Book C: Transposition Worksheet 1 - Solutions and Commentary

Introduction

Transposition involves taking an equation for one variable, and changing it so that it becomes in terms of another variable in the equation. Transposition is handy, because it enables us to have a single equation, which can be transposed into the form we need.

Algebra involves the operations required in order to achieve successful transposition.

The equals sign in an equation may be considered a balance point. The balance between the left-hand and right hand sides of an equation must be maintained at all times.

A typical algebraic equation is of the form \( A = B \), where \(A\) the left-hand side, and \(B\) is the right-hand side.

The following operations may be performed on an equation of this form. Notice how the same operation must be performed on both sides. This maintains the balance. The rules may be combined with each other in any order to perform a set of operations.

The use of the double-ended arrow \(\leftrightarrow\) means the rule works both ways, at least under some sets of conditions.

• Horizontal flip: \( A = B \leftrightarrow B = A \).
  
  
• Multiplication of both sides: \( A = B \leftrightarrow C A = C B \). This works for any \(C \neq 0\). The value \(C\) does not have to be constant.
  
  
• Division of both sides: \( A = B \leftrightarrow \frac{A}{C} = \frac{B}{C} \). This works for any \(C \neq 0\). The value \(C\) does not have to be constant.
  
  
• Negation of both sides: \( A = B \leftrightarrow -A = -B \). This is a special case of multiplication or division.
  
  
• Addition to both sides: \( A = B \leftrightarrow A + C = B + C \). The value \(C\) does not have to be constant.
  
  
• Subtraction from both sides: \( A = B \leftrightarrow A - C = B - C \). The value \(C\) does not have to be constant.
  
  
• Moving a variable from one side to another (addition): \( A + C = B \leftrightarrow A = B - C \). This is a special case of the subtraction from
  both sides.
  
  
• Moving a variable from one side to another (subtraction): \( A - C = B \leftrightarrow A = B + C \). This is a special case of the addition from both sides.
  
  
• Moving a variable from one side to another (multiplication): \( AC = B \leftrightarrow A = \frac{B}{C} \). This is a special case of division of both sides.
  
  
• Zero-balance: \( A = B \leftrightarrow A - B = 0 \).
  
  
• Moving a variable from one side to another (division): \( \frac{A}{C} = B \leftrightarrow A = BC \). This is a special case of the multiplication of both sides.
  
  
• Implicit division by 1: \( A = B \leftrightarrow \frac{A}{1} = \frac{B}{1} \). This is a special case of the multiplication of both sides.
  
  
• Vertical Flip (type 1): \( A = B \leftrightarrow \frac{1}{A} = \frac{1}{B} \).
  
  
• Vertical Flip (type 2): \( \frac{A}{B} = \frac{C}{D} \leftrightarrow \frac{B}{A} = \frac{D}{C} \).

Many equations use Greek letters, to increase the number of available symbols for use. Some Greek letters and their pronunciation are shown below.

• \(\alpha\): alpha
• \(\beta\): beta
• \(\epsilon\): epsilon
• \(\phi\): phi
• \(\pi\): pi (said "pie"), most famously used as the ratio of a circle's circumference to diameter.
• \(\mu\): mu, also used to denote "micro" (10^-6) on a unit.

Question 1

The equation is:

\(I = \frac{V}{R} \)

We need to solve for \( R \).

Solution

The simplest way to transpose this equation is to multiply both sides by \( R \), then divide both sides by \( I \).

\(I = \frac{V}{R} \rightarrow IR = V \rightarrow R = \frac{V}{I} \)

Question 2

The equation is:

\(P = VI \)

We need to solve for \( I \).

Solution

The simplest way to transpose this equation is to horizontally flip the equation, then divide both sides by \( V \).

\(P = VI \rightarrow VI = P \rightarrow I = \frac{P}{V} \)

Question 3

The equation is:

\(I = \frac{V}{R} \)

We need to solve for \( V \).

Solution

The simplest way to transpose this equation is to multiply both sides by \( R \), then horizontally flip the equation.

\(I = \frac{V}{R} \rightarrow IR = V \rightarrow V = I R \)

Question 4

The equation is:

\(W = \frac{Z}{V} \)

We need to solve for \( Z \).

Solution

The simplest way to transpose this equation is to horizontally flip the equation, then multiply both sides by \( V \).

\(W = \frac{Z}{V} \rightarrow \frac{Z}{V} = W \rightarrow Z = WV \)

Question 5

The equation is:

\( \beta = \mu \alpha \)

We need to solve for \( \alpha \).

Solution

The simplest way to transpose this equation is to horizontally flip the equation, then divide both sides by \( \mu \).

\(\beta = \mu \alpha \rightarrow \mu \alpha = \beta \rightarrow \alpha = \frac{\beta}{\mu} \)

Question 6

The equation is:

\( \epsilon = \frac{\phi}{\pi} \)

We need to solve for \( \pi \).

