Use Omni's power dissipation calculator to determine the power dissipated in series and parallel resistor circuits. Just enter the applied voltage and the resistances of the resistors. The calculator will calculate the equivalent resistance of the circuit, total current through the circuit, and the power dissipated by the resistors. Continue reading to know what power dissipation is and the formula to calculate the power dissipated by resistors. You will also find an example of how to find the power dissipated in a circuit.
According to Ohm's law, when an electric current passes through a resistor, a voltage drop occurs across its end. This means that some electric potential energy is lost when the electric charge moves from one end of the resistor to the other. This potential energy is converted to heat, and we call this conversion (or loss) dissipation. Thus, power dissipation is the loss of electric potential energy (per unit time) in the form of heat by an electrical device when a current flows through it. Although power dissipation is an undesirable effect, there are instances where this phenomenon is of great use, for example, in electric heaters.
To calculate the power P dissipated by a resistor R when a current I passes though it, we will use Joule's law of heating: P = I2 * RWe can also substitute Ohm's law into the above equation to get: P = V * I…or: P = V2 / RYou can use any of the three formulas mentioned above to calculate the power dissipated by resistors. The power dissipated by a resistor appears in the form of heat, i.e., the resistor gets heated when current flows through it. The maximum power that a resistor can dissipate without burning is called the power rating of a resistor.
In a series circuit (see figure 1), the circuit's total resistance (or equivalent resistance) is the sum of the individual resistances. Req = R1 + R2 + ... + RnIf we connect a voltage source V, to this circuit, the same current I will flow sequentially through all the resistors. We can calculate the total current in a series circuit using Ohm's law: I = V / Req Fig. 1: A series combination of resistors. (Source: wikimedia.org)The power dissipated by each resistor is: P1 = I2 * R1, P2 = I2 * R2, ......,Pn = I2 * Rn The total power dissipated by the circuit is: P = P1 + P2 + .... + PnP = I2 * Req ...and the total power supplied by the voltage source (e.g., a battery) is: P = I * V
Let us consider a circuit consisting of parallel combination of resistors, R1, R2, ....., Rn (see figure 2). We can determine the total resistance in a parallel circuit using the formula: 1 / Req = 1 / R1 + 1 / R2 + ..... + 1 / Rn If we connect a voltage source V across this combination, the total current in a parallel circuit is: I = V / Req Fig. 2: A parallel combination of resistors. (Source: wikimedia.org)In parallel combination of resistors, the voltage drop across each resistor is the same. Hence we can calculate the power dissipation across each resistor as: P1 = V2 / R1, P2 = V2 / R2, ...., Pn = V2 / Rn The total power dissipated by the circuit is: P = P1 + P2 + .... + PnP = V2 / Req
Let us see how to calculate power dissipation in a series circuit using the power dissipation calculator. We will consider a circuit consisting of three resistors connected in series across a battery with a voltage output of 12.0 V. Let the three resistors be R1 = 2 Ω, R2 = 6 Ω, and R3 = 1 Ω.
You can also use this calculator just to calculate the total current in a parallel circuit or a series circuit.
To find the power dissipated in a series circuit, follow the given instructions:
To find the power dissipated in a parallel circuit, follow these steps:
A parallel circuit dissipates more power than a series circuit. For the same set of resistors, the total resistance of the parallel circuit is always lower than the total resistance in a series circuit. Since the power dissipated in a circuit is inversely proportional to the resistance of the circuit for a constant voltage source, a parallel connection will dissipate more power than a series circuit.
In a series circuit, the highest value resistor dissipates the most power. This is because the same current sequentially flows through each resistor in a series connection, and the power dissipated is directly proportional to the resistance. In a parallel connection, the power dissipated is inversely proportional to the resistance, as the voltage drop is the same across each resistor. Hence, the lowest value resistor dissipates the most power.
To determine the heat dissipation from power consumption, multiply the power consumed (in watts) by the time (in seconds) for which the current flows.
To calculate the power rating of any electrical appliance, use the given steps:
Do you know that we call power consumed by resistors the active power and power consumed by inductors or capacitors the reactive power? Read more about active and reactive power on our three phase calculator.
