A circuit is set up as shown - Scottish Highers Physics - Question 23 - 2019

Using Ohm's law, $V = I R$, the total current (I) from the power supply can be calculated as follows:

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

Substituting the values gives:

$I = \frac{6V}{2.0 \Omega} = 3.0 A$.

In a parallel circuit, the voltage across each resistor is the same. Therefore, the voltage across the 3.0Ω resistor is 6V. Using Ohm's law to find the current (I_3) through the 3.0Ω resistor:

$I_3 = \frac{V}{R_3} = \frac{6V}{3.0 \Omega} = 2.0 A.$

The power (P) dissipated in the resistor can be calculated using the formula:

$P = I^2 R$

Substituting the values gives:

$P = (2.0 A)^2 \cdot 3.0 \Omega = 4 \cdot 3 = 12 W.$ Thus, the power dissipated in the 3.0Ω resistor is 12 W.