In this example we’ll calculate the power dissipation of a **1 kΩ resistor** with a **constant current of 30 mA** flowing through it, using our science and engineering Python library *UliEngineering*.

In order to install *UliEngineering* (a Python3-only library), run

sudo pip3 install -U UliEngineering

Now we can compute the resistor power dissipation using `power_dissipated_in_resistor_by_current()`

from UliEngineering.EngineerIO import auto_print from UliEngineering.Electronics.Resistors import * # Just compute the value: power = power_dissipated_in_resistor_by_current("1 kΩ", "30 mA") # power = 0.9 # Print value: prints: prints "900 mW" auto_print(power_dissipated_in_resistor_by_current, "1 kΩ", "30 mA")

Since the result is `900 mW`

, you can deduce that you need to use a resistor with a power rating of at least one Watt.

Note that you can pass both numbers (like `0.03`

) or strings (like `30 mA`

or `0.03 A`

) to most *UliEngineering* functions. SI prefixes like `k`

and `M`

are automatically decoded

If you know the voltage across the resistor, you can use `power_dissipated_in_resistor_by_voltage()`

. Let’s assume there is `1V`

dropped across the resistor:

from UliEngineering.EngineerIO import auto_print from UliEngineering.Electronics.Resistors import * # Just compute the value: power = power_dissipated_in_resistor_by_voltage("1 kΩ", "30 mA") # power = 0.001 # Print value: prints: prints "1.00 mW" auto_print(power_dissipated_in_resistor_by_voltage, "1 kΩ", "30 mA")

So in this case the power dissipation is extremely low – only `1.00 mW`

– and won’t matter for most practical applications.

In many cases, you can also pass *NumPy* arrays to *UliEngineering* functions:

from UliEngineering.EngineerIO import format_value from UliEngineering.Electronics.Resistors import * import numpy as np # Compute the value: resistors = np.asarray([100, 500, 1000]) # 100 Ω, 500 Ω, 1 kΩ power = power_dissipated_in_resistor_by_voltage(resistors, "30 mA") # power = 0.001 # power = [9.0e-06 1.8e-06 9.0e-07]