 An ever-present challenge in electronic circuit design is selecting suitable components that not only perform their intended task but also will survive under foreseeable operating conditions. A big part of that process is making sure that your components will stay within their safe operating limits in terms of current, voltage, and power. Of those three, the “power” portion is often the most difficult (for both newcomers and experts) because the safe operating area can depend so strongly on the particulars of the situation. In what follows, we’ll introduce some of the basic concepts of power dissipation in electronic components, with an eye towards understanding how to select components for simple circuits with power limitations in mind. — STARTING SIMPLE — Let’s begin with one of the simplest circuits imaginable: A battery hooked up to a single resistor: Here, we have a single 9 V battery, and a single 100 ? (100 Ohm) resistor, hooked up with wires to form a complete circuit. Easy enough, right? But now a question: If you want to actually build this circuit, how “big” of a 100 ? resistor do you need to use to make sure that it doesn’t overheat? That is to say, can we just use a “regular” ¼ W resistor, like the one shown below, or do we need to go bigger? To find out, we need to be able to calculate the amount of power that the resistor will dissipate.
Power Rule: P = I × V
Aside: To understand this, we need to remember what current and voltage physically represent. Electric current is the rate of flow of electric charge through the circuit, normally expressed in amperes, where 1 ampere = 1 coulomb per second. (The coulomb is the SI unit of electric charge.) Voltage, or more formally, electric potential, is the potential energy per unit of electric charge —across the circuit element in question. In most cases, you can think of this as the the amount of energy that is “used up” in the element, per unit of charge that passes through. Electric potential is normally measured in volts, where 1 volt = 1 joule per coulomb. (The joule is the SI unit of energy.) So, if we take a current times a voltage, that gives us the amount of energy that is “used up” in the element, per unit of charge, times the number of those units of charge passing through the element per second: 1 ampere × 1 volt = 1 ( coulomb / second ) × 1 ( joule / coulomb ) = 1 joule / second The resulting quantity is in units of one joule per second: a rate of flow of energy, better known as power. The SI unit of power is the watt, where 1 watt = 1 joule per second. Finally then, we have 1 ampere × 1 volt = 1 watt Back to our circuit! To use the power rule (P = I × V), we need to know both the current through the resistor, and the voltage across the resistor. First, we use Ohm’s law ( V = I × R ), to find the current through the resistor. Therefore, the current through the resistor is:
I = V / R = 9 V / 100 ? = 90 mA Then, we can use the power rule ( P = I × V ), to find the power dissipated by the resistor. Therefore, the power dissipated in the resistor is:
P = I × V = 90 mA × 9 V = 0.81 W So can you go ahead and use that 1/4 W resistor? No, because it would likely fail from overheating. A 1 W resistor typically comes in a much larger physical package, like the one shown here: (A 1 W, 51 ? resistor, for size comparison.) Because a 1 W resistor is much larger physically, it should be able to handle dissipating a higher amount of power, with its higher surface area and wider leads. (It may still get very hot to the touch, but it should not get hot enough that it fails.) Here’s an alternate arrangement that works with four 25 ? resistors in series (which still adds up to 100 ?). In this case, the current through each resistor is still 90 mA. But, as there is only one quarter as much voltage across each resistor, there is only one quarter as much power dissipated in each resistor. For this arrangement, one only needs the four resistors to be rated for 1/4 W.
Aside: Working through that example. Because the four resistors are in series, we can add their values together to get their total resistance, 100 ?. Using Ohm’s law with that total resistance gives us the 90 mA current again. And again, since the resistors are in series, the same current (90 mA) must flow through each, back to the battery. The voltage across each 25 ? resistor is then V = I × R, or 90 mA × 25 ? = 2.25 V. (To double check that this is reasonable, note that the voltages across the four resistors sum up to 4 × 2.25 V = 9 V.) The power across each individual 25 ? resistor is P = I × V = 90 mA × 2.25 V ? 0.20 W, a safe level for use with a 1/4 W resistor. Intuitively, it also makes sense that if you divide up a 100 ? resistor into four equal parts, each should dissipate one quarter of the total power. — ON BEYOND RESISTORS — For our next example, let’s consider the following situation: Suppose that you have a circuit that takes input from a 9 V power supply, and has an onboard linear regulator to step the voltage down to 5 V, where everything actually runs. Your load, on the 5 V end, could be as high as 1 A. What does the power look like in this situation? The regulator essentially acts like a big variable resistor, that adjusts its resistance as needed to maintain a consistent 5 V output. When the output load is a full 1 A, the output power delivered by the regulator is 5 V × 1 A = 5 W, and the power input to the circuit by the 9 V power supply is 9 W. The voltage dropped across the regulator is 4 V, and at 1 A, that means that 4 W is dissipated by the linear regulator— also the difference between the power input and the power output. In each part of this circuit, the power relationship is given by P = I × V. Two parts — the regulator and the load — are places where power is dissipated. And in the part of the circuit across the power supply, P = I × V describes the power input to the system— the voltage increases as the current travels across the power supply. Additionally, it is worth noting that we have not said what kind of load is pulling that 1 A. Power is being consumed, but that does not necessarily mean that it is being converted into (just) heat energy— it could be powering a motor, or powering a set of battery chargers for example.
