I'm having a little trouble understanding how to do this problem. So an oscilloscope measures a battery voltage of 1.5V, and when a 1M(ohm) resistor is put in series, the oscilloscope only measures 0.75V.

I'm a little thrown by the term 'input resistance'. Is this the same as input impedance here, or something else?


You are measuring DC (the frequency is 0Hz). So the input resistance is equal to the impedance in this case.

You are forming a resistive voltage divider. You know the voltage of the source, you know the value of one resistor and you know the measured voltage. That is all you need to calculate the input impedance of your scope (or the impedance of the scope+probe).


simulate this circuit – Schematic created using CircuitLab


Input resistance is just that, the DC resistance, in most scopes it is 1 Mohm. When mentioning resistance we ignore anything which is not (or does not behave like) a resistor. So capacitors and inductors are ignored as these are reactive elements.

As Chupacabras writes in his answer, resistance is for the behavior at DC, so the signal frequency is zero.

Input impedance also takes non-resistive / reactive components into account like input capacitance. So if I would say "the input impedance of this oscilloscope is 1 M ohm" I sort of imply that there is no (zero) input capacitance or that it is so small that we can ignore it.

It would be more accurate to say "the input impedance of this oscilloscope is 1 M ohm in parallel with 5 pF"* then for any frequency you can determine what the input impedance will be.

  • \$\begingroup\$ There is a reason that high-resistance probes needs such awkward compensation circuitry :) The actual reactance of a 10:1 probe, by the way, can be LOWER than that of a properly terminated 50 Ohm 10:1 once you go near VHF territory.... \$\endgroup\$ – rackandboneman Apr 12 '17 at 9:06

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