Electric energy – problems and solutions

25 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

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.

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

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…

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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

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)

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

6. A circuit has a resistance of \(5\Omega\) and a current of \(2A\). Find the electric power.
Solution: \( P = IV = 2A \times 5\Omega = 10W \)

7. A \(10V\) battery is connected to a \(2\Omega\) resistor. Calculate the current.
Solution: \( I = \frac{V}{R} = \frac{10V}{2\Omega} = 5A \)

8. A light bulb rated at \(60W\) operates on a \(120V\) supply. Find the resistance.
Solution: \( R = \frac{V^2}{P} = \frac{120^2}{60} = 240\Omega \)

9. Calculate the energy consumed by a \(100W\) bulb in \(2\) hours.
Solution: \( E = Pt = 100W \times 2h = 200Wh = 0.2kWh \)

10. A \(20\mu F\) capacitor is charged to \(50V\). Find the stored energy.
Solution: \( E = \frac{1}{2} CV^2 = \frac{1}{2} \times 20 \times 10^{-6}F \times 50^2 = 25mJ \)

11. Determine the current in a \(10mH\) inductor when the rate of change of current is \(5A/s\).
Solution: \( V = L\frac{di}{dt} = 10 \times 10^{-3}H \times 5A/s = 50mV \)

12. Find the power loss in a transmission line with a resistance of \(8\Omega\) and a current of \(3A\).
Solution: \( P = I^2R = 3^2 \times 8\Omega = 72W \)

13. A transformer has a primary voltage of \(120V\) and a secondary voltage of \(240V\). If the primary current is \(2A\), find the secondary current.
Solution: Secondary current = \( \frac{{\text{{Primary voltage}}}}{{\text{{Secondary voltage}}}} \times \text{{Primary current}} = \frac{{120V}}{{240V}} \times 2A = 1A \)

14. Calculate the reactance of a \(50Hz\), \(0.2H\) inductor.
Solution: \( X_L = 2\pi fL = 2\pi \times 50 \times 0.2 = 62.83\Omega \)

15. Find the total capacitance of three \(10\mu F\) capacitors connected in series.
Solution: \( \frac{1}{{C_{\text{{total}}}}} = \frac{1}{{C_1}} + \frac{1}{{C_2}} + \frac{1}{{C_3}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}, C_{\text{{total}}} = \frac{10}{3} \mu F \approx 3.33\mu F \)

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16. A \(0.1F\) supercapacitor is charged to \(5V\). Find the energy stored.
Solution: \( E = \frac{1}{2} CV^2 = \frac{1}{2} \times 0.1 \times 5^2 = 1.25J \)

17. Calculate the power factor of a circuit with a real power of \(60W\) and an apparent power of \(75VA\).
Solution: Power factor = \( \frac{{\text{{Real Power}}}}{{\text{{Apparent Power}}}} = \frac{{60W}}{{75VA}} \approx 0.8 \)

18. Find the equivalent resistance of two \(4\Omega\) resistors in parallel.
Solution: \( \frac{1}{{R_{\text{{eq}}}}} = \frac{1}{{R_1}} + \frac{1}{{R_2}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}, R_{\text{{eq}}} = 2\Omega \)

19. Determine the impedance of a circuit with a resistance of \(5\Omega\) and a reactance of \(12\Omega\).
Solution: \( Z = \sqrt{R^2 + X^2} = \sqrt{5^2 + 12^2} \approx 13.34\Omega \)

15. Calculate the RMS value of a sinusoidal voltage with a peak value of \(120V\).
Solution: \( V_{\text{{RMS}}} = \frac{{V_{\text{{peak}}}}}{\sqrt{2}} \approx 84.85V \)

16. Find the power dissipated in a \(5\Omega\) resistor with a current of \(4A\).
Solution: \( P = I^2R = 4^2 \times 5\Omega = 80W \)

17. Calculate the frequency of a \(25mH\) inductor with a reactance of \(50\Omega\).
Solution: \( f = \frac{{X_L}}{{2\pi L}} = \frac{50}{{2\pi \times 25 \times 10^{-3}}} \approx 31.83Hz \)

18. Determine the current in a \(20\Omega\) resistor with a \(100V\) supply.
Solution: \( I = \frac{V}{R} = \frac{100V}{20\Omega} = 5A \)

19. Find the total inductance of two \(10mH\) inductors connected in parallel.
Solution: \( \frac{1}{{L_{\text{{total}}}}} = \frac{1}{{L_1}} + \frac{1}{{L_2}} = \frac{1}{10} + \frac{1}{10} = \frac{1}{5}, L_{\text{{total}}} = 5mH \)

20. Calculate the energy consumed by a \(1500W\) iron used for \(0.5\) hours.
Solution: \( E = Pt = 1500W \times 0.5h = 750Wh = 0.75kWh \)

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