How can I obtain the inductor current waveform on the oscilloscope? If L=1 henry then integral of voltage across the inductor can give me the current but if the oscilloscope can't take the integral or L is not equal to 1 Henry what can I do in that case ? My circuit diagrams is below. The voltage source is 8V peak-peak and 50Hz square wave.

Any suggestions?

In second case another resistor connected anf probe of signal generator in that case I think the circuit is not s standard parallel RLC anymore any suggestions for that one? Two circuit diagrams added as image.

first and second circuit

I need to do this in real life laboratory. I want to know the oscilloscope connections, techniques etc. but a theoretic explanation would also be nice.

  • 3
    \$\begingroup\$ Are you asking how to do this in theory, on a simulator such as the one you have shown, or in real life? Please edit your question with this new information. \$\endgroup\$
    – TimWescott
    Dec 20, 2021 at 18:07
  • \$\begingroup\$ Real life explanation needed but theoric or simulation explanations also be nice. Thanks! \$\endgroup\$
    – Hawx741
    Dec 20, 2021 at 19:23
  • 2
    \$\begingroup\$ If you want to measure a current, why not use a current probe or measuring a current with a shunt resistor that you know (from the differential equations for RLC) will have negligible effect? These are both fairly standard ways of measuring current when you have an oscilloscope. \$\endgroup\$
    – nanofarad
    Dec 20, 2021 at 19:40

1 Answer 1


Generally speaking, to measure current you would need break the circuit and insert some sort of device that is intended to measure current. A common DMM might work, but you asked about using an oscilloscope.

Oscilloscopes are designed to measure voltage, not current, so there is no way to simply hook up the probes and measure current. Exceptions exist, but they are just clever ways to convert current to a voltage reading. In any case, it is always necessary to convert current to voltage if you want to use a 'scope to measure current.

So-called current probes are available that plug into a 'scope where the ordinary voltage probe would go. They are non-contact in the sense that they consist of a loop of magnetic material that goes around a conductor and uses the Biot-Savart law to generate a voltage that is proportional to the current through the conductor. They work well over a range of frequencies that does not include DC.

Another way is to insert some kind of current to voltage converter in to the circuit where you want to make the measurement. There are sophisticated options, but the most common and least expensive current to voltage converter is a simple resistor. According to Ohm's Law, the voltage across a resistor depends on the current through it: V=IR.

If you want to use an oscilloscope to measure the voltage drop across a resistor, then there are (at least) two things to keep in mind:

  1. The resistor needs to have one end connected to ground so that the 'scope probe can be connected to the other end. That's how 'scopes work. They measure the voltage at the probe relative to ground. A 'scope can be use to measure the voltage across a resistor at some arbitrary part of a circuit not referenced to ground, but it's tricky business and ends up being either imprecise or expensive. So just find a place in the circuit to insert your resistor where one end of it is connected to ground and put your scope probe at the other end. Sometimes there's already a suitable resistor in the circuit that you can use. Unfortunately, that's not the case here.

  2. The other thing to keep in mind is that if you add a resistor to the circuit, then it will alters its behavior. Consequently, you need to choose a resistor small enough that its effect is insignificant. I'll let you decide how small is small enough. Impress your teacher: Use a range of resistors and plot I vs R. In an ideal world, measured I should approach the actual value as R gets smaller and smaller.

  • \$\begingroup\$ Some oscilloscope clip-on current probes also measure DC+AC. \$\endgroup\$
    – qrk
    Dec 20, 2021 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.