How To Measure Internal Resistance With A Multimeter

Measuring a lithium-ion cell’s internal resistance is super strait forward.: all you have to do is get three readings and do a bit of math. First, record the cell’s open-circuit voltage with no load attached. Next, apply a known load (or any practical load) and immediately note both the loaded voltage and the current flowing. Finally, subtract the loaded voltage from the open-circuit voltage to find the millivolts of drop, then divide that millivolt drop by the load current, and this will give you the cell’s internal resistance in milliohms.

This single figure succinctly captures all the tiny ohmic losses hiding inside the metal, electrolyte, and connections of your cell. With that number, you can quantify efficiency under real-world loads.

Understanding Inefficiencies and Ohmic Losses

Nothing is 100 % efficient—there are always losses in any energy conversion or transfer. By identifying the processes that create those losses, we can calculate them rather than just wonder where they occur. In the case of extracting energy from a lithium-ion cell, most losses stem from ohmic resistance. Every conductor, even the most “ideal” ones, has some finite resistance determined by its material properties and geometry.

What Determines Resistance?

• Material: Copper and nickel paths of identical dimensions exhibit very different resistances—copper being far more conductive.
  
  
• Geometry: A wide, flat plane of material has much lower resistance than a long, narrow strip of the same metal.
  
  
• Cell Construction: In a battery, resistance arises from the casing, current collectors, electrode materials, electrolyte chemistry, and their interconnections—some in series, others in parallel.
  
  

Ultimately, though, only the total internal resistance matters. We can measure it indirectly with any standard multimeter by comparing the cell’s voltage under no load to its voltage under load.

Preparing to Measure Internal Resistance

Regardless of the method you choose, you must:

1. Measure the open-circuit voltage of the cell (no load attached).
   
   
   
2. Measure the loaded voltage immediately after connecting your load.
   
   
   

The difference between these two readings, divided by the load current, yields the millivolt-per-amp drop, which directly corresponds to milliohms of internal resistance.

Method 1: Known Resistive Load

If you have a purely resistive load of known value, you don’t need to measure current directly—just apply Ohm’s law.

1. Select your load resistor(s).
   
   
   
   • Example: Two 6 Ω resistors in parallel → 3 Ω total.
     
     
     
2. Record the unloaded voltage.
   
   
   
   • Example: 4.100 V.
     
     
     
3. Connect the 3 Ω load and record the loaded voltage.
   
   
   
   • Example: 4.010 V.
     
     
     
4. Calculate current using Ohm’s law
    I=VloadedRload=4.010 V3 Ω=1.337 A I = \frac{V_{\text{loaded}}}{R_{\text{load}}} = \frac{4.010\,\text{V}}{3\,\Omega} = 1.337\,\text{A}I=Rload​Vloaded​​=3Ω4.010V​=1.337A
5. Find the voltage drop
    ΔV=Vopen−Vloaded=4.100 V−4.010 V=0.090 V=90 mV \Delta V = V_{\text{open}} - V_{\text{loaded}} = 4.100\,\text{V} - 4.010\,\text{V} = 0.090\,\text{V} = 90\,\text{mV}ΔV=Vopen​−Vloaded​=4.100V−4.010V=0.090V=90mV
6. Compute millivolts per amp (mV/A):
    ΔVI=90 mV1.337 A≈67.3 mV/A \frac{\Delta V}{I} = \frac{90\,\text{mV}}{1.337\,\text{A}} \approx 67.3\,\text{mV/A}IΔV​=1.337A90mV​≈67.3mV/A
    Since 1 mV/A = 1 mΩ, the cell’s internal resistance is approximately 67.3 mΩ.
   
   
   

Method 2: Measured Current Load

If you lack a known resistive load but can measure current, follow these steps:

1. Connect your load (e.g., a motor or other device) to the cell.
   
   
   
2. Measure the open-circuit voltage.
   
   
   
   • Example: 4.100 V.
     
     
     
3. Under load, simultaneously measure:
   
   
   
   • Loaded voltage (e.g., 4.010 V).
     
     
     
   • Load current with your multimeter in series (e.g., 1.38 A).
     
     
     
4. Calculate voltage drop:
    ΔV=4.100 V−4.010 V=90 mV \Delta V = 4.100\,\text{V} - 4.010\,\text{V} = 90\,\text{mV}ΔV=4.100V−4.010V=90mV
5. Compute millivolts per amp:
    90 mV1.38 A≈65.2 mV/A \frac{90\,\text{mV}}{1.38\,\text{A}} \approx 65.2\,\text{mV/A}1.38A90mV​≈65.2mV/A
    Hence, the internal resistance is 65.2 mΩ.
   
   
   

Interpreting Your Results

• Millivolts per amp = Milliohms. Once you have the mV/A figure, you’re done.
  
  
  
• Efficiency guideline: Stay at or above 90 % efficiency under your expected load. In practical terms, ensure the voltage doesn’t sag by more than 10 % under typical current draw.
  
  
  

Even with a basic $9 multimeter, you can achieve accuracy sufficient for most portable-electronics projects. More expensive meters simply yield tighter tolerances—but the principle remains the same.

Conclusion

Measuring the internal resistance of a lithium-ion cell is straightforward:

1. Capture unloaded vs. loaded voltage.
   
   
   
2. Determine either load resistance or load current.
   
   
   
3. Compute the millivolt drop per amp—your direct measure of internal resistance in milliohms.
   
   
   

By mastering this simple two-step approach, you’ll gain a clear, quantitative insight into where and how much energy is lost inside your battery, empowering smarter design and troubleshooting of portable-power systems.