In: Physics
You have a 40- lightbulb and a 100-W bulb.
1. Find the resistances of the 40-W bulb. Take that ΔV = 120 V.
2. Find the resistances of the 100-W bulb. Take that ΔV = 120 V.
3. If the two bulbs are connected separately, which of them will be brighter?
4. If the two bulbs are connected in series to a battery source, which of them will be brighter?
5. If the two bulbs are connected in parallel to a battery source, which of them will be brighter?
1.
Since, Power(P) = V^2/R
here, for 40-W bulb,
P1 = 40 W
V1 = 120 V
So, resistance(R1) = V1^2/P1 = 120^2/40
R1 = 360 ohm
2.
Similarly,
here, for 100-W bulb,
P2 = 100 W
V2 = 120 V
So, resistance(R2) = V2^2/P2 = 120^2/100
R2 = 144 ohm
3.
If the two bulbs are connected separately
P2>P1
So, 100 W bulb will be Brighter.
4.
Now, Since both are in series.
So, Net resistance of series will be,
Rnet = R1 + R2
Rnet = 360+144= 504 ohm
From ohm's law,
Current in circuit = I = V/Rnet = 120/(504)
I = 0.238 Amp
Now Power dissipated by Bulb 1,
P1 = I^2*R1 = 0.238^2*(360)
P1 = 20.39 W
Power dissipated by Bulb 2,
P2 = I^2*R2 = 0.238^2*(144)
P2 = 8.16 W
Since, P1>P2
So, 40 W bulb will be Brighter.
5.
Now, Since both are in Parallel.
So, Voltage across both bulb will remain same.
V1 = V2 = V = 120 V
then,Power dissipated by Bulb 1,
P1 = V1^2/R1 = 120^2/(360)
P1 = 40 W
Power dissipated by Bulb 2,
P2 = V^2/R2 = 120^2/(144)
P2 = 100 W
Since, P2>P1
So, 100 W bulb will be Brighter.
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