To determine the current and the power dissipated in the 146.6-Ω resistor, we can follow these steps:
Step 1: Calculate the equivalent resistance of the parallel resistors.
To find the equivalent resistance of two resistors connected in parallel, we can use the formula:
1/Req = 1/R1 + 1/R2
Given R1 = 61.5 Ω and R2 = 146.6 Ω, we can calculate Req:
1/Req = 1/61.5 + 1/146.6
1/Req = 0.0163 + 0.0068
1/Req = 0.0231
To find Req, take the reciprocal of both sides:
Req = 1/0.0231
Req ≈ 43.30 Ω
Step 2: Calculate the total resistance in the circuit.
Since the parallel group is connected in series with a 16.0-Ω resistor, we can calculate the total resistance (RT) by summing these resistances:
RT = Req + 16.0
RT ≈ 43.30 + 16.0
RT ≈ 59.30 Ω
Step 3: Calculate the current (I).
The current (I) can be determined using Ohm's Law:
I = V / RT
Given V = 13.6 V and RT ≈ 59.30 Ω:
I = 13.6 / 59.30
I ≈ 0.229 A
Therefore, the current flowing through the circuit is approximately 0.229 A.
Step 4: Calculate the power dissipated in the 146.6-Ω resistor (P).
We can calculate the power dissipated using the formula:
P = I^2 * R
Given I ≈ 0.229 A and R = 146.6 Ω:
P = (0.229)^2 * 146.6
P ≈ 7.20 W
Therefore, the power dissipated in the 146.6-Ω resistor is approximately 7.20 W.