Welcome back — missed you this week. To find the energy dissipated by the $4\Omega$ resistor, we need to determine the power dissipated by it and then multiply by the time. Step 1: Identify the circuit configuration. The circuit consists of two main parallel branches connected to the 60V battery. • The top branch contains a $3\Omega$ resistor in series with a parallel combination of $4\Omega$ and $12\Omega$ resistors. • The bottom branch contains a $6\Omega$ resistor. Step 2: Calculate the equivalent resistance of the parallel combination of the $4\Omega$ and $12\Omega$ resistors. Let's call this $R_{p1}$. $$R_{p1} = \frac{R_1 \times R_2}{R_1 + R_2}$$ $$R_{p1} = \frac{4\Omega \times 12\Omega}{4\Omega + 12\Omega} = \frac{48\Omega^2}{16\Omega} = 3\Omega$$ Step 3: Calculate the total resistance of the top branch. Let's call this $R_{top}$. The $3\Omega$ resistor is in series with $R_{p1}$. $$R_{top} = 3\Omega + R_{p1} = 3\Omega + 3\Omega = 6\Omega$$ Step 4: Calculate the current flowing through the top branch. Let's call this $I_{top}$. The voltage across the top branch is the battery voltage, $V = 60\text{ V}$. $$I_{top} = \frac{V}{R_{top}} = \frac{60\text{ V}}{6\Omega} = 10\text{ A}$$ Step 5: Calculate the voltage across the parallel combination ($R_{p1}$). Let's call this $V_{p1}$. This voltage is the product of the current through the top branch and the equivalent resistance of the parallel part. $$V_{p1} = I_{top} \times R_{p1} = 10\text{ A} \times 3\Omega = 30\text{ V}$$ This is the voltage across both the $4\Omega$ and the $12\Omega$ resistors. Step 6: Calculate the power dissipated by the $4\Omega$ resistor. Let's call this $P_{4\Omega}$. We can use the formula $P = \frac{V^2}{R}$. $$P_{4\Omega} = \frac{V_{p1}^2}{4\Omega} = \frac{(30\text{ V})^2}{4\Omega} = \frac{900\text{ V}^2}{4\Omega} = 225\text{ W}$$ Step 7: Calculate the energy dissipated by the $4\Omega$ resistor in 2 seconds. Energy $E = P \times t$. $$E_{4\Omega} = P_{4\Omega} \times t = 225\text{ W} \times 2\text{ s} = 450\text{ J}$$ The energy dissipated by the $4\Omega$ resistor in 2 seconds is $450\text{ J}$. The correct option is B. $\boxed{\text{B}}$ Send me the next one 📸