# Thevenin/Kirchhoff('s) laws.

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#1

Question as above.

My working:
Break each loop into its own current:

Simultaneous equations, solve:

Solutions:
and

Is this right? If it is, how do I do part (d)? Thevenin's theorem is for working out the simplified circuit of Figure 5, why would I use it to work out the short-circuit current through ?
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6 years ago
#2
(Original post by halpme)

Question as above.

My working:
Break each loop into its own current:

Simultaneous equations, solve:

Solutions:
and

Is this right? If it is, how do I do part (d)? Thevenin's theorem is for working out the simplified circuit of Figure 5, why would I use it to work out the short-circuit current through ?

How did you arrive at the second equation: ?

The solutions are not correct.

figure 7 is missing

Part c) using Thévenin:

Thévenin's theorem is based on superposition and relies on simplifying the circuit connected to the load, to that of a voltage source with an internal resistance which (when connected to the load) forms a potential divider.

1) open circuit the load resistor between points a and b.

2) calculate the current flowing through R2 for the new circuit.

3) calculate pd across R2.

4) the p.d. across points ab is the same as VR2 since these are in parallel.

5) replace the voltage source V1 with a short circuit and calculate the equivalent circuit resistance between points ab with GL omitted

6) redraw the circuit using the voltage calculated in 3) as the source with an internal resistance calculated in 5)

7) reconnect GL to the new voltage source as described by 6) and calculate the current.

ANSWER = 3.43 mA (2 d.p.)
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#3
(Original post by uberteknik)
How did you arrive at the second equation: ?
Using mesh currents, where the first loop has a clockwise current and the second loop has a clockwise current of and has a resistance value of because it has a value of 1mS which is the reciprocal of an Ohm, I think?

And on closer inspection, it should have a value of ... Apologies, not sure if this is the cause of the problem?
(Original post by uberteknik)
The solutions are not correct.

figure 7 is missing
They must be using a typesetter and the figure count has updated as they've taken a section out of the exam paper. Figure 7 is figure 5. Apologies.

(Original post by uberteknik)
Part c) using Thévenin:

Thévenin's theorem is based on superposition and relies on simplifying the circuit connected to the load, to that of a voltage source with an internal resistance which (when connected to the load) forms a potential divider.

1) open circuit the load resistor between points a and b.

2) calculate the current flowing through R2 for the new circuit.

3) calculate pd across R2.

4) the p.d. across points ab is the same as VR2 since these are in parallel.

5) replace the voltage source V1 with a short circuit and calculate the equivalent circuit resistance between points ab with GL omitted

6) redraw the circuit using the voltage calculated in 3) as the source with an internal resistance calculated in 5)

7) reconnect GL to the new voltage source as described by 6) and calculate the current.

ANSWER = 3.43 mA (2 d.p.)
With the new resistance value added, I get an value of 24mA. Why can I not use mesh currents to analyse this circuit, because the nodes a and b are shorted anyway, it isn't as if the circuit is open. Can you only use mesh currents if there is a source on the other loop?
0
6 years ago
#4
(Original post by halpme)
Using mesh currents, where the first loop has a clockwise current and the second loop has a clockwise current of and has a resistance value of because it has a value of 1mS which is the reciprocal of an Ohm, I think?

And on closer inspection, it should have a value of ... Apologies, not sure if this is the cause of the problem?

They must be using a typesetter and the figure count has updated as they've taken a section out of the exam paper. Figure 7 is figure 5. Apologies.

With the new resistance value added, I get an value of 24mA. Why can I not use mesh currents to analyse this circuit, because the nodes a and b are shorted anyway, it isn't as if the circuit is open. Can you only use mesh currents if there is a source on the other loop?
1mS = 1000 ohms is correct.

You can use mesh but you have set up the loop equations incorrectly - specifically the second equation.

Nodes a and b are NOT shorted in Thévenin analysis.

Nodes a and b are opened.

The voltage source V1 is shorted.

Review Kirchoff's mesh method again more rigorously. They are simple errors but it's critical to get the methodology correct which means meticulous attention to detail i.e. current direction and hence +/- signs.
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#5
(Original post by uberteknik)
1mS = 1000 ohms is correct.

You can use mesh but you have set up the loop equations incorrectly - specifically the second equation.

Nodes a and b are NOT shorted in Thévenin analysis.

Nodes a and b are opened.

The voltage source V1 is shorted.

Review Kirchoff's mesh method again more rigorously. They are simple errors but it's critical to get the methodology correct which means meticulous attention to detail i.e. current direction and hence +/- signs.

