# Power In Electric Circuits

## Concept of power in electrical circuits. Calculation of consumed power in resistors. Power in series and parallel circuits. Subject expression and solved examples.

**Concept Of Power**

Meaning of the power in physics is amount of work that does at unit time.

Equation of the power is

P = | W | |

t |

P: Power (watt)

W: Amount of the work that is done (joule)

t: time (second)

using the equality of P = W/t

We can obtain the equality of W = P*t easily.

Electric energy is consumed as heat energy in resistances. When a current flow through a resistor, it is occur a power in the resistor. This power is dissipated form heat energy. The dissipated heat energy is calculates as joule. 1 joule is loss of electrical energy in1 second.

We will discuss the energy that consumed in resistors another page. In this page, we will discuss the power loss in resistors.

**Power In Resistors**

The power loss in a resistor is given as below,

P = V*I

Because of V = I*R

P = I*R*I

P = I^{2}*R

P = | V^{2} | |

R |

So we can find the power with both V * I, I^{2} * R or V^{2}/R equations.

The unit of power is Watt. Power of 1 watt is

Watt = Volt*Amper

1 Watt = 1 Volt *1 Amper

**Example:**

In the circuit shown in the figure, power in the resistor R_{1} is P_{1}, power in the R_{2} is P_{2}, power in the R_{3} is P_{3}, power in the R_{4} is P_{4}, power in the R_{5} is P_{5}.

Calculate the power P_{1}, P_{2}, P_{3}, P_{4} and P_{5}.

Calculate total power drawn from supply source.

**Solution:**

The power in a resistor can be found by two equation in the following

P = I^{2}*R

P = V*I

For find the power, we must find the equivalent resistance of the circuit. Then, we will find total current drawn from source. Finally, we will find power in each resistors.

R_{2,3} = 12 + 8

R_{2,3} = 20 Ω

R_{2,3} and R_{4} are parallel.

R_{2-4} = | 20*30 | |

50 |

R_{2-4} = 12 Ω

R_{1}, R_{2-4} and R_{5} series connected.

R_{1-5} = 12 + 12 +6

R_{1-5} = 30 Ω

The equivalent resistance is

R_{s} = 30 Ω

Main current of the circuit is

I_{s} = | 120 | |

30 |

I_{s }= 4 A

The current I_{1} that flow through R_{1} is main current.

I_{1} = 4 A

Power in the R1 is

P_{1} = 4^{2 }*12

P_{1} = 192 W

The current I_{1} is divided to into two arms at point A.

I_{2} = | I_{1}*R_{4} | |

R_{2} + R_{3} + R_{4} |

I_{2} = | 4*30 | |

50 |

I_{2} = 2,4 A

I_{4} = 4 – 2,4

I_{4} = 1,6 A

The current I_{3} is equal to the current I_{2}.

I_{2} = I_{3} = 2,4 A

Power in the resistor R_{2} is

P_{2} = (I_{2})^{2 }*R_{2}

P_{2} = 5,76*12

P_{2} = 69,12 W

Power in the resistor R3 is

P_{3} = (I_{3})^{2} *R_{3}

P_{3} = 5,76*8

P_{3} = 46,08 W

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

P_{4} = 2,56*30

P_{4} = 76,8 W

P_{5} = (I_{5})^{2}*R5

P_{5} = 16*6

P_{5} = 96 W

Total power drawn from the source is

P_{t} = P_{1} + P_{2} + P_{3} + P_{4} + P_{5}

P_{t} = 192 + 69,12 + 46,08 + 76,8 + 96

P_{t} = 480 W

**Electric Circuits Questions And Its Solutions**

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November 19 2018

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