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Can I use two or more of 0.25 W resistors to get a 0.5 W resistor? How do I make a 220Ω 0.5 W resistor using 0.25 W resistors (no matter what the value of the resistors). All I know is

V=I×RV=I×R (which we call Ohm's law)

P=I×VP=I×V (sometimes called Watt's law)

am gonna explain more about the project .. its a metal detector with a pulse induction and the problem was at the coil poles it must be a 220 ohm 0.5w and the battery was 12 v 70 mA this is the shematic of the project http://imgur.com/PNc4XG4

sorry for my bad english

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If you connect two 440Ω resistors in parallel you will have a 220Ω resistance, and the current will be split equally between the two, so they each dissipate half the total power.

(470Ω may be close enough, or you can put two 220Ω resistors in series to get 440Ω)

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  • \$\begingroup\$ and what about the 0.5w \$\endgroup\$ – ala khemiri Mar 28 '17 at 23:42
  • \$\begingroup\$ If the two 440Ω resistors in parallel are each 0.25W rating the net power rating of the two is 0.5W. \$\endgroup\$ – Michael Karas Mar 29 '17 at 0:12
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UPDATED. I observe that the problem is easy to solve when the resistors are all of the same resistance and same maximum power handling, but more complex when they are different. In the analysis below, I have gone through the various cases to provide what I think is a complete answer to the question for series, parallel, and combinations.

Unfortunately, the answer depends entirely on the specifications for each resistance, and the way they are connected together (series, parallel, or combinations). There is no one plug-and-play formula for any but some simple cases (mentioned below); analysis is required to ensure that you don’t burn up resistors.

Resistance in series

Resistors connected in series have a total resistance that is the sum of each of the resistors.

If there are n resistors with the same resistance R, the total resistance is therefore R*n.

Resistance in parallel

Resistors connected in parallel have a resistance calculated by:

  • Rparallel = 1/(1/R1 + 1/R2 + …)

If there are n resistors with the same resistance R, the total resistance is therefore R/n.

Ohm's Law

The relationship of voltage (E), current (I), and resistance (R) can be expressed in three algebraically equivalent ways:

  • E = I * R
  • R = E / I
  • I = E / R

Calculating Power Dissipation

For a resistance (single resistor R, or effective resistance of a network), the power is simply

  • P = E * I

or

  • P = E * E / R

(The latter substitutes an Ohm's Law equation for current, so that power can be determined using only the voltage and resistance.)

Power Handling for Series Resistors

Power handling must be computed for each resistor in order to determine whether its rated maximum is exceeded, or not. First, calculate the total resistance:

  • Rseries = R1 + R2 + …

Now determine the current for the series group:

  • I = E / Rseries

For a series of resistors, the current is the same through all resistors, but the voltage depends on the resistance of each resistance:

  • Er1 = I / R1
  • Er2 = I / R2
  • etc

Note that the sum of the Er values adds up to E. The power dissipated for each resistor is determined by:

  • Pr1 = Er1 * I
  • Etc

The total power dissipated by the series network is the sum of Pr1 + Pr2 etc. You have to be sure that none of the Pr values exceeds the rated maximum for the resistor.

If you have n resistors, each with the same resistance R and maximum power Pr, the total power is

  • Ptotal = n * (I / R) * I

(Note: (I/R) is the voltage drop of each resistor.)

But Ptotal is not all that useful, as it is the power dissipated for a given voltage. Increasing the voltage means having to check all resistors again to ensure that their individual maximum power dissipation is not exceeded.

Power handling for resistors in parallel

Assuming all resistors are connected in parallel, across the power source of voltage E, with resistors of different resistance and power capacity:

  • Pr1 = E * (E/R1)
  • etc

(Note: E/R1 is the current dissipated by resistor R1)

The total power dissipated by the parallel network is simply the sum of the power dissipated for each resistor. However, you must ensure that for each resistor, the power dissipated (as calculated above) does not exceed the maximum rating for that resistor.

For a parallel circuit of n resistors, each with the same resistance, and a maximum power dissipation Pr, the total maximum dissipation is

  • Pmax = n * Pr

Resistors in series/parallel combinations

Determine the resistance and power for each series or parallel group, then treat the result as a virtual replacement resistor. Continue to analyze the network until you are down to one replacement resistor. Check to make sure that no resistor has a power dissipation greater than the rated maximum.

