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
The resistors in parallel have a combined resistance R given by the formula
1/R = 1/59 + 1/116 = 0.02557
R = 39.1 ohms
Total circuit resistance = 15.0 + 39.1 = 54.1 ohms
39.1/54.1 is the fraction of the circuit voltage that is applied to the 116 ohm resistor.
Take it from there
To find the current in the 116.0-Ω resistor, we need to calculate the total resistance of the parallel group, and then use Ohm's Law to find the current flowing through it.
First, we need to find the total resistance of the parallel group. The formula for calculating the total resistance of two resistors in parallel is:
1/RTotal = 1/R1 + 1/R2
In this case, R1 is the resistance of the 59.0-Ω resistor and R2 is the resistance of the 116.0-Ω resistor. Plugging in the values:
1/RTotal = 1/59.0 + 1/116.0
Now, we can calculate the total resistance by taking the reciprocal of both sides:
RTotal = 1 / (1/59.0 + 1/116.0)
Simplifying this expression gives us:
RTotal = 37.02 Ω
Now that we have the total resistance, we can use Ohm's Law to find the current flowing through the parallel group. Ohm's Law states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R):
I = V / R
In this case, the voltage (V) is 15.0 V and the resistance (R) is 37.02 Ω. Plugging in these values:
I = 15.0 V / 37.02 Ω
Calculating this expression gives us:
I ≈ 0.405 A (rounded to three decimal places)
Therefore, the current in the 116.0-Ω resistor is approximately 0.405 A.
Now, let's move on to finding the power dissipated in the 116.0-Ω resistor.
The power (P) dissipated by a resistor can be calculated using the formula:
P = I^2 * R
In this case, we already know the current (I) flowing through the 116.0-Ω resistor, which is approximately 0.405 A, and the resistance (R) of the resistor, which is 116.0 Ω. Plugging in these values:
P = (0.405 A)^2 * 116.0 Ω
Calculating this expression gives us:
P ≈ 19.1 W (rounded to one decimal place)
Therefore, the power dissipated in the 116.0-Ω resistor is approximately 19.1 W.