Step 1: Understanding the Question:
The topic of this question is Current Electricity, analyzing the combination of cells.
We must compute the total short-circuit current capability between the output terminals A and B of the given network. Step 2: Key Formula or Approach:
For parallel branches, the equivalent EMF is $E_{eq} = \frac{\sum E_i/r_i}{\sum 1/r_i}$ and the equivalent internal resistance is $r_{eq} = \frac{1}{\sum 1/r_i}$. Step 3: Detailed Explanation:
Each of the top three rows consists of three cells in series.
For each active branch: $E_{branch} = 5 + 5 + 5 = 15 \text{ V}$ and $r_{branch} = 3 + 3 + 3 = 9 \ \Omega$.
Since we have three identical parallel branches, the equivalent system parameters are:
$E_{eq} = 15 \text{ V}$ and $r_{eq} = \frac{9}{3} = 3 \ \Omega$.
The short-circuit current that can be delivered between terminals A and B is:
\[ I = \frac{E_{eq}}{r_{eq}} = \frac{15 \text{ V}}{3 \ \Omega} = 5 \text{ A} \] Step 4: Final Answer:
The current is 5 A.
