A.singh18
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Hey so just needed to confirm whether I'm doing these questions right.

The circuit is labeled as D attached as P_20160320_145954.jpg
Question b is complete the answer is 5.3 mA

Question c is complete and the answer is 4.69mA

I have no idea how to do question D

All questions are shown in P_20160320_153730
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uberteknik
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(Original post by A.singh18)
Hey so just needed to confirm whether I'm doing these questions right.

The circuit is labeled as D attached as P_20160320_145954.jpg
Question b is complete the answer is 5.3 mA

Question c is complete and the answer is 4.69mA

I have no idea how to do question D

All questions are shown in P_20160320_153730
Hmmmm.

You have managed to answer them all incorrectly!

Voltage sources are modelled by:

i) Stating the EMF with a series internal resistance. i.e. the non-ideal voltage source will drop a p.d. across the internal resistance when placed in a series circuit with a load.

Current sources are modelled by:

ii) Stating the current source with a parallel resistance. i.e. the current source will share current with any load resistance also placed in parallel.

Using the above rules makes the answer to the first question c) and not d)


The application of Kirchoff's rule means:
For series circuits, the current must be identical at all points in the current path.
For parallel circuits, the p.d. across all parallel components sharing the same nodes must be identical.


Superposition requires voltage sources to be neutralised by replacing them with a short-circuit and then calculating the current through the load using the remaining current source.

Similarly, current sources are neutralised by replacing them with an open circuit and then calculating the p.d. developed across the load using the remaining voltage source. From that, the current through the load due to the voltage source alone can be calculated.

Current through R2 due to voltage source (current source open circuit):

I_{3k3} = \frac{V_s}{R_T} = \frac{17.6}{6500}= 2.71\rm mA since the current through the 3k3 load due to the voltage source must be the same as that through the total series resistance.

Current through R2 due to current source (voltage source short circuit):

\rm R_{parallel} = R2||R_{IS}||(R_{VS}+R1) = 1600\Omega

\rm V_{IS} = I_{IS}\rm \ x \ R_{parallel} = 5\rm x10^{-3} \rm \ x \ 1600 = 8V

Hence current through R2 due to current source:

\rm I_{3k3} = \frac{V_{IS}}{R2} = \frac{8}{3.3x10^{3}} = 2.424mA

Finally, the actual current through the load is calculated by summing the two answers above taking into account the direction of the currents.

\rm 2.71mA + 2.424mA = 5.134mA
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A.singh18
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Name:  P_20160320_203627.jpg
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Size:  493.7 KBThank you so much for the help. Just a few things I'm still confused on. So for part b, I 3.3 is the same as It? And I don't understand how you got r parallel. When I nullified the voltage source I got this
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sunni money
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(Original post by A.singh18)
Name:  P_20160320_203627.jpg
Views: 137
Size:  493.7 KBThank you so much for the help. Just a few things I'm still confused on. So for part b, I 3.3 is the same as It? And I don't understand how you got r parallel. When I nullified the voltage source I got this
you need to learn to quote people in your responses to them, otherwise they will not be aware of your reply.
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uberteknik
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(Original post by A.singh18)
Name:  P_20160320_203627.jpg
Views: 137
Size:  493.7 KBThank you so much for the help. Just a few things I'm still confused on. So for part b, I 3.3 is the same as It? And I don't understand how you got r parallel. When I nullified the voltage source I got this
Yes, it's the same as It because it's a series current provided by the voltage source. You need to calculate the total series resistance and use the source voltage, which gives the series current through the 3k3 load. i.e. the current in the series path must be the same at all points in the path. (think: where do the electrons flow?)


Nullified voltage source:

Because you incorrectly used the circuit modelled by answer d) when you should have used answer c).

i.e. the 105K is in parallel with the current source and not in series with it.

NB Read and understand this definition of current sources:"A current source provides a constant current, as long as the load connected to the source terminals has sufficiently low impedance. An ideal current source would provide no energy to a short circuit and approach infinite energy and voltage as the load resistance approaches infinity (an open circuit). An ideal current source has an infinite output resistance in parallel with the source. A real-world current source has a very high, but finite output resistance.



PS. Please use the "+quote" button when replying as the person/s responding to you will not know you have replied to them. Thanks.
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