Asked by ashley
A 75.0-W lamp is placed in series with a resistor and a 120.0-V source. If the voltage across the lamp is 24 V, what is the resistance R of the resistor?
Answers
Answered by
drwls
The voltage drop across the resistor is 120 - 24 = 96 V
The resistance R' of the lamp can be derived from its Wattage at 120 V:
75 W = V^2/R' = 14400/R'
R' = 192 ohms
Since the lamp gets 24/120 = 20% of the voltage drop, the other 80% of the voltage (96 V) is across the resistor.
With both in series, circuit current = lamp current = 24 V/192 ohms = 0.125 Amps
Resistor resistance = V(resistor)/I
= 96/(1/8) = 768 ohms
The resistance R' of the lamp can be derived from its Wattage at 120 V:
75 W = V^2/R' = 14400/R'
R' = 192 ohms
Since the lamp gets 24/120 = 20% of the voltage drop, the other 80% of the voltage (96 V) is across the resistor.
With both in series, circuit current = lamp current = 24 V/192 ohms = 0.125 Amps
Resistor resistance = V(resistor)/I
= 96/(1/8) = 768 ohms
Answered by
Henry
Vr = 120-24 = 96 V. = Voltage across the
resistor.
V*I = 75W.
24I = 75
I = 3.125A = Lamp current
R = Vr/I = 96/3.125 = 30.72 Ohms in series with lamp.
resistor.
V*I = 75W.
24I = 75
I = 3.125A = Lamp current
R = Vr/I = 96/3.125 = 30.72 Ohms in series with lamp.
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