Book "B": Worksheet 2: Power Formulae III Assignment - Answers and Commentary

Introduction

The three power formulas are:

• \( P = V I \), where P is the power (W), V is the voltage (V), and I is the current (A). NOTE: This formula is not \( P = VA \). The symbol \( A \) does not represent current!
  
  
• \( P = I^2 R \), where P is the power (W), I is the current through a resistance (A), and R is the value of that resistance (Ω).
  
  
• \( P = \frac{V^2}{R} \), where P is the power (W), V is the voltage across the resistance (V), and R is the value of that resistance (Ω).

These formulas can all be related to each other by Ohm's Law.

There is also an Ohm's Law and power "wheel", as shown below. This wheel has all the power, voltage, current and resistance formulas.

The quantities on the wheel are:

The quantities in the centre are the unknown. You can pick from any of four formulas on the outside. In general, you pick the formula that has the two quantities you do know.

Question 1

Four 100 watt lamps are connected in parallel to a 230 volt supply. Calculate the current drawn from the supply.

Solution

The question has given us \( P \) and \( V \), and is asking us for \( I \).

The picture of the bulbs merely confirms they are in parallel.

According to the formula wheel above, the appropriate formula is \( I = \frac{P}{V} \). The current is calculated by plugging this information into the formula.

\( I = \frac{P}{V} = \frac{400}{230} = \) 1.74 A (3sf) #.

In other words, the current draw is 1.74 A.

Question 2

This TPS cable has a cable resistance of 2.5 Ω.

1. The volt drop that occurs when 15 amps is flowing; and
2. The power dissipated in the cable.

Solution

The voltage drop is calculated using Ohm's Law as \( V = I R \).

\( V = I R = 15 \cdot 2.5 = \) 37.5 V #.

NOTE: In a "real" installation, the voltage drop should be kept at 11.5 V or less.

The power dissipation may be calculated using \( P = I^2 R \).

\( P = I^2 R = 15^2 \cdot 2.5 = 225 \cdot 2.5 = \) 562.5 W #.

Question 3

This kitchen contains electrical equipment as listed below:

• A 3 kW water heater;
• An 8 kW stove;
• Two 750 watt toasters;
• Ten 60 watt ceiling lights;
• Fifteen 100 watt batten holder lights;
• Two 1.5 kW infra-red heaters;
• A 1200 watt microwave oven;
• One 450 watt fridge / freezer.

Determine the total kilowatts and total current to this single phase 230 volt kitchen when everything is turned on at once.

Solution

This question is a matter of totalling up the power and/or current, and getting the current for the entire installation.

The information in tabulated form is shown below.

NOTE: When adding up power ratings of different devices, it is essential that all the units are consistent with each other.

A table of the loads is shown below.

All of the load ratings have been converted to watts. The reason is that watts are not too unwieldy in this situation, and dividing watts by volts gives amps directly.

The total power is 19250 W, or 19.25 kW #.

The current is calculated below.

\( I = \frac{P}{V} = \frac{19250}{230} = \) 83.7 A (3 sf) #.

Question 4

Question 4 is a question about completing a table.

Solution

Two versions of the table are shown. The first is with formulas, the second is with answers. The formulas are written solely in terms of what is known at the time of the question. It may be possible to use one answer to help calculate another. This is not shown here.

The calculation results are shown below. All answers are given to 4 sf.

NOTE: The printed answers for column 2 (the 25 W column) are wrong! The correct answers are shown below.

Question 5

A generator set provides a current to a load through cables having a total resistance of 0.9 ohms. Find by calculation (show all working):

1. The volt drop in the cables;
2. The voltage at the load terminals;
3. The total power generated by the generator set;
4. The power lost in the cables.

Solution

1. Cable Voltage Drop

The voltage drop in the cables is calculated using Ohm's Law. The inputs are \( R_{\mathrm{c}} = \) 0.9 Ω, where \( R_{\mathrm{c}} \) is the cable resistance; and \( I_{\mathrm{L}} = \) = 29.79 A, where \( I_{\mathrm{L}} \) is the load current. The voltage drop will be called \( V_{\mathrm{D}} \).

\( V_{\mathrm{D}} = I_{\mathrm{L}} R_{\mathrm{c}} = 29.79 \cdot 0.9 = \) 26.8 V (3 sf) #.

2. Load Terminal Voltage

The voltage drop at the load terminals is based on the total voltage drop in the cables. Remember that the cable resistance is 0.9 Ω, as any current that goes somewhere must have a return path. The terminal voltage will be called \( V_{\mathrm{L}} \), and the generator terminal voltage will be called \( E \).

\( V_{\mathrm{L}} = E - V_{\mathrm{D}} = 235 - 26.8 = \) 208.2 V (4 sf) #.

3. Total Generator Set Power

Assuming a purely resistive load (there will be more about this in AC Theory), the power output of the generator is \( E I_{\mathrm{L}} \). The generator must not only generate the load power, but also the power losses in the cables. The generator power will be called \( P_{\mathrm{gen}} \).

\( P_{\mathrm{gen}} = E I_{\mathrm{L}} = 235 \cdot 29.79 = \) 7000 W (3 sf) #.

4. Power Losses in the Cables

We know the load current, and cable resistance, so the most direct route will be \( P_{\mathrm{c}} = I_{\mathrm{L}}^2 R_{\mathrm{c}} \).

\( P_{\mathrm{c}} = I_{\mathrm{L}}^2 R_{\mathrm{c}} = 29.79^2 \cdot 0.9 = 887.444 \cdot 0.9 = \) 799 W (3 sf) #.