ANY combination of resistors can be analyzed as combinations of series and parallel networks. If the diagram you start with does not show obvious networks of those types, try redrawing it so that it does.

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  • \$\begingroup\$ ok ok ok i still didn't get it but can you have a look on this reddit.com/r/electronic_circuits/comments/40967n/… \$\endgroup\$ – ala khemiri Mar 29 '17 at 0:02
  • \$\begingroup\$ Resistors in series does not increase power handling? Are you sure about that? 2 x 100r in series instead of 1 x 200r each would dissipate half the power. \$\endgroup\$ – Colin Mar 29 '17 at 5:38
  • \$\begingroup\$ well am not sure that's why i asked for help \$\endgroup\$ – ala khemiri Mar 29 '17 at 7:40
  • \$\begingroup\$ @alakhemiri just use equations from your question (ohm's law, watt's law) \$\endgroup\$ – Chupacabras Mar 29 '17 at 7:51
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    \$\begingroup\$ @Colin__s you are quite right. I was a little tired when I wrote the previous answer. In thinking about what you said, I have replaced my answer with a methodology for analyzing any network of resistors, including combinations of series and parallel, and determining total resistance and power dissipation. I would be grateful if you checked my work for errors. \$\endgroup\$ – Mark Colan Mar 30 '17 at 14:44
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As previously mentioned, place TWO 0.25W resistors of DOUBLE the desired resistance in parallel.

P = I^2*R I=V/R.

Do this for each resistor. You'll find that each resistor is dissipating HALF of the power, but the two resistors pass the same current as ONE resistor of the specified resistance. Comment if you need help with the math and I will respond.

Edit: adding examples

Parallel example:

schematic

simulate this circuit – Schematic created using CircuitLab

We have voltage V and two resistors with the same resistance, R. Let's find the current through each resistor. I(R) = V/R I(total) = 2V/R P(R) = I^2R = V^2/R P(total) = 2*(I^2R) = 2*V^2/R Thus using TWO parallel resistors will dissipate twice the power. But also note the current is doubled. Fix this by doubling the resistor values.

Let's use real numbers now
Suppose you have a 10V DC source and you want 40mA of current.
R = V/I = 10/0.04 = 250Ohm.
P(R) = V^2/R = 100/250 = 0.4W ==> too much for 1/4W resistor
So now let's try parallel resistors. R1||R2 = 250 ==> R1 = R2 = 500Ohm.
I(R1) = I(R2) = 10/500 = 20mA.
I(R1) + I(R2) = 40mA.
P(R1) = P(R2) = 10^2/400 = 0.20W. This is less than 1/4 watt.
You do need to consider deratings for resistors, but this isn't considered here.

Alternatively, you can use series resistors. Instead of DOUBLING the resistors, we will HALVE their values.

schematic

simulate this circuit

I = V/(R+R). P(R) = (V/(R+R))^2*R = (V/2R)^2*R = V^2/(4R^2)R = V^2/4R So let's do the same example as above. I(R) = V/R I(total) = 2V/R P(R) = I^2R = V^2/R P(total) = 2(I^2R) = 2*V^2/R Thus using TWO parallel resistors will dissipate twice the power. But also note the current is doubled. Fix this by doubling the resistor values.

Let's use real numbers now
Suppose you have a 10V DC source and you want 40mA of current.
R = V/I = 10/0.04 = 250Ohm.
P(R) = V^2/R = 100/250 = 0.4W ==> too much for 1/4W resistor
So now let's try seriesresistors. R1+R2 = 250 ==> R1 = R2 = 125Ohm.
I(R1) = I(R2) = 10/250 = 40mA.
P(R1) = P(R2) = 0.04^2*125 = 0.20W. This is less than 1/4 watt.
You do need to consider deratings for resistors, but this isn't considered here.

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  • \$\begingroup\$ thank you @amwalker707 for the explanation .. please help me with the math \$\endgroup\$ – ala khemiri Mar 29 '17 at 7:39
  • \$\begingroup\$ Added the math. Sorry for the late response. \$\endgroup\$ – amwalker707 Mar 30 '17 at 13:02

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