Question
Question: The ammeter reading in the circuit below is 
We apply Kirchhoff’s current law at point O as follows,
I1+I2+I3=11 …… (2)
Using equation (1), we can write the above equation as,
2V+V+3V=11
⇒611V=11
⇒V=6V
Now, the current flowing through the ammeter is I2. Therefore, from equation (1), we have,
I2=V
Therefore, the current through the ammeter is,
I2=6A
So, the correct answer is option (C).
Additional information: Kirchhoff’s voltage current law states that at any junction in the circuit the total incoming current is equal to the total current leaving the junction. The expression of Kirchhoff’s current law can be given as,
Iin=Iout
Note:
Students should remember that in series circuits, the current remains constant. Therefore, the total current leaves the resistor R2 is equal to total current leaves ammeter. KCL can be applied at any point in the circuit to determine the value of unknown current in the circuit. If we determine the values of current through each branch, the sum of the current should be 11 A.