Answer :
The total emf of the battery is 6 V and the total resistance in the circuit is 6 "+". Using Ohm's law, the current delivered by the battery is calculated to be 1 A.
To find the current delivered by the battery, we need to consider both the total emf and the total internal resistance of the battery, as well as the equivalent resistance of the circuit. Since we have three identical cells of emf 2 V each connected in series, the total emf (E) of the battery will be:
E = 2 V + 2 V + 2 V
= 6 V
The total internal resistance (r) being the sum of all three cells' internal resistances, also in series, will be:
r = 1 "+" + 1 "+" + 1 "+" = 3 "+"
Since two identical resistors, each of resistance 6 "+", are connected in parallel, the equivalent resistance (R_eq) of these parallel resistors is given by:
1/R_eq = 1/6 "+" + 1/6 "+" = 1/3 "+"
R_eq = 3 "+"
The total resistance in the circuit is the sum of the equivalent resistance of the parallel resistors and the internal resistance of the battery:
R_total = R_eq + r = 3 "+" + 3 "+" = 6 "+"
Now, we can use Ohm's law to calculate the total current (I) delivered by the battery:
I = E / R_total
= 6 V / 6 "+"
= 1 A
Therefore, the current delivered by the battery is 1 A.
