# Energy In Electric Circuits

## Subject of electric circuits. Energy consumed in electric circuits. Energy consumed in circuits in series and parallel. The amount of energy consumed in any resistor. Questions and its solutions.

**Concept Of Energy In Electric Circuits**

If an amount current flows through any electric components, an amount energy loss occur in this component.

This energy loss is the energy dissipated to the environment in the form of heat.

The amount of the energy of dissipated is depends to the power in the electric components.

Mechanical work in physics was given in the following formula.

P = | Amount Of work done | |

Time elapsed |

The definition of electrical power is as following.

P = | Work done by electric | |

Time elapsed |

The energy that consumed from any resistor is depend to the power of this resistor. More power consumes more energy.

Power in electric circuits was given in the formulas following .

P = I^{2}.R

P = | V^{2} | |

R |

P = V*I

Since the energy is power consumed per unit time is

W = P.t

W = I^{2}*R*t

W = | V^{2}*t | |

R |

W = V*I*t

The power was ratio to the time elapsed of the work done.

P = | W | |

t |

The work done is equal to the multiplication of time and power. This multiplication also gives the energy consumed.

W = P*t

The greater the power of a resistor the more it many do the work. Of course, the greater the power of a resistor the more it many consumes energy.

The unit of energy consumed in the electric circuits is joule like others energy units.

1 Joule = 1 Watt*1 second

1 Watt = (1 A)^{2} * 1 Ohm

1 J = 1.A^{2 }*Ohm*second

**Concept Of The Horsepower**

Is called 1 horsepower to the power of 745.7 Watts. The horsepower is represented by hp.

The horsepower is taken 746 Watts approximately.

1 kW = 1000 Watt

1 kW = 1000/746

1 kW = 1,34 hp

** Example:**

In the circuit shown in the figure, R1 = 6 Ω, R2 = 4 Ω, R3 = 2 Ω, R4 = 4 Ω, R5 = 20 Ω, R6 = 5 Ω, R7 = 4 Ω, R8 = 16 Ω

The voltage of the supply source is 110 V.

A) Find the power in each resistor.

B) Find the total power drawn from the source.

C) Find the energy consumed in 10 minutes in each resistor.

D) Find the total energy consumed in 10 minutes in the circuit.

**Solution:**

First we will find the equivalent resistance of the circuit. Then we will find main current of the circuit.

Then we will calculate current flowing through each resistor.

After finding the current flowing through each resistor, we can do calculations the power and energy.

The equivalent resistance of the resistors R_{1}, R_{2} and R_{3} on upper arm.

R_{1-3} = 6 + 4 + 2

R_{1-3} = 12 Ω

The equivalent resistance of the lower arm resistors.

R_{5,6} = | 20*5 | |

25 |

R_{5,6} = 4 Ω

R_{4-7 }= 4 + 4 + 4

R_{4-7} = 12 Ω

The resistor R_{1-3} and the resistor R_{4-7} are parallel. The equivalent resistance of these resistors is

R_{1-7} = | 12 | |

2 |

R_{1-7 }= 6 Ω

The resistor R_{8} is connect to these resistors in series.

R_{1-8} = 6 +16

R_{1-8} = 22 Ω

The equivalent resistance of the circuit is

R_{s} = 22 Ω

The main current of the circuit is

Is = | 110 | |

22 |

Is = 5 A

Suppose, the current flowing through the resistors R_{1}, R_{2} and R_{3} is I_{1}.

I_{1} = | R_{4-7} *I_{s} | |

R_{4-7} + R_{1-3} |

I1 = | 5*12 | |

24 |

I_{1} = 2,5 A

Suppose, the current flowing through the arm which found the resistors R_{4-7} is I_{2}.

I_{2} = 5 – 2,5 = 2,5 A

The current flows through resistors R_{1}, R_{2} and R_{3} on upper arm is 2.5 A.

P_{R1} = (2.5)^{2}*6

P_{R1} = 37.5 W

P_{R2} = (2.5)^{2}*4

P_{R2} = 6.25*4

P_{R2} = 25 W

P_{R3} = (2.5)^{2}*2

P_{R3} = 12.5 W

The current flows through resistors R_{4-7} in lower arm is 2.5 A.

P_{R4 }= (2.5)^{2}*4

P_{R4} = 25 W

The current flows through resistor R_{5} is

I_{R5} = | 2.5*5 | |

25 |

I_{R5} = 0,5 A

The current flows through resistor R_{6} is

I_{R6} = 2,5 – 0,5 = 2 A

The power in the resistor R_{5} is

P_{R5 }= (0.5)^{2}*20

P_{R5} = 5

The power in the resistor R6 is

P_{R6} = 2^{2}*5

P_{R6 }= 20 W

The current flows through the resistor R_{7} is 2.5 A.

P_{R7} = 6.25*4

P_{R7} = 25 W

The current flows through the resistor R_{8} is 5 A.

P_{R8} = 5^{2} *16

P_{R8} = 400 W

B) The total power drawn from source is

P_{total }= 37.5 + 25 + 12.5 + 25 + 5 + 20 + 25 + 400

P_{total} = 550 W

Or

P_{total} = V*I

P_{total} = 110*5 = 550 W

C) The energy consumed in 10 minutes in each resistor

10 minutes = 10*60 = 600 seconds

W_{R1} = P_{R1}*600

W_{R1} = 37,5*600

W_{R1} = 22500 J = 22,5 kJ

W_{R2} = 25*600 = 15000 J = 15 kJ

W_{R3} = 12,5*600 = 7500 J = 7,5 kJ

W_{R4} = 25*600 = 15000 J = 15 kJ

W_{R5} = 5*600 = 3000 J = 3 kJ

W_{R6} = 20*600 = 12000 J = 12 kJ

W_{R7} = 25*600 = 15000 J = 15 kJ

W_{R8} = 400*600 = 240000 J = 240 kJ

D) The total energy consumed in 10 minutes in the circuit.

W_{total} = 22.5 + 15 + 7.5 + 15 + 3 + 12 + 15 + 240

W_{total} = 330 kJ

Or

W_{total }= P_{total}*600

W_{total} = 550*600 = 330 000 J

W_{total} = 330 kJ

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December 1 2018

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