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 = I2*R

P = V2
R




So we can find the power with both V * I, I2 * R or V2/R equations.

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

Watt = Volt*Amper

1 Watt = 1 Volt *1 Amper


Example:

circuitpower_s1i1


In the circuit shown in the figure, power in the resistor R1 is P1, power in the R2 is P2, power in the R3 is P3, power in the R4 is P4, power in the R5 is P5

Calculate the power P1, P2, P3, P4 and P5.

Calculate total power drawn from supply source.


Solution:

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

P = I2*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.


R2,3 = 12 + 8

R2,3 = 20 Ω

R2,3 and R4 are parallel.

R2-4  = 20*30
50




R2-4 = 12 Ω


R1, R2-4 and R5 series connected.

R1-5 = 12 + 12 +6

R1-5 = 30 Ω

The equivalent resistance is

Rs = 30 Ω


Main current of the circuit is

Is = 120
30



Is = 4 A

The current I1 that flow through R1 is main current.

I1 = 4 A

Power in the R1 is

P1 = 42 *12

P1 = 192 W


The current I1 is divided to into two arms at point A.

I2 = I1*R4
R2 + R3 + R4



I2 =4*30
50



I2 = 2,4 A

I4 = 4 – 2,4

I4 = 1,6 A

The current I3 is equal to the current I2.

I2 = I3 = 2,4 A

Power in the resistor R2 is

P2 = (I2)2 *R2

P2 = 5,76*12

P2 = 69,12 W


Power in the resistor R3 is

P3 = (I3)2 *R3

P3 = 5,76*8

P3 = 46,08 W


P4 = (I4)2*R4

P4 = 2,56*30

P4 = 76,8 W


P5 = (I5)2*R5

P5 = 16*6

P5 = 96 W


Total power drawn from the source is

Pt = P1 + P2 + P3 + P4 + P5

Pt = 192 + 69,12 + 46,08 + 76,8 + 96

Pt = 480 W


Electric Circuits Questions And Its Solutions



RISE KNOWLEDGE

November 19 2018

  • WRITE COMMENT
  • NAME SURNAME(or nick)
  • COMMENT
COPYRIGHT© ALL RIGHTS RESERVED