Perhaps you can help me with this question which is the same setup?
I get:

where and are the clockwise currents in the first and second loop respectively.
I end up with:

and
Simplifying to:

leaving

and .
Is this correct now?
0
6 years ago
#6
(Original post by halpme)

Perhaps you can help me with this question which is the same setup?
I get:

where and are the clockwise currents in the first and second loop respectively.
I end up with:

and
Simplifying to:

leaving

and .
Is this correct now?
What course / year are you studying?

You seem to be having difficulty delineating the different methods for solving the problems.

They all take their roots from Kirchoff's rules, but the methodology for arriving at the answers are different.

Mesh analysis does not use the same methodology as Thévenins theorem, which itself does not use the same methodology as Superposition theorem and none are the same as Norton's theorem etc.

The questions are asking you to solve each problem using the stated method.

0
6 years ago
#7
(Original post by halpme)

Question as above.

My working:
Break each loop into its own current:

Simultaneous equations, solve:

Solutions:
and

Is this right? If it is, how do I do part (d)? Thevenin's theorem is for working out the simplified circuit of Figure 5, why would I use it to work out the short-circuit current through ?
i'm also studying this, do you know of any books or anything which have lots of questions on this topic thanks
0
#8
(Original post by uberteknik)
What course / year are you studying?

You seem to be having difficulty delineating the different methods for solving the problems.

They all take their roots from Kirchoff's rules, but the methodology for arriving at the answers are different.

Mesh analysis does not use the same methodology as Thévenins theorem, which itself does not use the same methodology as Superposition theorem and none are the same as Norton's theorem etc.

The questions are asking you to solve each problem using the stated method.

I'm studying BEng Electronic Engineering (Year 1).

I seem to have mislead you here, I know the question asked to use superposition theorem, but I wanted to try it using mesh currents because you noticed that I was having a problem with that particular analysis method. Apologies.
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#9
(Original post by Ilovemaths96)
i'm also studying this, do you know of any books or anything which have lots of questions on this topic thanks
Any standard physics/electronics first year UG books have questions. If you have a library, there's almost an unlimited supply of circuit analysis questions. Online there are a lot too.
0
6 years ago
#10
(Original post by halpme)
I'm studying BEng Electronic Engineering (Year 1).

I seem to have mislead you here, I know the question asked to use superposition theorem, but I wanted to try it using mesh currents because you noticed that I was having a problem with that particular analysis method. Apologies.
Going back to the first question then:

So that we know we have the right answers, the current through GL calculated thus

R2 ¦¦ GL = 4000 ohms ¦¦ 1x10-3 Siemens =

So 3.429 mA is the check answer we need.

Draw the circuit and label the loops. At this stage, we do not know the direction of all the currents so it's vital to stick rigidly to whatever convention we choose until the answers are known.

Also place the voltage potentials (signs) developed across the components in accordance with the current convention chosen. NB once again some of these may not be correct at this stage but it is vital to stick with the chosen convention or the answers will be incorrect. Also note that because R2 is in parallel with GL then the voltage potential developed across the pair is the same and hence the polarity of the signs must also be the same.

I initially chose Loop 1 to follow the electron current flow direction based on the polarity of the supply. Loop 2 was chosen to flow in the same direction as Loop 1 through R2 although this does not matter so long as you stick with the sign convention you choose.

The diagram looks like:

Mesh equations following Kirchoff's voltage rules around each loop and paying attention to the potential signs and loop direction labelled on the diagram:

12 - 2000I1 - 4000(I1 + I2) = 0 (Loop 1)

12 - 2000I1 - 4000I1 - 4000I2 = 0 (expanding)

12 - 6000I1 - 4000I2 = 0 (collecting terms) ..........eq 1

4000(I1 + I2) + 1000I2 = 0 (Loop 2)

NB the sign convention used in Loop2 follows the loop arrows and labelled signs. i.e. because the loop arrow flows from +ve to -ve on GL, the KVL rule means the current is added not subtracted.

4000I1 + 4000I2 + 1000I2 = 0 (expanding)

4000I1 + 5000I2 = 0 (collecting terms)..................eq 2

6000I1 + 7500I2 = 0 (eq 2 x 3/2)

and adding this to eq 1 to eliminate I1 produces:

12 + 3500I2 = 0

3500I2 = -12

I2 = -12 / 3500 = -3.429 mA

The -ve sign tells us the current flow direction for loop 2 via GL is opposite to that shown by the loop arrow in that part of the circuit.

I2 = +3.429mA in accordance with the original electron flow convention chosen.

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