Answered

A string of 18 identical Christmas tree lights are connected in series to a 130 V source. The string dissipates 61 W.

What is the equivalent resistance of the light string?

Answer in units of Ω.


What is the resistance of a single light? Answer in units of Ω.


How much power is dissipated in a single light?

Answer in units of W.


One of the bulbs quits burning. The string has a wire that shorts out the bulb filament when it quits burning, dropping the resistance of that bulb to zero. All the rest of the bulbs remain burning.

What is the resistance of the light string now?

Answer in units of Ω.


How much power is dissipated by the string now?

Answer in units of W.

Answer :

asuwafohjames

Answer:

(a)  277.05 Ω

(b) 15.39 Ω

(c) 3.76 W

Explanation:

(a)

Applying,

P = V²/R.......................... Equation 1

Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string

Make R the subject of the equation

R = V²/P................... Equation 2

Given: V = 130, P = 61  W

Substitute into equation 2

R = 130²/61

R = 277.05 Ω

(b) The resistance of a single light is given as

R' = R/18 (since the light are connected in series and the are identical)

Where R' = Resistance of the single light.

R' = 277.05/18

R' = 15.39 Ω

(c)

Heat dissipated in a single light is given as

P' = I²R'..................... Equation 3

Where P' = heat dissipated in a single light, I = current flowing through each light.

We can calculate for I using

P = VI

make I the subject of the equation

I = P/V

I = 61/130

I = 0.469 A.

Also given: R' = 15.39 Ω

Substitute into equation 3

P' = 0.496²(15.39)

P' = 3.76 W

(a)The equivalent resistance of the light string is 277.05 Ω

(b)The power is dissipated in a single ligh15.39 Ω

(c)The resistance of the light string now3.76 W

Calculation of Power is dissipates

(a) P is = V²/R.......................... Equation 1

Where P is = Power dissipated by the string.

Then V = Voltage source,

After that R is = the equivalent resistance of the light string

Now Make R the subject of the equation

R is = V²/P................... Equation 2

Then Given: V = 130,

P = 61 W

After that Substitute into equation 2

Then R = 130²/61

Therefore, R = 277.05 Ω

(b) When The resistance of a single light is given as

R' is = R/18 (since the light are connected in series and are identical)

Now Where R' is = Resistance of the single light.

R' is = 277.05/18

Therefore, R' is = 15.39 Ω

(c) When Heat dissipated in a single light is given as

P' is = I²R'..................... Equation 3

Where P' is = heat dissipated in a single light,

Then I = current flowing through each light.

Now We can calculate for I using

P is = VI

Now we make I the subject of the equation

After that I = P/V

Then I = 61/130

I is = 0.469 A.

Also given: R' is = 15.39 Ω

Then Substitute into equation 3

P' is = 0.496²(15.39)

Therefore, P' is = 3.76 W

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