Question

The Current Ix in network is

​

Answer 1

Zero

Since ammeter has zero resistance,

the branch of 9 ohm resistor is at same potential therefore, will be short circuited.

Answer 2

Answer:

The answer is 1 if the ammeter has an internal resistance of 4.5 ohms.

Explanation:

Assuming the 2 A ammeter has an internal resistance R, Let us get an equation in terms of ix and R.

By using Kirchhoff's Voltage law (KVL) in the ammeter and 9ohm resistor circuit (ckt) we get

9×ix - 2×R =0

ix/R = 2/9 ------->>>>>> giving ix as 1 and R as 4.5

To verify the above conclusion let's apply KCL at the principal node where the 6ohm, 9ohm and 2A meet

we get the current through 6ohm resistor as 1A

Now apply KVL for 3v, 6ohm and 9ohm

3 + 6(1) -9(1) =0

9-9=0

HENCE PROVED!!!