Book "B": Worksheet 2: Power Formulae II 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

The colour bands on this resistor indicate that it has a nominal value of 1000 Ω. The physical size indicates that it has a power rating of 2 W. Calculate the maximum current this resistor can carry without risking it burning out.

Solution

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

The picture of the resistor itself, and an indication of its physical size, is not directly relevant. The power rating is based on the resistor design itself.

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

\( I = \sqrt{\frac{P}{R}} = \sqrt{\frac{2}{1000}} = \sqrt{0.002} = \) 0.0447 A (3sf) or 44.7 mA (3 sf) #.

In other words, the maximum current a 1000 Ω, 2 W resistor can pass is 44.7 mA.

Question 2

The resistor pictured below is rated at 500 watts and has a nominal value of 150 ohms. Calculate the power being dissipated by the resistor when 175 volts is applied across it.

Solution

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

The picture of the resistor itself is not directly relevant. The power rating is not relevant to this question, as it does not appear in the solution.

According to the formula wheel above, the appropriate formula is \( P = \frac{V^2}{R} \). The voltage is calculated by plugging this information into the formula.

\( P = \frac{V^2}{R} = \frac{175^2}{150} = \frac{30625}{150} = \) 204 W (3sf) #.

In other words, a 150 Ω resistor with 175 V across it will dissipate 204 W.

Question 3

2 heater elements can be wired to achieve 3 different power settings. Low, medium or high. Task: Wire each set of elements to achieve these three settings.

Solution

This question is referring to a three-heat switch setup.

A three-heat configuration uses two elements to achieve three distinct heat settings. The three settings are:

• Low always has both elements in series.
• Medium has one element only.
• High has both elements in parallel.

In terms of heat output as a percentage of rated heat output, the heat settings are as follows.

• Low is 25% of full output.
• Medium is 50% of full output.
• High is 100% of full output.

Three heat wiring normally consists of two elements with the same resistance \( R \). In terms of load resistance, the heat settings are as follows.

• Low is \( 2 R \).
• Medium is \( R \).
• High is \( \frac{R}{2} \).

Some acceptable wiring configurations are shown below.

There are also many wiring variations, nut as long as the electrical connections will ensure that those electrical connections will apply, the elements will work electrically as expected.

The next step is to calculate the power dissipation. The element resistances are given as 20 Ω. No power supply voltage is assumed, but the power supply voltage will be assumed to be 230 V.

The power can now be calculated.

The table below shows the calculations. All the power dissipation figures are calculated using the formula \( \frac{V^2}{R} \). All the power figures are rounded to four significant figures.

Question 3

These 6 resistors are all 10 ohms each.

Calculate the power dissipated by each resistor and the total power dissipated by the complete circuit (show all working).

Solution

The first step is to calculate the load resistance \( R_{\mathrm{L}} \), so that we can solve the circuit.

\( R_{\mathrm{L}} = R_1 + (R_2 + R_3) \parallel (R_4 + R_5 \parallel R_6) \)

All the resistors are the same value, so we will use the resistance \( R \) to represent this single resistance value. Note that \( R \) is 10 Ω.

\( R_{\mathrm{L}} = R + 2R \parallel (R + R \parallel R) = R + 2R \parallel (\frac{3 R}{2}) = R + \frac{6R}{7} = \frac{13}{7} R = \frac{130}{7} = \) 18.5714 Ω (6 sf).

The supply voltage is 215 V. The current draw is calculated below.

\( I_{\mathrm{L}} = \frac{V}{R_{\mathrm{L}}} = 215 \cdot \frac{7}{130} = \frac{301}{26} = \) 11.5769 A (6 sf).

The total power \( P_{\mathrm{TOT}} \) is calculated below in three different ways. Note that they all match, at least to the final significant figure.

Power Dissipation in Individual Resistors

The power dissipation of each individual resistor requires calculation.

There are two main methods of doing this:

• Calculate the current flow through each resistor, and use \( I^2 R \) to calculate the power dissipation.
• Calculate the voltage drop across each resistor, and use \( \frac{V^2}{R} \) to calculate the power dissipation.

Different methods are easier in different circumstances.

Power Dissipation in \( R_1 \)

The resistor \( R_1 \) carries the entire load current.

\( P_{\mathrm{R1}} = I_{\mathrm{L}}^2 \cdot R_1 = 11.5769^2 \cdot 10 = 134.025 \cdot 10 = \) 1340 W (4 sf) #.

Power Dissipation in \( R_2 \) and \( R_3 \)

\( R_2 \) and \( R_3 \) are in series and have the same resistance. They will therefore have the same power dissipation.

In order to calculate the power dissipation of \( R_2 \) and \( R_3 \), the voltage at the junction point of \( R_1 \), \( R_2 \) and \( R_4 \) needs to be known. This will be called \( V_A \).

\( V_A = V_{\mathrm{s}} - 11.5769 \cdot 10 = 215 - 115.769 = \) 99.2310 V (6 sf).

The current flow in \( R_2 \) and \( R_3 \) is calculated using Ohm's Law. Although not strictly required to calculate the power, the current is used in subsequent calculations.

\( I_A = \frac{V_A}{20} = \) 4.96155 A (6 sf).

The total power dissipation in the branch containing \( R_2 \) and \( R_3 \) can now be calculated.

\( P = V_A I_A = 99.231 \cdot 4.96155 = \) 492.340 W (4 sf).

The power dissipations in \( R_2 \) and \( R_3 \) are identical, therefore their power dissipations must be \( \frac{1}{2}P \)

\( P_{\mathrm{R2}} \) = 246.2 W (4 sf) #.

\( P_{\mathrm{R3}} \) = 246.2 W (4 sf) #.

Power Dissipation in \( R_4 \)

The incoming 11.5769 A load current is split between 4.96155 A going down \( R_2 \) and \( R_3 \), with the remainder going through \( R_4 \). The current going through R_4 will be termed \( I_4 \).

\( I_4 = 11.5769 - 4.96155 = \) 6.61535 A (6 sf).

The power dissipation in R_4 is calculated using \( P = I^2 R \).

\( P_{\mathrm{R4}} = I_4^2 R_4 = 6.61535^2 \cdot 10 = \) 437.6 W (4 sf) #.

Power Dissipation in \( R_5 \) and \( R_6 \)

\( R_5 \) and \( R_6 \) are in parallel and have the same resistance. They will therefore have the same power dissipation.

In order to calculate the power dissipation of \( R_5 \) and \( R_6 \), the voltage at the junction point of \( R_4 \), \( R_5 \) and \( R_6 \) needs to be known. This will be called \( V_B \).

\( V_B = V_A - 6.61535 \cdot 10 = 99.2310 - 66.1535 = \) 33.0775 V (6 sf).

\( P = \frac{V_B^2}{R} = \frac{33.0775^2}{10} = \frac{1094.12}{10} \) = 109.4 W (4 sf).

The power dissipations in \( R_5 \) and \( R_6 \) are identical.

\( P_{\mathrm{R5}} \) = 109.4 W (4 sf) #.

\( P_{\mathrm{R6}} \) = 109.4 W (4 sf) #.