Three resistors, 1.17, 2.84, and 3.65Ω, are connected in series across a 20.3-V battery. Find the power delivered to each resistor.
R=R₁+R₂+R₃=1.17+2.84+3.65 =7.66 Ω
I=U/R = 20.3/7.66=2.65 A
P₁ =I²R₁=2.65²•1.17=
P₂ =I²R₂=2.65²•2.84=
P₃=I²R₃=2.65²•3.65=
To find the power delivered to each resistor, we need to use Ohm's Law and the formula for power. Ohm's Law states that the current flowing through a resistor is equal to the voltage across the resistor divided by the resistance.
First, let's calculate the total resistance (R_total) of the series circuit. In a series circuit, the total resistance is simply the sum of the individual resistances:
R_total = R1 + R2 + R3
Given that R1 = 1.17Ω, R2 = 2.84Ω, and R3 = 3.65Ω, we can find the total resistance:
R_total = 1.17Ω + 2.84Ω + 3.65Ω = 7.66Ω
Next, we can use Ohm's Law to find the total current (I_total) flowing through the circuit:
I_total = V / R_total
Given that V = 20.3V and R_total = 7.66Ω, we can calculate the total current:
I_total = 20.3V / 7.66Ω ≈ 2.652A
Now, we can calculate the power (P) delivered to each resistor using the formula:
P = I^2 * R
For each resistor, we will use the total current I_total.
Power delivered to R1:
P1 = I_total^2 * R1
P1 = (2.652A)^2 * 1.17Ω ≈ 8.270W
Power delivered to R2:
P2 = I_total^2 * R2
P2 = (2.652A)^2 * 2.84Ω ≈ 20.039W
Power delivered to R3:
P3 = I_total^2 * R3
P3 = (2.652A)^2 * 3.65Ω ≈ 26.017W
Therefore, the power delivered to each resistor is approximately:
P1 = 8.270W
P2 = 20.039W
P3 = 26.017W