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what will be the resultant wattage when 4 resistors having individual resistances and wattage given as (2.4ohm,1W) , (5.1ohm,2W) , (10ohm,5W) , (22ohm,10W) connected in parallel across an DC voltage source? which resistor would dissipate more power in the resultant circuit? please help me on this question. i had spent lot of time trying to understand it in detail.

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  • \$\begingroup\$ This smells like a homework question. Just use basic math to calculate the flowing circuits and then compare... \$\endgroup\$
    – PlasmaHH
    Sep 11, 2015 at 11:46
  • \$\begingroup\$ Can you determine the power dissipation if its only one resistor connected to the power supply ? What is the consequence of all resistors being "in parallel with the power supply" ? \$\endgroup\$ Sep 11, 2015 at 11:47
  • \$\begingroup\$ it's a question asked in an examination last year, i was revising that paper again and struck with a doubt regarding wattage of resistor combinations in particular, not with the calculation of power dissipation rather the role of wattage in series and parallel connection of resistors. Thank u. \$\endgroup\$ Sep 12, 2015 at 3:46

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The fact that the resistors are connected in parallel is irrelevant except to establish that the same voltage is impressed across each resistor.

Likewise, a resistor's power rating has nothing to do with the actual power dissipated by a resistor, it's merely a rating based on how hot the resistor can get (how much work it can do / how much power it can dissipate) before its temperature rises to the point where it'll be damaged in some way.

What decides the actual dissipation is the product of the voltage across the resistor and the current through it.

For example, if we have a one ohm resistor and we place one volt across it, the current through it will be

$$ I =\frac{E}{R} = \frac{1V}{1 \Omega} = \text {1 ampere}$$

and it'll dissipate

$$ P = IE = 1V \times 1A = 1 \text{ watt}$$

If the resistor is rated to dissipate one watt, then that means that its temperature will rise to, say, 100C when it's dissipating one watt which is ,say, the highest temperature it can safely tolerate.

If the resistor had a power rating of two watts, however, it would still dissipate one watt, but it would be running cooler than if it were rated at one watt.

In that vein, then, since:

$$ P = \frac{E^2}{R}, $$

if we arbitrarily choose 10 volts, DC, as the excitation source, we can write for the first resistor:

$$ P = \frac{E^2}{R}=\frac{10V^2}{2.4\Omega} \approx 42\text{watts.} $$

Considering that the resistor is rated to dissipate one watt indicates that it will be destroyed in short order.

The calculation for the other three resistors is identical except for the values of resistance, so determining which one will dissipate the most power is easily determined by comparing the quotients, and the total power dissipated by the circuit will be the sum of the power dissipated by each resistor.

Finally, since there's no voltage specified in the problem, it might be an interesting exercise to work out, academically, the voltage which will cause the 2.4 ohm resistor to dissipate its rated power and use that voltage to work out the rest of the problem. :)

Like this:

Since $$ P =\frac{E^2}{R} $$

we can rearrange and solve for E like this:

$$ E = \sqrt {PR} = \sqrt {1W\times 2.4\Omega} \approx 1.55 \text{volts,} $$

We now have the voltage which will appear across all of the paralleled resistors, so it should be a simple matter of plugging numbers into formulas and watching answers fall out.

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  • \$\begingroup\$ firstly,thanks for the reply. i do know the formula for power P=V^2/R or I^2 *R my doubt was regarding the role of wattage in deciding the power dissipation of any resistor. \$\endgroup\$ Sep 12, 2015 at 3:32
  • \$\begingroup\$ Also how can we calculate the total wattage of the above equivalent circuit finally? \$\endgroup\$ Sep 12, 2015 at 3:40
  • \$\begingroup\$ @RamaKrishna: Please see my edited answer. \$\endgroup\$
    – EM Fields
    Sep 12, 2015 at 8:17
  • \$\begingroup\$ @RamaKrishna: There are two ways to calculate the total power dissipated, one being to calculate the power dissipated in each resistor and then to add up the watts, and the other is to calculate the equivalent resistance of all four resistances in parallel and then calculate the power dissipated as if they were a single resistor. \$\endgroup\$
    – EM Fields
    Sep 12, 2015 at 9:02
  • \$\begingroup\$ yes,but what would be the total wattage,i think it should be different from the total power dissipation, because wattage is the maximum possible power dissipation without burning out.isn't it? \$\endgroup\$ Sep 12, 2015 at 9:16
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When you know the voltage across a resistor, the power is given by:

$$P = \frac{V^2}{R}$$

Given that the voltage across all resistors in parallel is the same, what does this tell you about the relationship between resistance and power?

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  • \$\begingroup\$ yes, i can see the inverse proportionality between resistance and power dissipation. but, i am more confused regarding the role of wattage, in particular about what would be the final total wattage of the combination. \$\endgroup\$ Sep 12, 2015 at 3:49
  • \$\begingroup\$ @Rama Krishna (3:49) The way to approach that is to find out which resistor will hit its power limit first as the voltage increases. This would be the highest voltage you could safely apply to the set of them. Then calculate the total power all of them dissipate. \$\endgroup\$
    – plasmo
    Feb 24, 2017 at 16:18

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