Homework: DC circuit help

Question

I need help with the following problem:

Find range of resistance R4 and voltage U4 so that the range of current I is $$I \in [-3A,-1A]$$

E=5V, E4=6V, Ig=1A, Ig4=-3A, R1=2ohm, R2=3ohm, R3=5ohm

Firstly, I set current I=-3A. Setting potential V3=0 V, current I23=0.25 A.

Using Kirchhoff's law on node 2 I12=1.25 A. On node 1 I14=1.75 V.

Now potential of nodes are V1=5 V, V2=1.25 V, V3=0 V V4=1.5 V.

With potential of nodes method on node 4, R4=2.09 ohm. Then, U4=-4.5 V.

In my book's solution, it says that R4 starts from 0.22 ohm and U4 starts from -4.5 V.

Could someone check this?

Answer 1

By applying the superposition theorem to your circuit I have noticed that the current you are looking for can only be dependent from voltage source E, because when you are looking at another source, E is shorted, so your current gets 0A. So now it depends on E alone (all other sources shorted or open).

• Ig is open
• Ig4 is open
• E4 is shorted

So R1 gets in serial to R4 and R2 and R3 get just serial, too. Finally the serial resistance groups are in parallel.

R14=2 Ω + R4
R23=8 Ω
=> R14|23 = (16 Ω + 8R4)/(10 Ω + R4)
I=-(E / R14|23)
 =-(50 Ω + 5R4)/(16 Ω + 8R4) A
=> R4=(-50 ΩA - 16I Ω )/(8I + 5 A)

When I=-3 A:
R4=2/19 Ω ≈ 0.1 Ω
When I=-1 A:
R4=34/3 Ω ≈ 11.3 Ω

As R4 is effectively rising with I, it must be that R4∈[0.1 Ω; 11.3 Ω].

Hope this helped a bit, though late....