The reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ___ A.

after removing capacitor branch:(as in steady state capacitor is an open circuit)

• from middle node → one 8Ω to bottom
• from middle → 4Ω → right node
• right node → two 8Ω in parallel to bottom

step 1: right side

$$8||8=4\Omega$$

step 2: path via right:

$$4+4=8\Omega$$

so from middle node to bottom we have:

$$8\Omega\text{ (direct)}\parallel8\Omega\text{ (via right)}$$

R=4Ω

step 3: total resistance:

$$R_{total}=1+4=5Ω$$

step 4: current:

$$I=\frac{10}{5}=2A$$

current splits equally in two branches so answer is 1A