Understand how energy transmission works from a fundamental perspective
At a high level, electrical energy in the form of electromagnetic waves zips through the vacuum of space at the speed of light. But here on Earth, we have to shuffle energy along conductive materials such as copper wire. This physical route slows things down by a hefty amount!
TLDR we gotta push it real good!
To effectively discuss and explain energy transfer processes, it’s crucial to grasp these key concepts:
voltage (V = I x R): similar to pressure in a water pipe, voltage can represent how much “push” an electrical charge will have with higher amounts of voltage corresponding to faster traveling times.
current (I): the current rate of travel energy will move through a conductive material at, with higher currents representing a higher rate of “electrical flow”
resistance (R): the forces acting against the “electrical flow”, similar to friction
Note: typically as voltage increases, current increases as well because in most contexts the physical conduit that energy travel along will remain the same unless the material gets replaced (or worn down by long periods of time)
watt (aka power, P = V x I): represents the rate of “pushing”, or work being done, will occur—with higher levels of power having more “push”, a higher rate of “push”, or both.
Note: the relationship of voltage and power can also be represented as:
P = (I × R) × I
P = I^2 × R
I = V / R
current is directly proportionate to resistance (double one and the other halves)
doubling the current quadruples the power (therefore power will always be sensitive to current)
While P = V x I and P = I^2 × R operate in an algebraically equivalently manner:
P = V x I represents the total power transmitted through a system
P = I^2 × R represents the power lost as heat due to resistance, aka Joule heating
joule (aka energy, E = integral of P over time): the total amount of energy that has flowed through the system, with higher joules representing more energy
Step to the left, now step to the right
To provide context for what qualifies as “high” vs. “low” voltage, the average American household uses 110-volt power outlets for most appliances.
Power generation facilities typically produce electricity between 10-25 kV (averging 127 times greater than what powers your toaster).
Newly generated energy often travels 50-300 miles after production, necessitating stepped-up voltage levels for efficient transmission.
Feelings of powerlessness
Transmission lines provide a conduit for energy to travel through, but also resistance. The majority of power loss will be caused by this resistance.
Here’s the fancy workaround trick engineers use to maximize energy transfer:
Since power = voltage * current (I), increasing the voltage will result in a lower current if power remains constant
Power loss responds to current quadratically in power loss = current^2 × resistance , so a decrease in current leads to major efficiency improvement in power loss
example with 1,000,000 W of power, 10 Ω of resistance, and 10,000 V :
using I = P / V to calculate current, 1,000,000 W / 10,000 V = 100 A
using P = I^2 * R to calculate power loss, (100 A)^2 * 10 Ω = 100,000 W of power lost to resistance
example with 1,000,000 W of power, 10 Ω of resistance, and 100,000 V :
using I = P / V to calculate current, 1,000,000 W / 100,000 V = 10 A
using P = I^2 * R to calculate power loss, (10 A)^2 * 10 Ω = 1,000 W of power lost to resistance
import matplotlib.pyplot as plt# Provided definitionsdef calculate_current(power, voltage):return power / voltagedef calculate_power_lost(current, resistance):return (current**2) * resistance# Initializationpower_in_watts =1000000resistance_in_ohms =10low_voltage =10000high_voltage = low_voltage *10# Example 1 - power lost when using low_voltagelow_voltage_current = calculate_current(power_in_watts, low_voltage)power_lost_to_low_voltage = calculate_power_lost(low_voltage_current, resistance_in_ohms)# Example 2 - power lost when using high_voltagehigh_voltage_current = calculate_current(power_in_watts, high_voltage)power_lost_to_high_voltage = calculate_power_lost(high_voltage_current, resistance_in_ohms)# Data for plottingvoltages = [low_voltage, high_voltage]currents = [low_voltage_current, high_voltage_current]power_losses = [power_lost_to_low_voltage, power_lost_to_high_voltage]# Plotting the resultsfig, ax1 = plt.subplots()ax2 = ax1.twinx()ax1.plot(voltages, currents, 'g-', marker='o', label="Current (A)")ax2.plot(voltages, power_losses, 'b-', marker='o', label="Power Loss (W)")ax1.set_xlabel('Voltage (V)')ax1.set_ylabel('Current (A)', color='g')ax2.set_ylabel('Power Loss (W)', color='b')# Adjusting the scale of the power loss axis to show quadratic relationshipax2.set_yscale('log')# Adding an index to the side to show exact valuesfor i, (v, c, p) inenumerate(zip(voltages, currents, power_losses)): ax1.annotate(f'V: {v}\nI: {c:.2f} A\nP_loss: {p:.2e} W', xy=(v, c), xytext=(5, -10-30* i), textcoords='offset points', ha='center', va='center', fontsize=9, color='black', bbox=dict(boxstyle="round,pad=0.3", edgecolor="black", facecolor="lightgray"))# Centering the figure titleplt.title('Quadratic Relationship between Voltage, Current, and Power Loss', pad=20)plt.show()
TLDR a 10x increase in voltage results in a 100x less power lost because less current means less power lost to heat!