Final answer:
Using Kirchhoff's Rules, the reading of the third ammeter in the branch with E will be the sum of the currents in R1 and R2, which are 2 A and 3 A, thus giving a total of 5 A.
Step-by-step explanation:
Finding Current Using Kirchhoff's Rules
To find the magnitude of the current in the upper branch of the complex circuit, we must apply Kirchhoff's Rules. Since the circuit cannot be simplified using Ohm's law and the series-parallel techniques, we use Kirchhoff's junction and loop rules. Given that ammeter readings in the branches with resistors R1 and R2 are 2 A and 3 A respectively, we start by applying Kirchhoff's first rule (junction rule) at point a which gives us the equation I1 = I2 + I3. This implies that current I1, which is the sum of the currents in R1 and R2 branches, is equal to the current in the upper branch plus the current in the branch containing the voltage source E.
For the loop involving R1 and R2, we apply Kirchhoff's second rule (loop rule) to find the equation representing the sum of the voltage drops around the loop. This will help us find the unknown current in the upper branch which comprises of R1, R2, and E.
In the case that the current in the branch with the voltage source is upward and currents in the other two branches are downward, we can make a statement about the direction of the net current Ia according to the equation Ia = I1 + Ic. If Ia is the total net current upwards, it would be equal to the algebraic sum of currents in the two downward-resistor branches plus the ammeter's reading in the voltage source's branch. Based on the provided information, we can deduce the reading of the third ammeter, which is in the same branch as the voltage source E, by using the junction rule.
The reading of the third ammeter, which is in the branch with E, will be the sum of the currents recorded by the other two ammeters, as no current is lost or gained in a junction according to Kirchhoff's junction rule. Therefore, the reading is 5 A, which is the algebraic sum of 2 A and 3 A.