# Energy In Electric Circuits Test-II Solutions

## Calculations of the power and energy in series and parallel circuits. Calculating the power and energy in circuits created with bulb, motor and resistors. Solutions of the test-II questions.

**Solutions 1**

Let we find the current flowing through in electrical motor.

V = 380 V

R = 95 Ω

I = | 380 | |

95 |

I = 4 A

P = V*I

P = 380*4

P = 1520

1 hp was 746 Watt. Let we establish proportion.

1 hp 746 W

x 1520

x = | 1520 | |

746 |

x = 2.04 hp

The right answer is option E

**Solution 2**

First, let we calculate the heat energy given to the water by container K and L

For this, we must find the current flowing through each container.

The sum of the resistance values of the two container is

Rs = 18 + 26 = 44 Ω

Current of the circuit is

I_{s} = | 110 | |

44 |

I_{s} = 2,5 A

Power of the container K

P_{K} = (2.5)^{2 }*18

P_{K} = 6,25*18

P_{K} = 112.5 W

Power of the container L

P_{L} = (2.5)^{2}*26

P_{L} = 162.5 W

20 minutes = 20*60 = 1200 seconds

Given energy to the water by the container K in 1200 seconds is

W_{K} = 112.5*1200

W_{K} = 135000 J

Given energy to the water by the container L in 1200 seconds is

W_{L} = 162.5*1200

W_{L} = 195000 J

Now, let we calculate last temperatures of the waters in the containers.

Q_{K} = m*c*t

135000 = 600*4*(t_{2} – 10)

1350 = 24(t_{2} – 10)

56.25 = t_{2} – 10

t_{2} = 66.25 °C

T1 = 66.25 °C

The last temperature of water in the container K is 66.25 °C

Let we calculate last temperature of the water in the container L too.

QL = m*c*Dt

195000 = 600*4*(t_{2} – 10)

1950 = 24(t_{2} – 10)

81.25 = t_{2} – 10

t_{2} = 91.25 °C

T_{2} = 91.25 °C

The right answer is option C

**Solution 3**

The voltage of all three bulb are equal and 100 V.

Let we find the power of each lamp.

P_{L1} = | V^{2} | |

R_{L1} |

P_{L1} = | 110^{2} | |

160 |

P_{L1} = 75.6 W

P_{L2} = | 110^{2} | |

200 |

P_{L2} = 60.5 W

P_{L3} = | 110^{2} | |

220 |

P_{L3} = 55 W

Total power of all three bulb is

P_{total} = 75.6 + 60.5 + 55

P_{total} = 191.1 W

1 hour = 3600 seconds

W_{total} = 191.1*3600

W_{total} = 687960 J

W_{total} = 687.96 kJ

The right answer is option D

**Solution 4**

Let we find the power of the bulbs.

P_{L1} = | 110^{2} | |

220 |

P_{L1} = 55 W

P_{L2} = | 110^{2} | |

100 |

P_{L2} = 121 W

If the bulb L1 operates in 8 hours per a day, the amount of energy to be consumed in one day is

W_{day} = 55*8

W_{day} = 440 Wh = 0.44 kWh

The amount of energy to be consumed in 30 day is

W_{month} = 0.44*30

W_{month} = 13,2 kWh

1 kWh = 0.2 USD,

1 kWh 0,2 USD

13,2 kWh x

x = 2,64 USD

Cost L_{1} = 2.64 USD

If the bulb L_{2} operates in 8 hours per a day, the amount of the energy to be consumed in one day is

W_{day} = 121*8

W_{day} = 968 W

The amount of the energy to be consumed in 30 day is

W_{month} = 968*30

W_{month} = 29040

W_{month} = 29.04 kWh

1 kWh 0.2 USD

29.04 kWh y

y = 5.80 USD

y – x = 5.8 – 2.64

y – x = 3.16 USD

The right answer is option B

**Solution 5**

P = 5 hp

1 hp = 746 W, power of the appliance is

Wc = 746*5

Wc = 3730 W

P = V*I

3730 = 380*I

I = 9.8 A

The right answer is option A

**Solution 6**

First, we finds the equivalent resistance of the circuit. Then we find the main current (I_{s}) of the circuit.

We can find the currents I_{1} and I_{5} using the current Is.

Using the equailities of P = I^{2}*R and E = P*t we can find the value of E_{1} and E_{5}.

R_{3,4} = 6 Ω

R_{1,3,4} = | 6 | |

2 |

R_{1,3,4} = 3 Ω

R_{2,5} = | 12*4 | |

16 |

R_{2,5} = 3 Ω

R_{s} = R_{1,3,4} + R_{2,5} + R_{6}

R_{s} = 3 + 3 + 4

R_{s} = 10 Ω

I_{s} = | 60 | |

10 |

I_{s} = 6 A

I_{1} = | I_{s}*R_{3,4} | |

R_{3,4} + R_{1} |

I_{1} = | 6*6 | = 3A |

12 |

P_{R1} = 3^{2}*6

P_{R1} = 54 W

E_{1} = 54*t

I_{5} = | 6*4 | |

16 |

I_{5} = 1.5 A

P_{R5} = (1.5)^{2}*12

P_{R5} = 27 W

E_{5} = 27*t

E_{1} |
| |||||

E_{5} |

= 2

The right answer ip option C

**Solution 7**

Let we calculate the energy consumed in one day in the house.

Washing machine→ 2000 W*1 hour = 2000 Wh

W.M. = 2 kWh

Television → 30 W*5 hour = 150 Wh

TV = 0.15 kWh

Bulbs → 2*75 W*6 hours = 900 Wh

Bulbs = 0.9 kWh

Refrigerator → 100 W*8 hours = 800 Wh

Refrigerator = 0.8 kWh

The total energy consumed in one day.

T_{day} = 2 + 0.15 + 0.9 + 0.8

T_{day} = 3.85 kWh

The total energy consumed in thirty day.

T_{month} = 3.85*30

T_{month} = 115.5 kWh

Total cost = 115.5*0.2

Total cost = 23.1 USD

The right answer is option E

**Solution 8**

The equivalent resistance of the circuit is

R_{1,2} = 6 + 4 = 10 Ω

R_{3,4} = 3 + 12 = 15 Ω

R_{1-4} = | 10*15 | |

25 |

= 6 Ω

R_{s} = R_{1-4 }+ R_{5} = 6 + 4 = 10 Ω

Total power is

P = | V^{2} | |

R |

P = | 60^{2} | |

10 |

P = 360 W

Consumed energy is

W = P*t

15 minutes = | 15 | = 0.25 hours |

60 |

W = 360*0.25

W = 90 kWh

The right answer is option B

**Electrical Energy Test-2 Questions**

**Power Questions In Electric Circuits**

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December 15 2018

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