Question

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?

Answer 1

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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