Electric energy – problems and solutions

1. A 220 V – 5 A electric lamp is used for 30 minutes. How much energy does it require?

Solution :

Voltage (V) = 220 Volt

Electric current (I) = 5 Ampere

Time (t) = 30 minutes = 30 x 60 seconds = 1800 seconds

Electric power (P) :

P = V I = (220 Volt)(5 Ampere) = 1100 Volt Ampere = 1100 Watt = 1100 Joule/second

Electric energy = Electric power x time = (1100 Joule/second)(1800 second)

Electric energy = 1,980,000 Joule = 1,980 kiloJoule

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2. A 220 V – 60 W solder is used for 4 minutes. How much energy does it require.

Known :

Power (P) = 60 Watt = 60 Joule/second

Voltage (V) = 220 Volt

Time (t) = 4 minutes = 4 x 60 seconds = 240 seconds

Wanted: Electric power

Solution :

220 Volt – 60 Watt means the electric solder works well if the potential difference or voltage is 220 volts and has a power of 60 Watt = 60 Joule/second, means that electric solder using the energy of 60 Joules per second.

Electric energy = electric power x time interval = (60 Joule/second)(240 second) = 14,400 Joule.

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3. The energy used by the iron for 1 minute is 33 kJ, at a voltage of 220 volts. How large the current is in the iron.

Known :

Time interval (t) = 1 minute = 60 seconds

Energy (W) = 33 kiloJoule = 33,000 Joule

Voltage (V) = 220 Volt

Wanted : Electric current (I)

Solution :

Electrical power is the electrical energy used during a certain time interval.

P = W / t = 33,000 Joule / 60 seconds

P = 550 Watt

Electric current :

I = P / V = 550 / 220 = 2.5 Ampere

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4. Someone watches TV on average 6 hours each day. The TV is connected to a 220 Volt voltage so that the electric current flows through the TV is 0.5 Amperes. If the electric company charges $0.092 per kWh, then the cost of using electric energy for TV for 1 month (30 days) is…

Known :

Time interval = 6 hours x 30 = 180 hours

Voltage (V) = 220 Volt

Electric current (I) = 0.5 Ampere

Wanted : The cost per month

Solution :

Power of TV :

P = V I = (220 Volt)(0.5 Ampere) = 110 Volt Ampere = 110 Watt

Electric energy = electric power x time interval

Electric energy of TV = 110 Watt x 180 hours = 19800 Watt hours = 19.8 kilo Watt hours = 19.8 kilo Watt hours = 19.8 kWh

The cost of using electric energy for TV during 1 month :

19.8 kWh x $ 0.092 / kWh = $ 1.8216

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5. In a house there are 4 lamps 20 Watt, 2 lamps 10 Watt, 3 lamps 40 Watt, are used 5 hours every day. If the electric company charge 0.092 per kWh, then the cost of using electric energy during 1 month (30 days) is ….

Known :

4 lamps 20 Watt = 4 x 20 Watt = 80 Watt

2 lamps 10 Watt = 2 x 10 Watt = 20 Watt

3 lamps 40 Watt = 3 x 40 Watt = 120 Watt

Total power (W) = 80 Watt + 20 Watt + 120 Watt = 220 Watt

Time interval (t) = 5 hours x 30 = 150 hours

Wanted : The cost of using electric energy during 1 month (30 days)

Solution :

Electric energy = electric power x time interval = 220 Watt x 150 hours = 33,000 Watt hour = 33 kilo Watt hour = 33 kilo Watt hour = 33 kWh

The cost of using electric energy during 1 month (30 days)

(33 kWh) ( 0.092 / kWh) = $ 3.036