Electronic circuit

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From users who are members of MfEP Progress Quizzes :

Hanh Vo

said

Ready to use

3 years, 4 months ago

History

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Don Shearman 11 months, 4 weeks ago

Published this.

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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 as a copy of Reaction forces.

Name	Status	Author	Last Modified
Reaction forces	Ready to use	Don Shearman	08/05/2024 06:12
Electronic circuit	Ready to use	Don Shearman	17/04/2024 02:20
Blending Liquids	draft	Don Shearman	16/04/2024 03:07
Electronic circuit 2	draft	Don Shearman	17/04/2024 02:24

There are 3 other versions that do you not have access to.

Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

c	integer	7
a	integer	5
b	integer	6
d	integer	8
f	integer	12
factor1	number	5
factor2	number	8

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..10)

JME function reference

Description

Coefficent of V_1 in equation 2, random integer between 2 and 10

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`

→ Used by:

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap $V_1$ Number entry
  Add
2. Gap $V_2$ Number entry
  Add

Gap-fill

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

Analysis of a particular electronic circuit shows that the voltage drops across two sections of the circuit must satisfy the equations

$\begin{align} \simplify[unitFactor]{-{a}V_2 + {b}V_1} &= 0\\ \simplify[unitFactor]{{c}V_1 - {d}V_2 + {f}} &=0 \end{align}$ Find the values of

$V_1$ and

$V_2$ . Write your final answer in the box below and show full working on your handwritten notes. (Note: Leave answers in fraction form. For example to enter the fraction

$\frac32$ type 3/2.)

$V_1=$ $V_1$ V

$V_2=$ $V_2$ V

Advice

We need to solve

$\begin{align} \simplify[unitFactor]{-{a}V_2 + {b}V_1} &= 0\\ \simplify[unitFactor]{{c}V_1 - {d}V_2 + {f}} &=0\end{align}$ which can be written as

$\begin{align} \simplify[unitFactor]{{b}V_1-{a}V_2} &= 0\quad (1)\\ \simplify[unitFactor]{{c}V_1 - {d}V_2} &=\simplify{-{f}}\quad (2)\end{align}$

Using the method of elimination to eliminate $V_2$ we multiply equation $(1)$ by $\var{factor2}$ and equation $(2)$ by $\var{factor1}$ to get

$\begin{align}\simplify[unitFactor]{{b*factor2}V_1-{a*factor2}V_2} &= 0\quad (3)\\ \simplify[unitFactor]{{c*factor1}V_1 - {d*factor1}V_2} &=\simplify{-{f*factor1}}\quad (4)\end{align}$

Taking $(3)-(4)$ we obtain

$\simplify{({b*factor2}-{c*factor1})V_1={f*factor1}}$

and hence

$V_1=\simplify{{a*f/(b*d-a*c)}}$

Substituting this into $(1)$ gives us

$\simplify[unitFactor]{-{a}V_2 + {b*a*f/(b*d-a*c)}} = 0$ and hence

$V_2=\simplify{{b*f/(b*d-a*c)}}$

Thus the voltages are $V_1=\simplify{{a*f/(b*d-a*c)}}$ volts and $V_2=\simplify{{b*f/(b*d-a*c)}}$ volts.

(Note that this problem could also be solved by eliminating $V_1$ instead of $V_2$ or by the method of substitution, however the final answers will be the same.)

Give a worked solution to the whole question.

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Part	Test	Passed?
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Number entry
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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.

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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Checkpoint description

Don Shearman 11 months, 4 weeks 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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Electronic circuit

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Don Shearman 11 months, 4 weeks 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`