A 59.0-Ω resistor is connected in parallel with a 116.0-Ω resistor. This parallel group is connected in series with a 15.0-Ω resistor. The total combination is connected across a 15.0-V battery.
(a) Find the current in the 116.0-Ω resistor.
______A
(b) Find the power dissipated in the 116.0-Ω resistor.
______ W
R1 = 59 Ohms.
R2 = 116 Ohms.
R3 = 15 Ohms
Rt=R1*R2 / (R1 + R2) + R3 = Total Res.
Rt = 59*116 / (59 + 116) + 15,
To find the current in the 116.0-Ω resistor, we need to calculate the total resistance of the parallel combination first. The formula to find the total resistance of two resistors in parallel is:
1/RTotal = 1/R1 + 1/R2
Given the individual resistances R1 = 59.0 Ω and R2 = 116.0 Ω, we can substitute these values into the formula:
1/RTotal = 1/59.0 + 1/116.0
To simplify this calculation, we can find a common denominator:
1/RTotal = (116 + 59)/(59 * 116)
1/RTotal = 175 / 6844
Now, to find RTotal, we can take the reciprocal of both sides:
RTotal = 6844 / 175
RTotal ≈ 39.109 Ω
Now that we have the total resistance, we can calculate the current in the parallel combination using Ohm's Law:
I = V / R
Where V is the voltage and R is the resistance. In this case, the voltage is given as 15.0 V, so we substitute this value along with RTotal:
I = 15.0 / 39.109
I ≈ 0.383 A
Therefore, the current in the 116.0-Ω resistor is approximately 0.383 Amps (A).
To find the power dissipated in the 116.0-Ω resistor, we can use the formula:
P = I^2 * R
Where P is the power, I is the current, and R is the resistance. Given that the current is 0.383 A and the resistance is 116.0 Ω, we substitute these values into the formula:
P = (0.383)^2 * 116.0
P ≈ 16.207 W
Therefore, the power dissipated in the 116.0-Ω resistor is approximately 16.207 Watts (W).
R1 = 59 Ohms.
R2 = 116 Ohms.
R3 = 15 Ohms
Rt=R1*R2 / (R1 + R2) + R3 = Total Res.
Rt = 59*116 / (59 + 116) + 15,
Rt = 39.1 + 15 = 54.1 Ohms.
It = I3 = Eb / Rt = 15 / 54.1 = 0.28A
V1 = V2 = Eb - I3*R3,
V1 = V2 = 15 - 0.28*15,
V1 = V2 = 15 - 4.16 = 10.84 Volts.
a. I2 = V2 / R2 = 10.84 / 116 = 0.093A
b. P2 = V2*I2 = 10.84 * 0.093 = 1.0W.