When two batteries are connected in parallel across a load resistor, the voltage across the load is determined by the battery with the lower internal resistance because it can deliver more current to the load without significant voltage drop.
In this case, one of the batteries has an internal resistance of 1 ohm, and the other has an internal resistance of 2 ohms. Therefore, the battery with the 1 ohm internal resistance will provide a larger portion of the current to the load, resulting in a smaller voltage drop across its internal resistance.
To calculate the voltage across the load, we can use the formula for the total resistance in a parallel circuit:
1/R_total = 1/R1 + 1/R2
where R1 is the internal resistance of the first battery (1 ohm), and R2 is the internal resistance of the second battery (2 ohms).
1/R_total = 1/1 + 1/2 = 2/2 + 1/2 = 3/2
R_total = 2/3 ohms
Now, we can use Ohm's law to calculate the current flowing through the circuit:
I = E_total / R_total
where E_total is the total electromotive force (EMF) of the two batteries in parallel. Since the EMFs add up in parallel, E_total = 12V + 13V = 25V.
I = 25V / (2/3 ohms) = 75/2 A = 37.5 A
Now that we know the current, we can calculate the voltage drop across the internal resistance of the first battery (1 ohm):
V1 = I * R1 = (37.5 A) * (1 ohm) = 37.5 V
So, the voltage across the load resistor is:
V_load = E_total - V1 = 25V - 37.5V = -12.5V
Since the voltage across the load resistor is negative, it means that the batteries are providing less voltage than the load resistor. To find the absolute value of the voltage across the load, we take the magnitude:
|V_load| = |-12.5V| = 12.5V
So, the voltage across the load resistor is 12.5 volts.
Therefore, none of the given answer choices (A, B, C, D) are correct. The voltage across the load resistor is 12.5V.
