For the circuit shown, determine the current flowing through the branch AB.

The circuit contains three identical branches. Each branch consists of a 15 V cell connected in series with a 3 Ω resistor. These branches are connected between the same two nodes A and B.

The lower branch AB consists of two 3 Ω resistors connected in series.

Therefore:

R_AB = 3 + 3 = 6 Ω

The three upper branches are identical voltage sources with equal internal resistances. Since they are connected in parallel, their equivalent voltage remains:

V = 15 V

The equivalent resistance of the three 3 Ω resistors in parallel becomes:

1/R = 1/3 + 1/3 + 1/3

R = 1 Ω

Thus the circuit reduces to:

• 15 V source
• 1 Ω equivalent series resistance
• 6 Ω load resistance between A and B

Total resistance:

R_total = 1 + 6

R_total = 7 Ω

Using Ohm's Law:

I = V/R

I = 15/7 A

This current flows through the AB branch because the 6 Ω combination forms the external load of the equivalent source network.

The problem demonstrates how multiple identical sources connected in parallel can be replaced by a single equivalent source, greatly simplifying circuit analysis.

Therefore, the current through AB is:

I_AB = 15/7 A