Electronic circuit 2

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Don Shearman 11 months, 3 weeks ago

Published this.

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Don Shearman 2 years ago

Created this as a copy of Electronic circuit.

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

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

c	integer	8
a	integer	8
b	integer	2
d	integer	2
f	integer	8
factor1	number	4
factor2	number	1

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..10)

JME function reference

Description

Coefficent of I_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`

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap $I_1$ Number entry
  Add
2. Gap $I_2$ 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 currents across two sections of the circuit must satisfy the equations

$\begin{align} \simplify[unitFactor]{-{a}I_2 + {b}I_1} &= 0\\ \simplify[unitFactor]{{c}I_1 - {d}I_2 + {f}} &=0 \end{align}$ Find the values of

$I_1$ and

$I_2$ . 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.)

$I_1=$ $I_1$ amps

$I_2=$ $I_2$ amps

Advice

We need to solve

$\begin{align} \simplify[unitFactor]{-{a}I_2 + {b}I_1} &= 0\\ \simplify[unitFactor]{{c}I_1 - {d}I_2 + {f}} &=0\end{align}$ which can be written as

$\begin{align} \simplify[unitFactor]{{b}I_1-{a}I_2} &= 0\quad (1)\\ \simplify[unitFactor]{{c}I_1 - {d}I_2} &=\simplify{-{f}}\quad (2)\end{align}$

Using the method of elimination to eliminate $I_2$ we multiply equation $(1)$ by $\var{factor2}$ and equation $(2)$ by $\var{factor1}$ to get

$\begin{align}\simplify[unitFactor]{{b*factor2}I_1-{a*factor2}I_2} &= 0\quad (3)\\ \simplify[unitFactor]{{c*factor1}I_1 - {d*factor1}I_2} &=\simplify{-{f*factor1}}\quad (4)\end{align}$

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

$\simplify{({b*factor2}-{c*factor1})I_1={f*factor1}}$

and hence

$I_1=\var{rational(a*f/(b*d-a*c))}$

Substituting this into $(1)$ gives us

$\simplify[unitFactor]{-{a}I_2 + {b*a*f/(b*d-a*c)}} = 0$ and hence

$I_2=\simplify{{b*f/(b*d-a*c)}}$

Thus the currents are $I_1=\simplify{{a*f/(b*d-a*c)}}$ amps and $I_2=\simplify{{b*f/(b*d-a*c)}}$ amps.

(Note that this problem could also be solved by eliminating $I_1$ instead of $I_2$ 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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Number entry
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This question is used in the following exam:

Class Test 1 by Don Shearman in MfEP Progress Quizzes.

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Checkpoint description

Don Shearman 11 months, 3 weeks ago

Don Shearman 2 years ago

Error in variable testing condition

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Generated value: `integer`

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Electronic circuit 2

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Don Shearman 11 months, 3 weeks ago

Don Shearman 2 years ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`