Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 7.1Ω and 20.9 W, 29.5Ω and 11.4 W, and 13.6Ω and 12.5 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?
I^2 = W / Ω = 11.4 / 29.5
calculate I
the resistors add in series
power = I^2 * total R
No, current will be limited by the lowest safe current a resistor can handle.
So do the above, but for EACH resistor. Now, pick the lowest current of them all, that will be safe current for all resistors.
Power = I^2 *total R
resistor1: I= for 7.1Ω and 20.9 W,, I= 4.12amp
resistor2:
I= 29.5Ω and 11.4 W=.62amp
resistor3:13.6Ω and 12.5 W
I=13.6Ω and 12.5 W = .962 amps
so the safe current is .62amps
Power=.62^1*total R
missed part (a)
(a) voltage = [√(11.4/29.5)] * total R
To find the greatest voltage that the battery can have without one of the resistors burning up, we need to determine the maximum power dissipated by each resistor. Power is given by the formula:
P = (V^2) / R
(a) To find the maximum power dissipated by the first resistor, we use the given resistance and power rating:
P₁ = 20.9 W
R₁ = 7.1 Ω
Rearranging the power formula gives us:
V₁ = √(P₁ * R₁)
Substituting the given values, we get:
V₁ = √(20.9 * 7.1) = 10.704 V
The maximum voltage for the first resistor is 10.704 V.
(b) The total voltage across the circuit is equal to the sum of the voltage drops across each resistor in series. Therefore, we need to find the maximum voltage for the second and third resistors as well.
Using the same formula, we can calculate the maximum voltage for the second resistor:
P₂ = 11.4 W
R₂ = 29.5 Ω
V₂ = √(P₂ * R₂) = √(11.4 * 29.5) = 16.758 V
For the third resistor:
P₃ = 12.5 W
R₃ = 13.6 Ω
V₃ = √(P₃ * R₃) = √(12.5 * 13.6) = 16.162 V
To find the maximum battery voltage, we'll add up the voltages across each resistor:
V = V₁ + V₂ + V₃ = 10.704 V + 16.758 V + 16.162 V = 43.624 V
Therefore, the greatest voltage that the battery can have without one of the resistors burning up is 43.624 V.
Now, to determine the power delivered by the battery, we can use Ohm's Law:
P = (V^2) / R
Since the resistors are connected in series, their equivalent resistance (R_total) is the sum of their individual resistances:
R_total = R₁ + R₂ + R₃ = 7.1 Ω + 29.5 Ω + 13.6 Ω = 50.2 Ω
Using this value, we can calculate the power delivered by the battery:
P = (V^2) / R_total = (43.624^2) / 50.2 = 37.993 W
Therefore, the battery delivers approximately 37.993 W of power to the circuit.