A 67.6 ohms resistor is connected in parallel with a 103.8 ohms resistor. This parallel group is connected in series with a 20.8 ohms resistor. The total combination is connected across a 18.2-V battery. Find a) the current and b) the power dissipated in the 103.8 ohms resistor.
Given:
R1 = 67.6 Ohms.
R2 = 103.8 Ohms.
R3 = 20.8 Ohms.
E = 18.2 Volts.
a. R =( R1*R2)/(R1+R2) + R3 = (67.6*103.8)/(67.6+103.8) + 20.8 = 61.74 Ohms.
I = E/R = 18.2/61.74 = 0.295A.
b. V2 = E - I*R3 = 18.2 - 0.295*20.8 = 12.06 Volts.
P = V2^2/R2 = 12.06^2/103.8 = 1.40 W.
To find the current and power dissipated in the 103.8 ohms resistor, we can follow these steps:
Step 1: Find the equivalent resistance of the parallel combination.
In a parallel combination, the equivalent resistance (Rp) is given by the formula:
1/Rp = 1/R1 + 1/R2 + ...
Where R1 and R2 are the individual resistances.
In this case, R1 = 67.6 ohms and R2 = 103.8 ohms. Substituting these values into the formula, we get:
1/Rp = 1/67.6 + 1/103.8
To simplify, we can find a common denominator:
1/Rp = (103.8 + 67.6) / (67.6 * 103.8)
1/Rp = 171.4 / (7024.8)
1/Rp = 0.0244
To find Rp, we take the reciprocal of both sides:
Rp = 1 / 0.0244
Rp ≈ 41.0 ohms
So, the equivalent resistance of the parallel combination is approximately 41.0 ohms.
Step 2: Find the total resistance of the combination.
In a series combination, the total resistance (Rs) is simply the sum of the individual resistances.
In this case, Rs = 41.0 ohms + 20.8 ohms
Rs = 61.8 ohms
So, the total resistance of the combination is 61.8 ohms.
Step 3: Find the current.
Using Ohm's law (V = I * R), we can find the current (I) by dividing the voltage (V) by the total resistance (Rs).
In this case, V = 18.2 V and Rs = 61.8 ohms. Substituting these values into the formula, we get:
18.2 = I * 61.8
To find I, we divide both sides by 61.8:
I ≈ 18.2 / 61.8
I ≈ 0.294 A
So, the current flowing through the combination is approximately 0.294 A.
Step 4: Find the power dissipated in the 103.8 ohms resistor.
Using the formula for power (P = I^2 * R), we can find the power dissipated in the 103.8 ohms resistor.
In this case, I = 0.294 A and R = 103.8 ohms. Substituting these values into the formula, we get:
P = (0.294)^2 * 103.8
P ≈ 0.0266 * 103.8
P ≈ 2.76 W
So, the power dissipated in the 103.8 ohms resistor is approximately 2.76 W.