I understand that calculating voltage drop with two or more resistors in series is easy because we can apply ohms law and divide up the supply voltage relative to the resistance of each individual resistor.

I also understand that if we only have one component in a circuit, the voltage drop must be equal to the supply voltage.

What I find difficult to understand is what happens when we have a resistor and some other component (maybe an LED, or a motor , or something else).

How do we know what the voltage drop on each component will be, when there is usually not a resistance value specified for the component?

  • \$\begingroup\$ There are equations which are specific to a component type, which are describing the relationship between voltage drop and the current for this component. \$\endgroup\$
    – Eugene Sh.
    Feb 14, 2019 at 19:58

3 Answers 3


Welcome to electronics! Every challenge in this field starts with this question, or something very similar.

A resistor has a known voltage/current relationship, and different resistors all have the SAME relationship, just with a different scale constant (which we call 'resistance value'). Similarly, a battery, or motor, or capacitor, or transformer, can be characterized by a mathematical model.

Some of these models are complex (for a motor, it has to include inertia, work done on the load, internal wiring resistance, and possibly a half dozen other variables), but all will boil down to equations.

Then, with a system of five components, there are five or more equations, all simultaneously true. This sounds complex (because it is, in more senses than one), but it gets MORE complex when we come to the next step: solve all the equations simultaneously, There are mechanical methods for doing this, and shortcuts, but those are never perfect. The fact is that we can always solve such a system IF it can be built; no real device ever divides by zero, or does anything else that mathematics cannot handle.


It depends on the component, because inductors and capacitors voltages can vary over time. Using basic circuit theory can help to create a model of the physical world and predict what voltage and current for each device. For diodes, you can pull up the datasheet and look at the IV curve (there is a nice tutorial at spark fun that can show you how).

In some cases it can be useful to use a spice package or other circuit software (there is even one included on this site) to model the circuit to find the values. (for example, a motor with a fixed load can be modeled as an inductor and resistor, a power supply modeled as a voltage source)

Sometimes the easiest way is to measure the current and experimentally verify rather than calculate the current voltage relationship.

Usually what I do for LED's is get some way to vary the current, either with a power supply with a constant current, or with power supply with set voltage and a set of resistors.

  • \$\begingroup\$ For diode like things (diodes, LEDs, valves, transistors with some caveats) drawing load lines on a graph can be a useful graphical approach. Old school, but it works... \$\endgroup\$
    – Dan Mills
    Feb 14, 2019 at 22:08

in the mathematics, all of calculations are done in the complex numeric field. But we use a simple notation: (Simple) => 2+6=8 , (Complex) => (2+i0) + (6+i0) = (8+i0) where i is imaginary unit.

In Electronic/Electricity calculations, we can transfer a function in the time domain to a frequency domain.

A resistor have a Complex numeric model like this: just R But Capacitor have a relationship in the time domain equal to this: i=c*dv/dt And this capacitor have a complex function in the frequency domain: 1/SC where S is laplace transformation that is equal to p+jw.

you can calculate your system by ohms law in the frequency domain. this is useful for systems which have electrical motor. But for components like LED, you can determine its limits. For example, RED LED have a voltage drop about 1.8V at 7mA. if you change current from 7 to 14, the voltage will not change dramatically and it will be about previous value.


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