Question

answer the above questions​

Answer 1

Firstly calculating net equivalent resistance in the circuit : Req = {((24×24)/(24+24))+12} = 24 ohm

Current in the 12ohm resistor = 6/24 = 1/4

The difference in the readings of A1 and A2 is zero, as they are in series connected with the same wire, hence the value of current will remain constant all over the circuit.

Answer 2

In parallel two resistances 24ohm and 24ohm can be replaced by 12 ohm (since, 1/resultant resistance = 1/24 + 1/24). So, now, this 12ohm and the existing 12ohm resistors are in series. To get the resultant equivalent resistance we simply add , i.e., 12+12 = 24 ohm. So, the allover resistance in the circuit is 24 ohm and not (24+24+12) = 60 ohm [Don't do this mistake].

Now, Overall voltage (emf) = 6V

Overall resistance in the circuit = 24 ohm

Overall current = 6/24 Ampere = 1/4 Ampere

Current in 12ohm resistor = 1/4 Ampere

Now, clearly in the A1 and A2 the amount of current entering is 1/4 Ampere. So, no difference in the reading of ammeters A1 and A2