NOTE: Although mostly used with circle calculations, \(\pi\) here is to be treated like any other variable.

Solution

The simplest way to transpose this equation is to vertically flip the equation, horizontally flip the equation, then multiply both sides by \( \phi \).

\(\epsilon = \frac{\phi}{\pi} \rightarrow \frac{1}{\epsilon} = \frac{\pi}{\phi} \rightarrow \frac{\pi}{\phi} = \frac{1}{\epsilon} \rightarrow \pi = \frac{\phi}{\epsilon} \)

Question 7

The equation is:

\( \mu = \frac{\beta}{\alpha} \)

We need to solve for \( \beta \).

Solution

The simplest way to transpose this equation is to horizontally flip the equation, then multiply both sides by \( \alpha \).

\( \mu = \frac{\beta}{\alpha} \rightarrow \frac{\beta}{\alpha} = \mu \rightarrow \beta = \mu \alpha \)

Question 8

The equation is:

\( \cos \phi = \frac{A}{H} \)

We need to solve for \( A \).

NOTE: \(\cos \phi \) is treated like a single variable in this equation.

Solution

The simplest way to transpose this equation is to horizontally flip the equation, then multiply both sides by \( H \).

\( \cos \phi = \frac{A}{H} \rightarrow \frac{A}{H} = \cos \phi \rightarrow A = H \cos \phi \)

NOTE: Multipliers of a trigonometric function are normally put to the left, to avoid confusion between \(\cos \phi H \) and \(\cos(\phi H) \).

Question 9

The equation is:

\( X_L = 2 \pi f L \)

We need to solve for \( f \).

NOTE: \( 2 \pi f L \) is a group of multiplied variables. Variables can be "divided out" one by one or in groups as we please.

Solution

The first step is to get \( f \) on the left hand side.

\( X_L = 2 \pi f L \rightarrow 2 \pi f L = X_L \)

We can now get \( f \) on its own by dividing both sides by \( 2 \pi L \). These divisions can be done one-by-one, or all at once.

\( 2 \pi f L = X_L \rightarrow f = \frac{X_L}{2 \pi L} \)

Question 10

The equation is:

\( X_C = \frac{1}{2 \pi f C} \)

We need to solve for \( C \).

NOTE: \( 2 \pi f C \) is a group of multiplied variables. Variables can be "divided out" one by one or in groups as we please.

Solution

With this one, a good first step is to get \(C\) "on top".

\( X_C = \frac{1}{2 \pi f C} \rightarrow \frac{1}{X_C} = 2 \pi f C \)

Then we can get it on the left.

\( \frac{1}{X_C} = 2 \pi f C \rightarrow 2 \pi f C = \frac{1}{X_C} \)

Both sides can now be divided by \(2 \pi f \).

\( 2 \pi f C = \frac{1}{X_C} \rightarrow C = \frac{1}{2 \pi f X_C} \)

Question 11

The equation is:

\( \beta = \mu_0 \mu_r H \)

We need to solve for \( H \).

Solution

With this one, flip H to the left.

\( \beta = \mu_0 \mu_r H \rightarrow \mu_0 \mu_r H = \beta \)

Then, both sides can be divided by \( \mu_0 \mu_r \).

\( \mu_0 \mu_r H = \beta \rightarrow H = \frac{\beta}{\mu_0 \mu_r} \)

Question 12

The equation is:

\( Q = \frac{2 \pi f_0 L}{R} \)

We need to solve for \( R \).

Solution

With this one, flip R to the left.

\( Q = \frac{2 \pi f_0 L}{R} \rightarrow \frac{2 \pi f_0 L}{R} = Q \)

Flip both sides over, so \( R \) is on top.

\( \frac{2 \pi f_0 L}{R} = Q \rightarrow \frac{R}{2 \pi f_0 L} = \frac{1}{Q}\)

Multiply both sides by \( 2 \pi f_0 L \).

\( \frac{R}{2 \pi f_0 L} = \frac{1}{Q} \rightarrow R = \frac{2 \pi f_0 L}{Q} \)

Question 13

The equation is:

\( Q = m c \Delta t \)

We need to solve for \( \Delta t \).

NOTE: The symbol \( \Delta \) (Greek letter capital delta) means "change in" the variable to the right of it.

Solution

With this one, flip \( \Delta t \) to the left.

\( Q = m c \Delta t \rightarrow m c \Delta t = Q \)

Divide both sides by \( m c \).

\( m c \Delta t = Q \rightarrow \Delta t = \frac{Q}{mc} \)

Question 14

The equation is:

\( W = \frac{1}{2} CV^2 \)

We need to solve for \( C \).

Solution

With this one, flip \( C \) to the left.

\( W = \frac{1}{2} CV^2 \rightarrow \frac{1}{2} CV^2 = W \)

Divide both sides by \( \frac{1}{2} V^2 \).

