Derive an expression for the effective resistance of two resistors in parallel - Leaving Cert Physics - Question c - 2022

First, we need to calculate the effective resistance of resistors Y and Z in parallel:

$R_{YZ} = \frac{1}{\left(\frac{1}{Y} + \frac{1}{Z}\right)} = \frac{1}{\left(\frac{1}{6} + \frac{1}{3}\right)} = \frac{1}{\left(\frac{1 + 2}{6}\right)} = \frac{6}{3} = 2 \, \Omega$

Now, the total resistance in the circuit is:

$R_{total} = R_{X} + R_{YZ} = 1 + 2 = 3 \, \Omega$

Using Ohm’s Law (V = IR), the total current flowing through the circuit can be calculated as:

$I_{total} = \frac{V}{R_{total}} = \frac{12}{3} = 4 \, A$

Since the current divides at the junction of Y and Z, we will use the current division rule to find the current through X:

$I_X = I_{total} = 4 \, A$

To find the current through resistor Y, we again apply the current division principle. The current through Y (I_Y) is proportional to the resistance of Z:

$I_Y = I_{total} \times \frac{R_Z}{R_Y + R_Z} = 4 \times \frac{3}{6 + 3} = 4 \times \frac{3}{9} = 4 \times \frac{1}{3} = \frac{4}{3} \, A \approx 1.33 \, A$