Please Help!! Pretty sure my teacher made an impossible question on my test.

Papabravo

Joined Feb 24, 2006
18,400
So, what have you tried so far to get a solution?
Can you write a KCL expression for the three currents?
Can you write a pair of KVL expressions for the two loops?
There is a solution, and the problem is not impossible.
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hexreader

Joined Apr 16, 2011
547
You guys are over-thinking this.

Look very carefully and eventually you will spot a glaringly simple solution.

I will not give the solution away, as I would guess that the whole purpose of the question is to see who spots the easy way.

EDIT: after seeing next reply #4, maybe your teacher wants you to do this Papabravo's way. - hard to guess. Best to give easy way and hard way if you want full marks.
 
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Papabravo

Joined Feb 24, 2006
18,400
You guys are over-thinking this.

Look very carefully and eventually you will spot a glaringly simple solution.

I will not give the solution away, as I would guess that the whole purpose of the question is to see who spots the easy way.
Possibly, but the basic method never fails.
 

Thread Starter

rycooproberts

Joined May 4, 2022
13
Well I've already submitted my answers since they were during my test today. The semester's over and my teacher hasn't responded to my email so I'm trying to find someone else to help. If I build this in real life, the 16V supply combats the 10V supply (and possible damages it) since there's no diode and I don't know internal resistances of the supplies... so I really don't know how to get any of these numbers. It looks like there's a total of 6v that makes the loop and it travels through 3 resistors, but shares the 4 and 6 with the other supply. I mean, I got that 1 amp should go through I3 if you ignore the 16V supply, but only 5/8 of the current actually comes back through the loop. I mean, I have an associates degree in electronics and just can't see how to get these numbers. That's why I'm asking for help.
 

ronsimpson

Joined Oct 7, 2019
2,058
If I build this in real life, the 16V supply combats the 10V supply (and possible damages it)
This is not real life. The 10V supply can output current or take it in. It is a perfect 10V with 0 ohm resistance. Both supplies force their voltage on the world. Does that help?

Think about 6 + 4 ohms is 10 ohms. Next, 10 volts is forced across the 10 ohms. (don't over think! don't ask where the current is coming from!) 10V, 10 ohms, what is the current in the resistors?
 

MrSalts

Joined Apr 2, 2020
1,762
Well, the 10v source across r2+r3 is fixed at 10v, right? That means the voltage drop across R1 must be 6v, right?
which means the current thigh R1 (2 ohms) must be 3 amps = I1, right?

Now the fun part, if the voltage drop across R2 + R3 is 10v, and the resistance is 10 ohms, then I2 = 1 amp. That means I3 must be -2 amps.
 

ErnieM

Joined Apr 24, 2011
8,292
I spy with my little eye... a two ohm resistor with 16 volts on one side and 10 volts on the other.

Hrmmm, now what current could THAt be???
 

Thread Starter

rycooproberts

Joined May 4, 2022
13
This is not real life. The 10V supply can output current or take it in. It is a perfect 10V with 0 ohm resistance. Both supplies force their voltage on the world. Does that help?

Think about 6 + 4 ohms is 10 ohms. Next, 10 volts is forced across the 10 ohms. (don't over think! don't ask where the current is coming from!) 10V, 10 ohms, what is the current in the resistors?
I get that there are some assumptions, but we can't ignore the 16V power supply in the calculations. If I3 has 1 amp because of (10V/10ohm), then there is no return current in the 16V supply. lf we calculate them together (but separately,) then 16V/12ohm resistors in series gives 1.33Amp through the resistors and 10V gives 1 amp, but that makes 2.3amps going through I2. (2.3amps x 4ohm) means 9.2V across 4 ohm resistor and (2.3amps x 6ohm resistor) is 13.8V across the 6 ohm resistor and somehow we get 23V across the 10ohms worth of resistors. I mean, does that sound like the numbers I'm supposed to get? since neither power supply is even close to that?
 

Thread Starter

rycooproberts

Joined May 4, 2022
13
Well, the 10v source across r2+r3 is fixed at 10v, right? That means the voltage drop across R1 must be 6v, right?
which means the current thigh R1 (2 ohms) must be 3 amps = I1, right?

Now the fun part, if the voltage drop across R2 + R3 is 10v, and the resistance is 10 ohms, then I2 = 1 amp. That means I3 must be -2 amps.
The problem with that is having a positive voltage and negative current.
 

Thread Starter

rycooproberts

Joined May 4, 2022
13
I should also add that for all of the items requested (V1, V2, V3, and I3) the answers were in multiple choice form and were whole numbers.
 

MrChips

Joined Oct 2, 2009
26,077
So you have figured that R1 + R2 = 4 + 6 = 10Ω
And you have figured that I2 = 10V/10Ω = 1A

What is the voltage V1?
 

crutschow

Joined Mar 14, 2008
29,773
You need to assume that the voltage sources are ideal with no internal impedance, and that they can source or sink current as needed.

An example is a battery.
It can source current if a resistive load is applied, but will absorb current if a charger with an output voltage larger than the battery voltage is applied.
 
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