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Molodets [167]
3 years ago
8

A heat release of 4800 J for 8 minutes is released in the conductor during the flow of electricity. What is the current in the c

onductor? The conductor resistance is 1kΩ​
Physics
1 answer:
hram777 [196]3 years ago
7 0

Answer: 0.1 Amps

Explanation:

Given that,

heat released (H) = 4800 J

Time taken (T) = 8 minutes

Since the unit of time is seconds, convert 8 minutes to seconds

(1 minute = 60 seconds

8 minutes = 8 x 60 = 480 seconds)

current in the conductor (I) = ?

The conductor resistance (R) = 1kΩ​

Convert kilo-Ohms (kΩ​) to ohms (Ω​)

(If 1000Ω = 1kΩ

1kΩ = 1000Ω)

Now, apply the formula:

H = I²RT

4800J = I² x 1000Ω x 480 Seconds

4800J = 480000Ω•I²

Divide both sides by 480000Ω

4800J/480000Ω = 480000Ω•I²/ 480000Ω

0.01 = I²

To get the value of I, find the square root of I²

I = √0.01

I = 0.1 Amps

Thus, the current in the conductor is 0.1 Amps

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Wattage consumption of CFL per hour = 15

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PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

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<u>Calculate the EAC of bulb</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

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<u>Calculate the EAC of CFL</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, <em>consider this equation 2</em>

<u>Equate 1 and 2 to find the amount of C</u>

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

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