Physics, 24.06.2019 23:10 garrettadkins2002

A60-w, 120-v light bulb and a 200-w, 120-v light bulb are connected in series across a 240-v line. assume that the resistance of each bulb does not vary with current. (note: this description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-w, 120-v light bulb dissipates 25 w when connected to a 120-v line.) a. find the current through the bulbs. b. find the power dissipated in the 60 w bulb. c. find the power dissipated in the 200 w bulb. d. one bulb burns out very quickly. which one? 60-w bulb, 200-w bulb.

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Answer from: ARAYAMYHAND

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

Answer from: Quest

a and b sometimes have a positive value

For which optical devices does d sometimes have a positive value

Answer from: Quest

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A60-w, 120-v light bulb and a 200-w, 120-v light bulb are connected in series across a 240-v line. a...

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