What is the voltage u in the circuit? How to find it by superposition theorem? Or other methods are also possible. Can you write down the process in detail?

Solution: 2A current source is open when 1 and 2V voltage source act alone.

Main current: I = 2/[2+2 ∨ (2+2)] = 2/(2+4/3) = 2/(10/3) = 0.6 (a).

Parallel branch voltage: u1= 0.6× [2 ∨ (2+2)] = 0.6× 4/3 = 0.8 (v).

U'=0.8×2/(2+2)=0.4(V).

2. When 2A current source acts alone, 2V voltage source is short-circuited.

The two 2 Ω resistors on the left are connected in parallel, which is equivalent to 2 Ω 2 =1(Ω) resistor.

The terminal voltage of the current source, that is, the required voltage: u "= 2× [(2+1) ∑ 2] = 2× 3× 2/(3+2) = 2.4 (v).

3. superposition theorem: U=U'+U"=0.4+2.4=2.8(V).

Using Thevenin theorem. Disconnect the resistor R = 2Ω from the circuit, as shown in the following figure:

According to the current setting in the figure and KVL: 2× (I-2)+2i = 2, I= 1.5(A).

uoc = Uab = Uam+Umb = 2×2+2×I = 4+2× 1.5 = 7(V).

Short-circuit the voltage source and turn on the current source:

req = Rab = 2+2∑2 = 3(ω).

Thevenin: u = uoc× r/(req+r) = 7× 2/(3+2) = 2.8 (v).