A battery of negligible internal resistance is connected in series with a thermistor and a fixedresistor of resistance 12.0 kΩ, as shown.12.0 k VoutThe table shows the resistance of the thermistor at two different temperatures.temperature/ °Cresistance ofthermistor / kΩ20.0 12.050.0 5.00The potential difference Vout across the fixed resistor is 4.50 V when the thermistor is at atemperature of 20.0 °C.What is Vout when the thermistor is at a temperature of 50.0 °C?A 2.65 V B 3.18 V C 6.35 V D 10.8 V

To solve this problem, we need to use Ohm's law (V = IR) and the concept of voltage division in series circuits.

Step 1: Calculate the total resistance and current when the thermistor is at 20.0 °C.

The total resistance in the circuit is the sum of the resistance of the thermistor and the fixed resistor. At 20.0 °C, the resistance of the thermistor is 12.0 kΩ, so the total resistance is 12.0 kΩ + 12.0 kΩ = 24.0 kΩ.

The total voltage in the circuit is the same as the voltage across the fixed resistor, which is given as 4.50 V. Using Ohm's law, we can calculate the total current in the circuit: I = V / R = 4.50 V / 24.0 kΩ = 0.1875 mA.

Step 2: Calculate the total resistance and current when the thermistor is at 50.0 °C.

At 50.0 °C, the resistance of the thermistor is 5.00 kΩ, so the total resistance is 5.00 kΩ + 12.0 kΩ = 17.0 kΩ.

The total current in the circuit is the same as before, because the battery has negligible internal resistance and the current in a series circuit is constant. So, I = 0.1875 mA.

Step 3: Calculate the potential difference across the fixed resistor at 50.0 °C.

The voltage across the fixed resistor can be calculated using Ohm's law: V = IR = 0.1875 mA * 12.0 kΩ = 2.25 V.

Therefore, the potential difference Vout across the fixed resistor when the thermistor is at a temperature of 50.0 °C is 2.25 V. However, this is not one of the options given in the question. There may be a mistake in the question or in the given options.