Complete Question:
The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference.
Part A: What is the resistance of a 100 W bulb?
part B: Current flowing through a 100 W bulb?
part C: Find the resistance of a 40 W bulb
part D: Current flowing through a 40W bulb
Answer:
a)Resistance of a 100 W bulb, R = 144 ohms
b) Current flowing through a 100 W bulb, I =0.833 A
c)Resistance of a 40W bulb, R = 360 ohms
d)current in a 40W bulb, I = 0.33 A
Explanation:
a) Resistance of the 100 W bulb
Power, P = 100 W
Potential difference, V = 120 V
P = V²/R
R = V²/P
R = 120²/100
R = 144 ohms
b) the current flowing through the bulb
According to Ohm's law, V = IR
I = V/R
I = 120/144
I = 0.833 A
c) Find the resistance of a 40 W bulb
Since the voltage supplied is constant, P = V²/R
P = 40 W
V = 120 V
40 = 120²/R
R = 120²/40
R = 360 ohms
d)current flowing through the 40 W bulb
V = IR
I = V/R
I = 120/360
I = 0.33 A
Answer:
144 Ω
Explanation:
Power: This can be defined as the rate at which Energy is used up in an electric circuit. The S.I unit of power is Watt(W).
The expression for electric power is given as
P = V²/R............... Equation 1
Where P = power, V = Voltage, R = Resistance.
make R the subject of the equation
R = V²/P............. Equation 2
Given: V = 120 V, P = 100 W
Substitute into equation 2
R = 120²/100
R = 14400/100
R =144 Ω
