# Algebraic Equation Manipulation for Electric Circuits

## Mathematics for Electronics

• #### Question 1

The electrical resistance of a conductor at any temperature may be calculated by the following equation:

$$R_T = R_r + R_raT - R_raT_r$$

Where,

RT = Resistance of conductor at temperature T

Rr = Resistance of conductor at reference temperature Tr

α = Temperature coefficient of resistance at reference temperature Tr

Simplify this equation by means of factoring.

• #### Question 2

The equation for voltage gain (AV) in a typical non-inverting, single-ended opamp circuit is as follows:

$$A_V = \frac{R_1}{R_2} + 1$$

Where,

R1 is the feedback resistor (connecting the output to the inverting input)

R2 is the other resistor (connecting the inverting input to ground)

Suppose we wished to change the voltage gain in the following circuit from 5 to 6.8, but only had the freedom to alter the resistance of R2:

Algebraically manipulate the gain equation to solve for R2, then determine the necessary value of R2 in this circuit to give it a voltage gain of 6.8.

• #### Question 3

The equation for voltage gain (AV) in a typical inverting, single-ended opamp circuit is as follows:

 AV = R1 R2

Where,

R1 is the feedback resistor (connecting the output to the inverting input)

R2 is the other resistor (connecting the inverting input to voltage signal input terminal)

Suppose we wished to change the voltage gain in the following circuit from 3.5 to 4.9, but only had the freedom to alter the resistance of R2:

Algebraically manipulate the gain equation to solve for R2, then determine the necessary value of R2 in this circuit to give it a voltage gain of 4.9.

• #### Question 4

The following equations solve for the output voltage of various switching converter circuits (unloaded), given the switch duty cycle D and the input voltage:

 Vout = D Vin (Buck converter circuit)

 Vout = Vin 1 − D (Boost converter circuit)

 Vout = D Vin 1 − D (Inverting or Cuk converter circuit)

Manipulate each of these equations to solve for duty cycle (D) in terms of the input voltage (Vin) and desired output voltage (Vout). Remember that duty cycle is always a quantity between 0 and 1, inclusive.

• #### Question 5

Solve for n in the following equations:

Equation 1: −56 = −14n

Equation 2: 54 − n = 10

Equation 3: 4/n = 12

Equation 4: 28 = 2 − n

• #### Question 6

The formula for calculating total resistance of three series-connected resistors is as follows:

$$R = R_1+R_2+R_3$$

Algebraically manipulate this equation to solve for one of the series resistances (R1) in terms of the other two series resistances (R2 and R3) and the total resistance (R). In other words, write a formula that solves for R1 in terms of all the other variables.

• #### Question 7

Manipulate this equation to solve for resistor value R1, given the values of R2 and Rparallel:

$$R_{parallel} = \frac{R_1 R_2}{R_1 + R_2}$$

Then, give an example of a practical situation where you might use this new equation.

• #### Question 8

The formula for calculating total resistance of three parallel-connected resistors is as follows:

$$R = \frac {1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$$

Algebraically manipulate this equation to solve for one of the parallel resistances (R1) in terms of the other two parallel resistances (R2 and R3) and the total resistance (R). In other words, write a formula that solves for R1 in terms of all the other variables.

• #### Question 9

The power dissipation of a transistor is given by the following equation:

$$P = I_C (V_{CE}+ \frac{V_{BE}}{\beta})$$

Manipulate this equation to solve for beta, given all the other variables.

• #### Question 10

The decay of a variable over time in an RC or LR circuit follows this mathematical expression:

$$e^{-{\frac{t}{τ}}}$$

Where,

e = Euler’s constant ( ≈ 2.718281828)

t = Time, in seconds

τ = Time constant of circuit, in seconds

For example, if we were to evaluate this expression and arrive at a value of 0.398, we would know the variable in question has decayed from 100% to 39.8% over the period of time specified.

However, calculating the amount of time it takes for a decaying variable to reach a specified percentage is more difficult. We would have to manipulate the equation to solve for t, which is part of an exponent.

Show how the following equation could be algebraically manipulated to solve for t, where x is the number between 0 and 1 (inclusive) representing the percentage of original value for the variable in question:

$$x=e^{-{\frac{t}{τ}}}$$

Note: the “trick” here is how to isolate the exponent $$-\frac{-t}{τ}$$. You will have to use the natural logarithm function!

• #### Question 11

Voltage and current gains, expressed in units of decibels, may be calculated as such:

 AV(dB) = 10 log( AV(ratio) ) 2

 AI(dB) = 10 log( AI(ratio) ) 2

Another way of writing this equation is like this:

 AV(dB) = 20 logAV(ratio)

 AI(dB) = 20 logAI(ratio)

What law of algebra allows us to simplify a logarithmic equation in this manner?

• #### Question 12

Solve for the value of x in the following equations:

 10x = 80 x =

 3 = 15 x x =

• #### Question 13

Solve for the value of x in the following equations:

 5x = 15 x =

 6 = x 2 x =

• #### Question 14

Solve for the value of a in the following equations:

Equation 1: a − 4 = 10

Equation 2: 30 = a + 3

Equation 3: −2a = 9

Equation 4: a/4 = 3.5

• #### Question 15

Solve for the value of x in the following equations:

$$\frac {x+5}{2} = 20 \ \ \ \ \ \ \ \ \ \ \ x =$$

$$6 = \sqrt{x-2} \ \ \ \ \ \ \ \ \ \ \ x =$$

• #### Question 16

Solve for the value of x in the following equations:

$$2(x+5) = 36 \ \ \ \ \ \ \ \ \ \ \ x =$$

$$3 = \sqrt{2-x}\ \ \ \ \ \ \ \ \ \ \ x =$$

• #### Question 17

Manipulate each of these equations to solve for a:

$$\frac{b-a}{c}=d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{a+b}=c^2d$$

• #### Question 18

Manipulate each of these equations to solve for a:

$$\frac{a-b}{c}=d^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b+a^2 = \frac{c}{d}$$

• #### Question 19

Calculate all currents in this DC circuit:

Hint: it may help you to set up the necessary equation by labeling the current through the lower resistor as I and the current through the upper resistor as I + 0.005.

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1 Comment
• B
bjm999 July 14, 2020

Question 14, Equation 2 - the ‘+’ symbol is missing between ‘a 3’.
Question 17, right-hand equation - the square-root should only be over the left side of the equation.
Question 19, Hint - the ‘+’ symbol is missing between ‘I 0.005’.

They do seem to be correct in the equivalent questions in the PDF version though.

Like.