Aside: — WHEN THERE ISN’T A SIMPLE “POWER” SPECIFICATION — Next, a slightly more challenging part: making sure that your regulator can handle the power. While resistors are clearly labeled with their power capacity, linear regulators are not always. In our regulator example above, let’s further suppose that we’re using a L7805ABV regulator from ST (datasheet here). (Photo: A typical TO-220 case, the type typically used for medium-power linear regulators) The L7805ABV is a 5 V linear regulator in a TO-220 package (similar to the one shown above), that is rated for 1.5 A output current and up to 35 V input voltage. Naively, you might guess that you can hook this right up to 35 V input, and expect to get 1.5 A of output, meaning that the regulator would be radiating 30 V * 1.5 A = 45 W of power. But this is a tiny plastic package; it actually can’t handle that much power. If you look in the datasheet under the “Absolute maximum ratings” section, to try and find how much power it can handle, all that it says is “Internally limited” — which is anything but clear on its own. It does turn out that there is an actual power rating, but it’s usually somewhat “hidden” within the datasheet. You can figure it out by looking at a couple of related specifications: • TOP, Operating junction temperature range: -40 to 125 °C • RthJA, Thermal resistance junction-ambient: 50 °C/W • RthJC, Thermal resistance junction-case: 5 °C/W The Operating junction temperature range, TOP, specifies how hot the “junction”– the active part of the regulator integrated circuit –can be allowed to get before it goes into thermal shutdown. (The thermal shutdown is the internal limit that makes the regulator power “Internally limited”.) For us, that’s a maximum of 125 °C. The thermal resistance junction-ambient RthJA (Often written as ?JA), tells us how hot the junction gets when (1) the regulator is dissipating a given amount of power and (2) the regulator is sitting in open air, at a given ambient temperature. Suppose that we need to design our regulator to only work under modest commercial conditions, that will not exceed 60 °C. If we need to keep the junction temperature under 125 °C, then the maximum temperature rise that we can allow is 65°C. If we have RthJA of 50 °C/W, then the maximum power dissipation that we can allow is 65/50 = 1.3 W, if we are to prevent the regulator from going into thermal shutdown. That’s well below the 4 W that we would expect with a 1 A load current. In fact, we can only tolerate 1.3 W / 4 V = 325 mA of average output current without sending the regulator into thermal shutdown. This, however, is for the case of the TO-220 radiating to ambient air– almost a worst-case situation. If we can add a heat sink or otherwise cool the regulator, we can do much better. The opposite end of the spectrum is given by the other thermal specification: the thermal resistance junction-case, RthJC. This specifies how much temperature difference you can expect between the junction and the outside of the TO-220 package: only 5 °C/W. This is the relevant number if you can quickly remove heat from the package, for example if you have a very good heat sink hooked up to the outside of the TO-220 package. With a big heat sink and perfect coupling to that heat sink, at 4 W, the junction temperature would rise only 20 °C above the temperature of your heat sink. This represents the absolute minimum heating that you can expect under ideal conditions. Depending on the engineering requirements, you can start from this point to build a full power budget, to account for the thermal conductivity of every element of your system, from the regulator itself, to the thermal interface pad between it and the heat sink, to the thermal coupling of the heat sink to the ambient air. You can then verify the couplings and relative temperatures of each component with a spot-reading non-contact infrared thermometer. But often, it’s a better choice to re-evaluate the situation and see if there’s a better way to go about this. For the present situation, one might consider moving to a surface mount regulator that offers better power handling capability (by using the circuit board as a heat sink) or it may be worth looking into adding a power resistor (or zener diode) before the regulator to drop most of the voltage outside the regulator package, easing the load on it. Or better yet, seeing if there’s a way to build your circuit without the lossy linear regulator stage. — AFTERWORD — We have covered the basics of understanding power dissipation in a few simple, dc circuits. The principles that we have gone over are quite general, and can be used to help understand power consumption in most types of passive elements and even most types of integrated circuits. There are real limitations, however, and one could spend a lifetime learning the nuances of power consumption, particularly at lower currents or high frequencies where small losses that we have neglected become important. In ac circuits, many things behave very differently, but the power rule still holds in most circumstances: P(t) = I(t) × V(t) for time-varying current and voltage. And, not all regulators are all that lossy: Switching power supplies can convert (for example) 9 V dc to 5 V dc with 90% or higher efficiency– meaning that with good design, it may only take about 0.6 A at 9 V to produce 5 V at 1 A. But that’s a story for another time. |
Power dissipated by resistor formula
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