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*Ask Swami Candela: How is Voltage Drop Like I-Squared-R Loss?*

*Ask Swami Candela: How is Voltage Drop Like I-Squared-R Loss?*

July 5, 2015

Dear Swami Candela,

What are the differences and similarities between voltage drop and I-squared-R loss? I know that they are calculating something different, but they seem linked in more ways than they are not.

Signed,

At a Loss

Dear Loss,

The universe connects many dots unbeknownst to us. In this case, voltage drop and losses due to I-squared-R are related by Ohm's law (V = I x R). If start with the power formula (P = V x I) and replace V using Ohm's law, you get P = (I x R) x I, which is the same as I x I x R, or I squared times R. That's the formula for the I-squared-R losses in a system. If you multiply it by the time, then it measures the energy lost to heat.

Voltage drop is straight up Ohm's law: V (drop) = I x R, or in other words, the voltage drop is the amount of current flowing through a circuit or component times the resistance of that circuit or component. Some people are thrown by the word "drop" so if it confuses you, just ignore it. It's only a descriptor.

All circuits are closed loops (if they're on and working). One of the laws of circuits is that the voltages around the loop always add to 0 volts. The supply raises the voltage and the circuit elements drop the voltage. In a perfect world, all the voltage is dropped across the load, but in the real world, the wiring has resistance and therefore, if there is any current, then it causes the voltage to drop in proportion to the current and the resistance. That means less voltage is available at the load. As a consequence, some of the power (I-squared-R of it) is lost due to heat in the wiring. Efficiency engineers refer to is as the I-squared-R losses. Losses also occur due to voltage drop across dimmers, connectors, transformers, inductors, capacitors, and anything else with resistance.

Excessive voltage drop can be caused by a very long run of cable between the supply and the load, or by wiring that is too small in diameter. These will also cause I-squared-r losses due to the resistance of the cable and the current going through it. As a consequence, less voltage will reach the load, so it will do less work. The cable is doing some work converting electricity to heat, which takes away from the work that the load is doing. That's how they are related.

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