Final answer:
To calculate the power delivered to each resistor in the circuit, we can use Ohm's Law and Joule's Law. By finding the current flowing through the circuit and using the equation P = IV, we can calculate the power dissipated by each resistor. In this case, the power dissipated in both R₁ and R₂ is 220.5 W.
Step-by-step explanation:
In order to calculate the power delivered to each resistor, we need to find the current flowing through the circuit. Since the resistors are connected in series, they have the same current flowing through them. To find this current, we can use Ohm's Law, V = IR, where V is the voltage and R is the resistance. In this case, V = 21.0 V and R₁ = 2.00 Ω, so the current flowing through the circuit is I = V / R₁ = 21.0 V / 2.00 Ω = 10.5 A.
Now that we have the current, we can calculate the power dissipated by each resistor using Joule's Law, P = IV, where P is the power, I is the current, and V is the voltage.
For the first resistor, R₁, the power dissipated is P₁ = IV = (10.5 A) * (21.0 V) = 220.5 W.
For the second resistor, R₂, which has twice the resistance of R₁, the power dissipated is P₂ = IV = (10.5 A) * (21.0 V) = 220.5 W. So, the power dissipated in R₂ is also 220.5 W.