Answer:
We have a battery here, composed of two cells joined in series, which is supplying current to an external resistance. The voltage of each cell is given as 0.6 volts and the resistance is 3 Ohms. In order to solve the problem, we need to calculate three things: the current flowing in the external resistance, the thermal potential difference and the lost voltage.
First, let's calculate the current flowing in the external resistance. Using Ohm's Law, we can find the current as I = V/R, where V is the total voltage of the battery (i.e. 2*0.6=1.2V) and R is the external resistance, which is given as 3 Ohms. Therefore, I = 1.2/3 = 0.4 amps.
Next, let's calculate the thermal potential difference. This is the amount of heat generated by the current flowing through the external resistance, and is given by the formula P = I^2*R, where P is the power, I is the current, and R is the resistance. Plugging in the values, we get P = 0.4^2*3 = 0.48 watts. Since we know that power is equal to voltage times current (P = VI), we can rearrange the formula to get V = P/I, which gives us V = 0.48/0.4 = 1.2 volts.
Finally, we need to calculate the lost voltage. This is the voltage drop that occurs across each cell due to internal resistance. We can use the formula V_lost = I*R_int, where R_int is the internal resistance. Since we know the current and the resistance of the external load, we can use the total voltage of the battery to find the internal resistance. Recall that the total voltage of the battery is 1.2V. Therefore, V_lost = I*R_int, or R_int = V_lost/I. We know that the voltage drop across each cell is equal, so we can divide the lost voltage by 2 to get the voltage drop across each cell. Therefore, V_cell = V_lost/2 = (0.4)*(R_int/2). Plugging in the values, we get V_cell = 0.4*(1.2-0.4*3)/2 = 0.06 volts.
In summary, the current flowing in the external resistance is 0.4 amps, the thermal potential difference is 1.2 volts, and the lost voltage across each cell is 0.06 volts.