A 40.0 ohm resistor, 20.0 ohm resistor, and a 10.0 ohm resistor are connected in series across a 220-V supply. Which resistor is the warmest? a. 40.0 ohm b. 20.0 ohm c. 10.0 ohm d. they are all the same
isn't Joule heating current^2 * R ?
they all have the same current...so...
To determine which resistor is the warmest, we can use the concept of power dissipation. The power dissipated by a resistor can be calculated using the formula:
P = I^2 * R
Where P is the power in watts, I is the current in amperes, and R is the resistance in ohms.
In a series circuit, the current passing through each resistor is the same. Therefore, we can calculate the power dissipated by each resistor and compare them.
First, let's calculate the current passing through the circuit using Ohm's Law:
V = I * R_total
Where V is the voltage in volts, I is the current in amperes, and R_total is the total resistance in ohms.
For the given circuit:
V = 220 V (as mentioned in the question)
R_total = 40.0 ohm + 20.0 ohm + 10.0 ohm = 70.0 ohm
Now, we can rearrange the equation to solve for current, I:
I = V / R_total
I = 220 V / 70.0 ohm
I ≈ 3.14 A
Now, we can calculate the power dissipated by each resistor:
For the 40.0 ohm resistor:
P_40 = I^2 * R_40
P_40 = (3.14 A)^2 * 40.0 ohm
P_40 ≈ 394.24 W
For the 20.0 ohm resistor:
P_20 = I^2 * R_20
P_20 = (3.14 A)^2 * 20.0 ohm
P_20 ≈ 197.12 W
For the 10.0 ohm resistor:
P_10 = I^2 * R_10
P_10 = (3.14 A)^2 * 10.0 ohm
P_10 ≈ 98.56 W
Comparing the power dissipated by each resistor, we can see that the 40.0 ohm resistor has the highest power dissipation. Therefore, the answer is (a) 40.0 ohm, it is the warmest resistor among the three.