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Electronics (Superposition) watch

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1. Right, here is a mechanical engineer, who is being forced to do basic electronics with an exam tomorrow.

There are 2 question and i'm not even going to attempt the other one because it has complex numbers in and e's and stuff so... here is one of the likely questions tomorrow.

How would you do it?

Any help appreicated and rewarded.

Left: Calculate the current drawn from the ideal 9V source shown.
Right: Obtain an expression for the current flowing through resistor R5
Attached Images

2. I'll have a go, should be able to do it. What's that mark at the bottom of the left one?
3. (Original post by Nylex)
What do you need to find out? Current in one of the resistors?
sorry, updated the post.

i'm guessing superposition, i dont really like nodal anaysis.
4. (Original post by Ollie)
sorry, updated the post.

i'm guessing superposition, i dont really like nodal anaysis.
I've edited my post accordingly .

Not sure what nodal analysis is, but I can do superposition.
5. (Original post by Nylex)
I'll have a go, should be able to do it. What's that mark at the bottom of the left one?
it might well be my scribbles imitating the earth sign.
6. Oh ok. So far, I've turned off the voltage source and got a current through that branch due to the current source as 2.54 mA (going downwards). The hard bit is turning off the current source.

Edit:

Ok, I replaced the voltage source with a short circuit. Then I calculated the currents in the 3 K and 10 K resistors using the current splitter rule. I then said the current from the 10 K resistor is split between the 6 K and the other 3 K resistor. Used the current splitter rule again to work out the current in the 6 K and added it to the current from the 3 K I'd already worked out. Hopefully that's right.
7. (Original post by Nylex)
Oh ok. So far, I've turned off the voltage source and got a current through that branch due to the current source as 2.54 mA (going downwards). The hard bit is turning off the current source.
am i right in saying, you turn off the voltage source by making it a short circuit and turn off the current source by making it an opn circuit?
8. (Original post by Ollie)
am i right in saying, you turn off the voltage source by making it a short circuit and turn off the current source by making it an opn circuit?
9. (Original post by Ollie)
am i right in saying, you turn off the voltage source by making it a short circuit and turn off the current source by making it an opn circuit?
Yes. Edited my posted and explained what I did.
10. current splitter rule,

is this Kirchoff,

Iin =Iout?

How did you work out the current through the 3k3 and the 10 at the start. it doesn't act like a potential divider does it?
11. (Original post by Ollie)
current splitter rule,

is this Kirchoff,

Iin =Iout?

How did you work out the current through the 3k3 and the 10 at the start. it doesn't act like a potential divider does it?
The current splitter rule is:

I2 = IR1/(R1 + R2)

Notice that the current in one branch depends on the resistance in the other branch.

The resistors don't have the same current through them, so they have to be in parallel.

Edit: I'm still stuck when turning the current source off .

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