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## A utility company charges 8.2 cents/kWh. If a consumer operates a 60-W light bulb continuously for one day, how much is …

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A utility company charges 8.2 cents/kWh. If a consumer operates a 60-W light bulb continuously for one day, how much is the consumer charged?

A 60-W incandescent lamp is connected to a 120-V source and is left burning continuously in an otherwise dark staircase. Determine: (a) the current through the lamp. (b) the cost of operating the light for one non-leap year if electricity costs 9.5 cents per kWh.

A Copper trace in a circuit board is 150 cm long with a square cross section of 1 mm x 1 mm. What is the resistance of the trace? (Coper: ? = 1.72 x 10-8 Mn)

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2023-03-15T10:00:11+00:00
2023-03-15T10:00:11+00:00 1 Answer
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## Answer ( 1 )

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To calculate the cost of operating a 60-W light bulb continuously for one day, we first need to calculate the energy consumption of the bulb in one day.

60 W = 0.06 kW

In one hour, the bulb will consume 0.06 kWh of energy. In one day (24 hours), the energy consumption will be:

0.06 kWh x 24 hours = 1.44 kWh

The cost of 1 kWh of energy is 8.2 cents, so the cost of operating the 60-W light bulb continuously for one day will be:

1.44 kWh x 8.2 cents/kWh = 11.808 cents

Therefore, the consumer will be charged 11.808 cents for operating the 60-W light bulb continuously for one day.

(a) To find the current through the lamp, we can use Ohm’s law:

V = IR

where V is the voltage, I is the current, and R is the resistance. The voltage of the source is given as 120 V. The resistance of the lamp can be calculated using the formula:

P = IV

where P is the power (in watts) and V is the voltage. The power of the lamp is 60 W. Therefore, the resistance of the lamp is:

R = V^2 / P = 120^2 / 60 = 240 ?

Now we can use Ohm’s law to find the current through the lamp:

I = V / R = 120 / 240 = 0.5 A

Therefore, the current through the lamp is 0.5 A.

(b) To find the cost of operating the light for one non-leap year, we need to first calculate the energy consumption of the lamp in one year.

In one hour, the lamp will consume 0.06 kWh of energy. In one day (24 hours), the energy consumption will be:

0.06 kWh x 24 hours = 1.44 kWh

In one non-leap year (365 days), the energy consumption will be:

1.44 kWh x 365 days = 525.6 kWh

The cost of 1 kWh of energy is 9.5 cents, so the cost of operating the lamp for one non-leap year will be:

525.6 kWh x 9.5 cents/kWh = 49.872 dollars

Therefore, the cost of operating the light for one non-leap year will be 49.872 dollars.

To calculate the resistance of the trace, we can use the formula:

R = ? x L / A

where R is the resistance, ? is the resistivity of copper (1.72 x 10^-8 ?m), L is the length of the trace (150 cm = 1.5 m), and A is the cross-sectional area of the trace (1 mm x 1 mm = 10^-6 m^2).

Plugging in the values, we get:

R = (1.72 x 10^-8 ?m) x (1.5 m) / (10^-6 m^2) = 25.8 m?

Therefore, the resistance of the trace is 25.8 m?.