Need some homework help (Voltage dividers with two sources)

[m_c]

It's probably sleep deprivation, but I'm currently having a total mental block over this specific part for an assignment I need to get finished.

I need to calculate the voltage at point D.
I know what it is (14V), as I've ran it through a circuit simulator, and I think that it most likely involves some form of Kirchoff's law, but I'm drawing a complete blank.

I highly suspect I'm missing the blindingly obvious at this point :/

So anybody care to point me in the correct direction?

 

[Dragnet]

This made me dredge up my minimal remembrance of Kirchoff! Try superposition theorem. If there are two sources in a circuit, we can treat them individually then add the results together. Take each power source individually. If the 16V supply were grounded, what would VD be? Vice versa, then sum. Both are positive in this instance with reference to ground, so you don't have other signs to worry about.

 

[Dr.Al]

Superposition is one way you could deal with it, but I'd probably just use KCL as you've suggested.

Current (call it i1) through the left resistor is (E-D)/R. Current (call it i2) through top resistor is (C-D)/(2R). Current (call it i3) through bottom resistor is (D/2R).

KCL gives i1 + i2 = i3

Substitute in the above equations and solve for D, gives (2E+C)/4.

 

[Red'n'Black]

Another shortcut you could use:

Point D looks like 12v with R in series (using Thevenin). So now you have that on one side, and 16v in series with R on the other. So as the two Rs are the same, the voltage is the average of 12 and 16, or 14v.

 

[Tangledfeet]

 

[m_c]

Thanks guys

Superposition is one way you could deal with it, but I'd probably just use KCL as you've suggested.

Current (call it i1) through the left resistor is (E-D)/R. Current (call it i2) through top resistor is (C-D)/(2R). Current (call it i3) through bottom resistor is (D/2R).

KCL gives i1 + i2 = i3

Substitute in the above equations and solve for D, gives (2E+C)/4.

This is what I tried last night, but managed to make a complete pigs ear off it!
My biggest issue, is it's about two years since I studied the applied science module, and none of the other modules until now have used any of it.

 

[Seadog]

It's over 50 years for me, which is why I didn't have a clue

 

[Dr.Al]

It's over 50 years for me, which is why I didn't have a clue

It's the sort of thing I ask people in interviews (for electronics design engineer roles) so it would have been a bit embarrassing if I couldn't do it!

 

[Dragnet]

Superposition is one way you could deal with it, but I'd probably just use KCL as you've suggested.

Current (call it i1) through the left resistor is (E-D)/R. Current (call it i2) through top resistor is (C-D)/(2R). Current (call it i3) through bottom resistor is (D/2R).

KCL gives i1 + i2 = i3

Substitute in the above equations and solve for D, gives (2E+C)/4.

Great shout - I'd forgotten about the current laws. I initially got bogged down with equivalent circuits before remembering superposition, but your way is much more elegant and doesn't need thirds of a parallel resistor...yes, I covered a few post-its and also had to rack my brains to remember how fractions work. Good workout, m_c, thank you!