9. Define (i) potential difference, (ii) resistance - Leaving Cert Physics - Question 9 - 2005

Resistance is defined as the ratio of the potential difference across a conductor to the current flowing through it. It can be expressed as:

$R = \frac{V}{I}$

where $R$ is the resistance, $V$ is the potential difference, and $I$ is the current.

To find the total resistance ($R_{total}$) of the parallel resistors, we can use the formula for resistors in parallel:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$

Plugging in the values $R_1 = 750 \Omega$ and $R_2 = 300 \Omega$:

$\frac{1}{R_{total}} = \frac{1}{750} + \frac{1}{300}$

Calculating gives:

$\frac{1}{R_{total}} = \frac{1}{750} + \frac{1}{300} = \frac{1}{150} \\ \therefore R_{total} = 600\,\Omega$

Using Ohm's Law, we can find the current flowing through the 750 Ω resistor. The total voltage across the circuit is 6 V:

$I_{total} = \frac{V}{R_{total}} = \frac{6}{600} = 0.01\,A$

Now using the voltage across the 750 Ω resistor:

$V_{750} = V_{total} \times \frac{750}{R_{total}} = 6 \times \frac{750}{600} = 3\,V$

Now to find the current through the 750 Ω resistor:

$I_{750} = \frac{V_{750}}{R_1} = \frac{3}{750} = 0.004\,A = 4\,mA$

As the temperature increases, the resistance of the thermistor decreases due to the increased energy provided to charge carriers. This results in more charge carriers being available for conduction, allowing the current to flow more easily.

As the resistance of the thermistor decreases, the total resistance of the circuit decreases. This results in a larger current flowing through the circuit, which increases the voltage or potential at point A.