When two batteries are connected in parallel, the voltage across each battery is the same, and the total current through the circuit is the sum of the currents through each branch.
Let's call the current through battery A I_A and the current through battery B I_B. Then the total current through the circuit is:
I = I_A + I_B
Using Ohm's Law, we can find the current through each battery in terms of the terminal voltage and internal resistance:
I_A = (V_A - V_R) / R_A
I_B = (V_B - V_R) / R_B
where V_A and V_B are the terminal voltages of batteries A and B, V_R is the voltage across the 80 ohm resistor, R_A and R_B are the internal resistances of batteries A and B, and I_A and I_B are the currents through batteries A and B, respectively.
Since the voltage across the 80 ohm resistor is the same as the voltage across the batteries, we can write:
V_R = V_A = V_B
Substituting this into the equations for I_A and I_B, we get:
I_A = (V_A - V_R) / R_A = (V_B - V_R) / R_A
I_B = (V_B - V_R) / R_B = (V_A - V_R) / R_B
Adding these equations, we get:
I = I_A + I_B =[(V_A - V_R) / R_A] + [(V_B - V_R) / R_B]
Substituting V_R = V_A = V_B, we get:
I = [(V_A - V_A) / R_A] + [(V_A - V_A) / R_B] = 0
This means that there is no net current flowing through the circuit, and the currents through batteries A and B are equal and opposite.
To find the terminal voltage of the batteries, we can use Ohm's Law again:
V_A = I_A * (R_A + R) = (V_A - V_R) / R_A * (R_A + R)
V_B = I_B * (R_B + R) = (V_B - V_R) / R_B * (R_B + R)
Substituting V_R = V_A = V_B and I_A = -I_B, we get:
V = V_A = V_B = I_A * (R_A + R) / 2 = -I_B * (R_B + R) / 2
Plugging in the given values, we get:
V = (-5 + 3)/(5+3+80) * 100 Volts = -0.74 Volts (approximate)
The negative sign indicates that the direction of the voltage is opposite to the direction of the current flow in the circuit.
Therefore, the current through each battery isapproximately zero, and the terminal voltage of each battery is approximately -0.74 volts.