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What is equivalent Resistor of two this different Resistor put them in series or in parallel if they have a different wattage and different resistance?
In practice somtimes we have a resistor with \$\frac{1}{4}\$watt , \$\frac{1}{2}\$watt, \$1\$watt or etc...
But let`s say that wattage of resistor is \$\frac{3}{4}\$watt or \$\frac{3}{5}\$watt, or etc...

So what is the equivalent Resistor of two this "\$w_1\$ watt \$r_1\$ resistance" and "\$w_2\$ watt \$r_2\$ resistance" resistor put them in series or in parallel?

We know that if they have the same wattage \$w\$ watt and \$r_1\$, \$r_2\$ resistance then the equivalent Resistor might be:

1) in Series \$w\$ watt \$r_1\$ + \$r_2\$ resistance
2) in parallel \$2w\$ watt \$\frac{1}{r_1}\$ + \$\frac{1}{r_2}\$ resistance

But what about my question?

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Forget the wattage of the resistors to begin with, that's causing confusion. Do the analysis, combine resistors together first.

You can combine resistors in series or parallel. When in series the resistors add together. When in parallel, the calculation is a little more complex. Here are the two formulae you need.enter image description here

When you combine resistance values together, the power doesn't come into it.

In practise, it's your job as a designer to ensure that you don't execeed the power rating for the resistor so you have to calculate the power dissipation of the resistor and ensure it doesn't exceed that stated for the resistor, so you need to work out the current through the resistor and what voltage is dropped across the resistor. You need to know Ohms Law and Power formula (R=V/I , W =VI), re-arrange, combine if necessary and apply.

If you have a problem with actual power exceeding the maximum power rating for the resistor then there are things you can do to get around that: you can use multiple resistors in series, the current through each is the same but the voltage is shared across the resistors, thereby reducing the actual power dissipated by each resistor.

Alternatively putting resistors in parallel will reduce the current but keep the voltage the same. But you have to ensure the the combined resistance of the multiple resistors (whether in series or parallel) achieves the value you want to achieve.

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Two 1 watt resistors in series can be regarded as a 2 watt resistor if the individual resistance values are the same. If the resistance values are different then this is not true.

If one resistance is much bigger than the other then the power rating of both combined in series tends towards the power rating of the higher resistance resistor because it will always have a bigger volt drop across it.

For resistors in parallel, the power rating of the combined parallel pair tends towards the one with the least resistance because this resistor always takes most of the current.

Only resistors of equal resistance can have their individual wattages added to yield the combined wattage.

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In series, the current is the same through each resistor so you can use P = \$I^2R\$ for each resistor Ri, and of course the total resistance is the sum of the Ri. Power for each resistor must not exceed the rating for each resistor, but it's perfectly legitimate to mix wattages. For example, a 1/4-W 10 ohm resistor can handle 158mA but at that current, a 100 ohm resistor would dissipate 2.5W.

In parallel, the voltage is the same across each resistor so you can use P = \$E^2/R\$, and ensure that each resistor power rating is not exceeded. A 100 ohm 1/4W resistor can handle 5V, but a 10 ohm resistor would dissipate 2.5W. Of course the parallel resistance is found from the reciprocal of the sum of the reciprocals of each value.

Rx = \$\frac{1}{1/R1+..+1/Ri}\$

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First and foremost, resistors have resistance.

Given a Resistor \$R_1\$ and another one \$R_2\$, the total resistance of both is

$$R_{series} = R_1 + R_2$$ $$R_{parallel} = R_1 || R_2 = \frac{R_1 R_2}{R_1 + R_2}$$

You can show that \$R_{series} > R_{parallel}\$: $$R_1 + R_2 > \frac{R_1 R_2}{R_1 + R_2}$$ $$(R_1 + R_2)² > R_1 R_2$$ $$R_1² + 2 R_1 R_2 + R_2² > R_1 R_2$$ $$R_1² + R_1 R_2 + R_2² > 0$$


The amount of power that they consume depends on the voltage that you apply or the current that runs through them.

$$R = \frac{U}{I}$$

$$P = U * I = R * I² = \frac{U²}{R}$$

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