A 24V lamp has a resistance of 8.0 ohms. What resistance must be placed in series with the lamp if it is to be used in a 117V line?
I think I understand how resistors work but this HW question is completely over my head..
I = 24/8 = 3A = Lamp current.
Vr = 117-24 = 93 V. = Voltage across the
resistor.
R = Vr/I = 93/3 = 31 Ohms.
To determine the resistance that must be placed in series with the lamp, you can use Ohm's Law and the concept of voltage division.
Ohm's Law states that the current passing through a circuit is equal to the voltage across the circuit divided by the resistance:
I = V / R
In this case, the given voltage across the circuit is 24V, and the resistance of the lamp is 8.0 ohms. Using Ohm's Law, we can calculate the current passing through the lamp:
I = 24V / 8.0 ohms
= 3.0 A
Now, let's consider the 117V line. To find the resistance that must be placed in series with the lamp, we need to determine the voltage across this resistance, which can be found using voltage division. The formula for voltage division is:
V_resistance = V_total * (R_resistance / (R_lamp + R_resistance))
Where:
V_resistance is the voltage across the resistance
V_total is the total voltage (117V)
R_resistance is the resistance we are trying to find
R_lamp is the resistance of the lamp (8.0 ohms)
Now, plug in the known values and solve for R_resistance:
V_resistance = 117V * (R_resistance / (8.0 ohms + R_resistance))
Next, we can substitute the current passing through the lamp, which we calculated earlier, into the equation using Ohm's Law:
V_resistance = (24V / 3.0 A) * (R_resistance / (8.0 ohms + R_resistance))
Simplifying this equation, we can cross-multiply:
V_resistance * (8.0 ohms + R_resistance) = (24V / 3.0 A) * R_resistance
Expand:
8.0V_resistance * ohms + V_resistance * R_resistance = (24V / 3.0 A) * R_resistance
Combine like terms:
V_resistance * R_resistance - (24V / 3.0 A) * R_resistance = -8.0V_resistance * ohms
Factor out R_resistance:
R_resistance * (V_resistance - (24V / 3.0 A)) = -8.0V_resistance * ohms
Divide both sides by (V_resistance - (24V / 3.0 A)):
R_resistance = (-8.0V_resistance * ohms) / (V_resistance - (24V / 3.0 A))
Finally, substitute the known values and calculate the resistance:
R_resistance = (-8.0V_resistance * ohms) / (V_resistance - (24V / 3.0 A))
R_resistance = (-8.0 * 24V * ohms) / (V_resistance - (24V / 3.0 A))
R_resistance = -192V * ohms / (V_resistance - (24V / 3.0 A))
To solve this problem, you need to apply Ohm's Law and the concept of power in electrical circuits.
First, let's identify the known values:
- Voltage across the lamp (Vlamp) = 24V
- Resistance of the lamp (Rlamp) = 8.0 ohms
- Voltage of the line (Vline) = 117V
The lamp and the series resistor will share the same current when connected in series. Using Ohm's Law (V = IR), we can find the current flowing through the lamp:
I = Vlamp / Rlamp
I = 24V / 8.0 ohms
I = 3.0 A (Amperes)
Now, using the voltage across the line and the current flowing through the circuit, we can determine the resistance needed for the series resistor.
Vline = IRtotal
117V = I (Rlamp + Rseries)
117V = 3.0 A (8.0 ohms + Rseries)
To solve for Rseries, rearrange the equation:
Rseries = (Vline - Vlamp) / I
Rseries = (117V - 24V) / 3.0 A
Rseries = 93V / 3.0 A
Rseries = 31.0 ohms
Therefore, a resistance of 31 ohms must be placed in series with the lamp to use it in a 117V line.