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

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

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

Now, the total resistance in the circuit is:

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

Using Ohm's Law, we calculate the total current (I) in the circuit:

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

Now, we can find the current through resistor X, since the total current splits at the junction. The voltage across X is equal to the total voltage:

$V_X = I \times R_X = 4 \, A \times 1 \, \Omega = 4 \, V$

Using Ohm's Law again:

$I_X = \frac{V_X}{R_X} = \frac{4}{1} = 4 \, A$

Using the effective resistance calculated for resistors Y and Z (2 Ω) and as stated before, the voltage across Y and Z is:

$V_{YZ} = V - V_X = 12 \, V - 4 \, V = 8 \, V$

Now, we can calculate the current through resistor Y:

Using Ohm's Law:

$I_Y = \frac{V_{YZ}}{Y} = \frac{8 \, V}{6 \, \Omega} = \frac{4}{3} \, A \approx 1.33 \, A$