (R1) is at +20 V and the left-hand lead of the 2 Ω resistor (R2) is at +5 V (E1). We can use the potential divider rule to find the open-circuit voltage, Voc, across the terminals as follows: 𝑉𝑡𝑧 = 𝑉𝐴𝐴 = 𝐸1 + 𝐸2

2 𝑅2 = 5 + 15 = 5 + 5 = 10 V 𝑅1 + 𝑅2 2+4

Next, replace each voltage source with a short circuit and calculate the resistance ‘looking in’.

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4×2 8 𝑅1 𝑅2 𝑅𝑧𝑧 = 𝑅 + 𝑅 = 4 + 2 = 6 = 1.33 Ω 1 2 Draw the Thévenin circuit, replace the load (8 Ω resistor), and calculate the voltage across the load and the current flowing through it. 𝑉 𝑡𝑡𝑡𝑙 = 10 𝑉𝑡𝑡𝑡𝑙 =

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8 = 8.57 V 1.33 + 8

10 = 1.07 A 1.33 + 8

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