A 30.343 V emf is placed across a series combination of three resistors (10 Ω, 15.632 Ω,
and 64 Ω).
At what rate is heat generated in the
15.632 Ω resistor?
Answer in units of W
solve for current: current=emf/total resistance.
now, heat rate= current^2 * Resistance
To determine the rate at which heat is generated in the 15.632 Ω resistor, we first need to consider Ohm's Law, which states that the current flowing through a resistor is directly proportional to the voltage across it and inversely proportional to its resistance.
We can use Ohm's Law, along with the power formula, to calculate the rate of heat generation in the resistor:
1. First, find the total resistance of the series combination of resistors by adding their individual resistances:
Total resistance (R_total) = 10 Ω + 15.632 Ω + 64 Ω = 89.632 Ω
2. Use Ohm's Law to calculate the current (I) flowing through the circuit:
Current (I) = Voltage (V) / Total resistance (R_total)
Current (I) = 30.343 V / 89.632 Ω
3. Substitute the calculated value of current (I) back into Ohm's Law to find the voltage drop across the 15.632 Ω resistor:
Voltage drop (V_drop) = Current (I) * Resistance (R)
Voltage drop (V_drop) = I * 15.632 Ω
4. Finally, use the power formula, which states that power (P) is equal to the product of current (I) and voltage drop (V_drop), to calculate the rate of heat generation in the 15.632 Ω resistor:
Power (P) = I * V_drop
Substitute the values obtained in steps 2 and 3 into the power formula and calculate the power in watts (W).
To determine the rate at which heat is generated in the 15.632 Ω resistor, we can use the formula for power in a resistor:
P = (V^2) / R
where P is the power (or rate of heat generation), V is the voltage across the resistor, and R is the resistance of the resistor.
First, we need to find the voltage across the 15.632 Ω resistor. In a series circuit, the total voltage is divided among the resistors based on their relative resistance values.
Given that the total voltage across the combination of resistors is 30.343 V, we can calculate the voltage across the 15.632 Ω resistor using the voltage divider rule:
V_15.632 = V_total * (R_15.632 / (R_10 + R_15.632 + R_64))
V_15.632 = 30.343 V * (15.632 Ω / (10 Ω + 15.632 Ω + 64 Ω))
V_15.632 = 30.343 V * (15.632 Ω / 89.632 Ω)
V_15.632 ≈ 5.269 V
Now, we can use the formula for power to calculate the rate of heat generation in the 15.632 Ω resistor:
P_15.632 = (V_15.632^2) / R_15.632
P_15.632 = (5.269 V)^2 / 15.632 Ω
P_15.632 ≈ 1.771 W
Therefore, the rate at which heat is generated in the 15.632 Ω resistor is approximately 1.771 W.