Voltage drop formula

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From users who are members of MfEP Progress Quizzes :

Hanh Vo

said

Ready to use

3 years, 4 months ago

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Don Shearman 11 months, 1 week ago

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Hanh Vo 3 years, 4 months ago

Gave some feedback: Ready to use

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Don Shearman 3 years, 8 months ago

Created this.

Name	Status	Author	Last Modified
Voltage drop formula	Ready to use	Don Shearman	14/05/2024 05:36
Parallel inductors formula	Ready to use	Don Shearman	06/05/2024 05:49
Parallel resistors formula	Ready to use	Don Shearman	06/05/2024 05:53
Heat flow formula	Ready to use	Don Shearman	06/05/2024 05:47

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Question statement

A battery with emf $E$ and internal resistance $r$ is connected across an external resistor of resistance $R$ . The voltage drop across the resistor can be shown to be $\displaystyle{V=\frac{ER}{R+r}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

E	integer	9
V	integer	4
R	integer	30

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(5..12)

JME function reference

Description

Battery emf in volts, random integer from 5 to 12

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Number entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Rearrange the formula to make $r$ the subject. (Note: to enter a fraction such as $\frac{a}{b}$ , type a/b.)

$r=$ Gap 0. (You must xhow full working on your handwritten working for this part to get full marks)

If $E=\var{E}$ volts, $V=\var{V}$ volts and $R=\var{R}$ ohms, what is the value of $r$ ?

$r=$ Gap 1. ohms (answer to 2 decimal places)

Advice

For the first part we have

$\begin{align}V&=\frac{ER}{R+r}\quad\text{multiply both sides by }(R+r)\\V(R+r)&=ER\quad\text{expand brackets}\\VR+Vr&=ER\quad\text{subtract }VR\text{ from both sides}\\Vr&=ER-VR\quad\text{factorise RHS}\\Vr&=R(E-V)\quad\text{divide both sides by }V\\r&=\frac{R(E-V)}{V}\end{align}$

Substituting $E=\var{E}$ volts, $V=\var{V}$ volts and $R=\var{R}$ ohms into the above formula gives

$\begin{align}r&=\frac{R(E-V)}{V}\\&=\frac{\var{R}(\var{E}-\var{V})}{\var{V}}\\&=\frac{\var{R*(E-V)}}{\var{V}}\\&\approx\simplify{{precround(R*(E-V)/V,2)}}\end{align}$

Hence $r$ is $\simplify{{precround(R*(E-V)/V,2)}}$ ohms (correct to two decimal places).

Give a worked solution to the whole question.

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This question is used in the following exams:

SCC Class Test 1 by Hanh Vo in SCC PrelimMaths.
Don's copy of Class Test 1 for testing by Don Shearman in Don's workspace.
SCC Final Exam by Hanh Vo in SCC PrelimMaths.

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Checkpoint description

Don Shearman 11 months, 1 week ago

Hanh Vo 3 years, 4 months ago

Don Shearman 3 years, 8 months ago

Error in variable testing condition

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Generated value: `integer`

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Voltage drop formula

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Don Shearman 11 months, 1 week ago

Hanh Vo 3 years, 4 months ago

Don Shearman 3 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`