\( \frac{1}{2} CV^2 = W \rightarrow C = \frac{2 W}{V^2} \)

Question 15

The equation is:

\( L = \frac{\mu_0 \mu_r N^2 A}{l} \)

We need to solve for \( N \).

Solution

With this one, flip \( N \) to the left.

\( L = \frac{\mu_0 \mu_r N^2 A}{l} \rightarrow \frac{\mu_0 \mu_r N^2 A}{l} = L \)

Divide both sides by \( \mu_0 \mu_r A \).

\( \frac{\mu_0 \mu_r N^2 A}{l} = L \rightarrow \frac{N^2}{l} = \frac{L}{\mu_0 \mu_r A} \)

Multiply both sides by \( l \).

\( \frac{N^2}{l} = \frac{L}{\mu_0 \mu_r} \rightarrow N^2 = \frac{Ll}{\mu_0 \mu_r A} \)

We now have an implicit equation for \( N \), because we have a function for \(N^2\). We can solve this by taking the square root of both sides, in order to reverse the squaring of \( N \).

\( N^2 = \frac{Ll}{\mu_0 \mu_r} \rightarrow N = \sqrt{\frac{Ll}{\mu_0 \mu_r A}} \)

Question 16

The equation is:

\( R_1 = R_0 \cdot (1 + \alpha_0 t_1) \)

We need to solve for \( R_0 \).

Solution

With this one, flip \( R_0 \) to the left.

\( R_1 = R_0 \cdot (1 + \alpha_0 t_1) \rightarrow R_0 \cdot (1 + \alpha_0 t_1) = R_1 \)

Divide both sides by \( 1 + \alpha_0 t_1 \). Note that division does not distribute through brackets, so the variables cannot be separated out and divided one by one.

\( R_0 \cdot (1 + \alpha_0 t_1) = R_1 \rightarrow R_0 = \frac{R_1}{1 + \alpha_0 t_1} \)

Question 17

The equation is:

\( f_0 = \frac{1}{2 \pi}\sqrt{\frac{1}{LC}} \)

We need to solve for \( L \).

Solution

With this one, flip \( L \) to the left.

\( f_0 = \frac{1}{2 \pi}\sqrt{\frac{1}{LC}} \rightarrow \frac{1}{2 \pi}\sqrt{\frac{1}{LC}} = f_0 \)

\( L \) is inside a square root, so we square both sides to "free" it.

\( \frac{1}{2 \pi}\sqrt{\frac{1}{LC}} = f_0 \rightarrow \frac{1}{4 \pi^2}\cdot\frac{1}{LC} = f_0^2 \)

The two fractions on the left can be "merged" together into one fraction.

\( \frac{1}{4 \pi^2}\cdot\frac{1}{LC} = f_0^2 \rightarrow \frac{1}{4 \pi^2 LC} = f_0^2 \)

We can now flip both sides over.

\( \frac{1}{4 \pi^2 LC} = f_0^2 \rightarrow 4 \pi^2 LC = \frac{1}{f_0^2} \)

We can now divide both sides by \( 4 \pi^2 C\).

\( 4 \pi^2 LC = \frac{1}{f_0^2} \rightarrow L = \frac{1}{4 \pi^2 f_0^2 C} \)

This answer may also be thought of as \( L = \frac{1}{(2 \pi f_0)^2 C} \).

Question 18

The equation is:

\( \frac{R_2}{R_1} = \frac{1 + \alpha_0 t_2}{1 + \alpha_0 t_1} \)

We need to solve for \( t_2 \).

Solution

With this one, flip \( t_2 \) to the left.

\( \frac{R_2}{R_1} = \frac{1 + \alpha_0 t_2}{1 + \alpha_0 t_1} \rightarrow \frac{1 + \alpha_0 t_2}{1 + \alpha_0 t_1} = \frac{R_2}{R_1} \)

Multiply both sides by \( 1 + \alpha_0 t_1 \). Notice the brackets on the right. That is because the right hand side is being multiplied by the entirety of \( 1 + \alpha_0 t_1 \), not just the parts one-by-one.

\( \frac{1 + \alpha_0 t_2}{1 + \alpha_0 t_1} = \frac{R_2}{R_1} \rightarrow 1 + \alpha_0 t_2 = \frac{R_2}{R_1} \cdot (1 + \alpha_0 t_1) \)

Subtract 1 from both sides to get rid of the 1 on the left.

\( 1 + \alpha_0 t_2 = \frac{R_2}{R_1} \cdot (1 + \alpha_0 t_1) \rightarrow \alpha_0 t_2 = \frac{R_2}{R_1} \cdot (1 + \alpha_0 t_1) - 1 \)

Divide both sides by \( \alpha \). Note that the entirety of the right hand side is divided by \(\alpha\), not just part of it.

\( \alpha_0 t_2 = \frac{R_2}{R_1} \cdot (1 + \alpha_0 t_1) - 1 \rightarrow t_2 = \frac{\frac{R_2}{R_1} \cdot (1 + \alpha_0 t_1) - 1}{\alpha_0} \)