A 61.5-Ω resistor is connected in parallel with a 146.6-Ω resistor. This parallel group is connected in series with a 16.0-Ω resistor. The total combination is connected across a 13.6-V battery. Find (a) the current and (b) the power dissipated in the 146.6-Ω resistor.
R1 = 61.5 Ohms.
R2 = 146.6 Ohms.
R3 = 16 Ohms.
V = 13.6 Volts.
a. Rt = R1*R2/(R1+R2) + R3
Rt = (61.5*146.6)/(61.5+146.6) + 16 =
59.32 Ohms. = Total resistance.
I = V/Rt = 13.6/59.32 = 0.229 Amps.
b. V2 = I*(R1*R2)/(R1+R2)=0.229 * 43.32 = 9.92 Volts. = Voltage across R2.
P2 = V2^2/R2 = 9.92^2/146.6=0.671 Watts.
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.
To find the current and power dissipated in the 146.6-Ω resistor, we need to follow these steps:
Step 1: Calculate the equivalent resistance of the parallel group.
The formula to find the equivalent resistance of two resistors connected in parallel is given by:
1/Req = 1/R1 + 1/R2
Substituting the values:
1/Req = 1/61.5 + 1/146.6
Now, let's calculate Req using this equation.
Step 2: Calculate the total resistance in the circuit.
Since the parallel group is connected in series with a 16.0-Ω resistor, the total resistance (Rt) can be found by adding the equivalent resistance (Req) to the resistance of the 16.0-Ω resistor.
Rt = Req + 16.0
Step 3: Calculate the current flowing through the circuit.
Using Ohm's Law, we can calculate the current (I) in the circuit.
I = V / Rt
where V is the voltage supplied by the battery (13.6 V).
Step 4: Calculate the power dissipated in the 146.6-Ω resistor.
The formula to find power (P) dissipated in a resistor is given by:
P = I^2 * R
Now, let's calculate the current and power.
(a) Calculate the current:
Substitute the values of V and Rt in the formula:
I = 13.6 / Rt
(b) Calculate the power dissipated in the 146.6-Ω resistor:
P = I^2 * 146.6
Now you can plug in the values and solve for I and P to find the answers.