The circuit diagram shows a resistor and a thermistor connected in series with a 6 V battery - Leaving Cert Physics - Question d - 2014

To find the total resistance (R_total) in a series circuit, the resistances add up:

$R_{total} = R_1 + R_2$

Here, we have:

• Resistor = 50 Ω
• Thermistor = 450 Ω

Thus,

$R_{total} = 50 \, ext{Ω} + 450 \, ext{Ω} = 500 \, ext{Ω}$

Using Ohm's law, we can find the current (I) in the circuit:

$I = \frac{V}{R_{total}}$

Substituting the values:

$I = \frac{6 \, ext{V}}{500 \, ext{Ω}} = 0.012 \, ext{A} = 12 \, ext{mA}$

The potential difference (V_R) across the 50 Ω resistor can be calculated using Ohm's law:

$V_R = I imes R$

Substituting the current we found earlier:

$V_R = 0.012 \, ext{A} imes 50 \, ext{Ω} = 0.6 \, ext{V}$

If the temperature were increased, the resistance of the thermistor would generally decrease (assuming it's a negative temperature coefficient thermistor). This would lead to a decrease in the total resistance of the circuit, potentially allowing more current to flow through the circuit.