Question

Figure below shows a network of resistors, cells, and a capacitor at steady state.

What is the current through the resistance 4 $\Omega$?

Answer 1

7/9 A

Answer 2

We first note that at DC steady state the capacitor is “open‐circuited” (i.e. it does not pass DC) so only the resistor–voltage‐source network matters. One can “redraw” the loop so that it is a single closed series circuit containing
 • two 2 Ω resistors, one 1 Ω resistor, and one 4 Ω resistor (total resistance Rₜ = 2 + 2 + 1 + 4 = 9 Ω)
 • three cells whose algebraic “net EMF” turns out to be 6 V + 3 V – 2 V = 7 V (that is, one of the sources opposes the other two).

Then by applying Kirchhoff’s Voltage Law (KVL) around the loop we have

  Net EMF = I·Rₜ  ⟹  I = (7 V)/(9 Ω) = 7/9 A.

Thus the current in the entire loop – and in particular the branch containing the 4 Ω resistor – is 7/9 A.

Below is a minimal mermaid diagram of the simplified loop:

(Note: The polarities of the voltage sources are taken so that the algebraic sum is 7 V in the loop.)

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Core minimal explanation:
At DC, the capacitor is open. The remaining loop has a total resistance of 9 Ω and a net EMF of 7 V (taking into account the proper polarities). Thus by KVL, the current is I = 7/9 A.