# Thevenin/Kirchhoff('s) laws.

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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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#2

(Original post by

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 ?

**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 V

_{R2}since these are in parallel.

5) replace the voltage source V

_{1}with a short circuit and calculate the equivalent circuit resistance between points ab with G

_{L}omitted

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

7) reconnect G

_{L}to the new voltage source as described by 6) and calculate the current.

ANSWER = 3.43 mA (2 d.p.)

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

(Original post by

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 V

5) replace the voltage source V

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

7) reconnect G

ANSWER = 3.43 mA (2 d.p.)

**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 V

_{R2}since these are in parallel.5) replace the voltage source V

_{1}with a short circuit and calculate the equivalent circuit resistance between points ab with G_{L}omitted6) redraw the circuit using the voltage calculated in 3) as the source with an internal resistance calculated in 5)

7) reconnect G

_{L}to the new voltage source as described by 6) and calculate the current.ANSWER = 3.43 mA (2 d.p.)

**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?

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#4

(Original post by

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?

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

**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?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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(Original post by

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.

**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?

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#6

(Original post by

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?

**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?

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.

http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

http://www.allaboutcircuits.com/vol_1/chpt_10/8.html

http://www.allaboutcircuits.com/vol_1/chpt_10/7.html

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#7

**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 ?

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reply

(Original post by

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.

http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

http://www.allaboutcircuits.com/vol_1/chpt_10/8.html

http://www.allaboutcircuits.com/vol_1/chpt_10/7.html

**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.

http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

http://www.allaboutcircuits.com/vol_1/chpt_10/8.html

http://www.allaboutcircuits.com/vol_1/chpt_10/7.html

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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(Original post by

i'm also studying this, do you know of any books or anything which have lots of questions on this topic thanks

**Ilovemaths96**)i'm also studying this, do you know of any books or anything which have lots of questions on this topic thanks

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#10

(Original post by

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.

**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.

So that we know we have the right answers, the current through G

_{L}calculated thus

R

_{2}¦¦ G

_{L}= 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 R

_{2}is in parallel with G

_{L}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 R

_{2}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 - 2000I

_{1}- 4000(I

_{1}+ I

_{2}) = 0 (Loop 1)

12 - 2000I

_{1}- 4000I

_{1}- 4000I

_{2}= 0 (expanding)

12 - 6000I

_{1}- 4000I

_{2}= 0 (collecting terms) ..........eq 1

4000(I

_{1}+ I

_{2}) + 1000I

_{2}= 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 G

_{L}, the KVL rule means the current is added not subtracted.

4000I

_{1}+ 4000I

_{2}+ 1000I

_{2}= 0 (expanding)

4000I

_{1}+ 5000I

_{2}= 0 (collecting terms)..................eq 2

6000I

_{1}+ 7500I

_{2}= 0 (eq 2 x 3/2)

and adding this to eq 1 to eliminate I

_{1}produces:

12 + 3500I

_{2}= 0

3500I

_{2}= -12

**I**= -12 / 3500 =

_{2}**-3.429 mA**

The -ve sign tells us the current flow direction for loop 2 via G

_{L}is opposite to that shown by the loop arrow in that part of the circuit.

I

_{2}= +3.429mA in accordance with the original electron flow convention chosen.

The two answers now concur.

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