- Materials & Industrial Technology
How much Power Dissipate Or Loss In Resistors?
Today I got the honor to write over quite interesting and important topic of electrical engineering i.e. “Power dissipation in resistors”. Before going into the topic we should know the basic terminologies used in above prescribed article.
Power is defined as rate of transforming energy from one form to another form. The S.I. unit for it is Watt.
A resistor is a passive component of a circuit that resists the flow of electrical current. The S.I. unit for resistor is Ohm.
Now we can go towards our main article. The above mentioned definitions will be quite beneficial in understanding the article clearly. We can derive the relation that how much a certain power dissipated across the resistor with the help of figure (1).When the current is flowing through high resistance electric appliances like bulb, heater and electric iron the energy lost by the electrons at the time of collision appears in the form of light and heat. The relation for power loss across a resistor is derived as follow:
Let a charge ‘Q’is made to pass through a resistor AB such that the voltage across its two ends is ‘V’, in ‘t’ time. The energy lost bythe charge Q while passing through the resistor is ‘VQ’. This energy is taken by the atoms of the resistor and so increases their temperature.
i.e. Energy = V Q
Time = t
But from above we know that power is transformation of energy or rate of change or energy or rate of loss of energy so
Power = Energy / time
P = V Q / t
Q/ t = I (Electric Current)
P = V I (I)
But also by Ohm’s Law
V = I R (II)
So (I) becomes
P = I R * I
P = I2R (III)
Also by Ohm’sLaw
I = V / R
So (II) becomes
P =(V / R)2 * R
P =V2 / R2 * R
P = V2 / R (IV)
Equation (IV)describes the amount of power dissipated across certain resistor. For example if ‘2’ Ampere of current is applied across the resistor of ‘5’ ohm, then the power dissipated across the resistor can be calculated using above equations.Equation (II) gives the value of voltage i.e. 10 Volts, further using equation(IV) we can calculate the amount of power loss i.e. 20 watt.