Final answer:
The emf of the batteries in the circuit is 26.00 V and the polarity shown is correct.
Step-by-step explanation:
The circuit shown in the figure consists of two batteries connected in series with an external resistor. The ammeter reads a current of 1.30 A in the direction shown. To find the emf of the batteries, we can use Ohm's Law. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance.
In this case, the current through the external resistor is 1.30 A and the resistance is 10.00 Ω. Therefore, the voltage across the resistor is V = (1.30 A)(10.00 Ω) = 13.00 V.
Since the two batteries are connected in series, the total emf of the circuit is equal to the sum of the individual emfs. Therefore, the emf of the batteries can be calculated as:
Emf = 13.00 V + 13.00 V = 26.00 V
As for the polarity of the batteries, the positive terminal of one battery is connected to the negative terminal of the other battery. This ensures that the emfs of the batteries are additive. Therefore, the polarity shown in the figure is correct.