IB MYP 4-5 Physics- Power transmission- Study Notes - New Syllabus
IB MYP 4-5 Physics-Power transmission- Study Notes
Key Concepts
- Power transmission
Power Transmission
Power Transmission
Power in an Electrical Circuit
Electric power is the rate at which electrical energy is transferred.
\( P = VI \)
- Since \( V = IR \), we can also write:
\( P = I^2 R \) (useful for calculating power loss in wires)
Energy Loss in Transmission
When electricity flows through transmission cables, some energy is lost as heat due to the resistance of the wires.
- Power loss in the cables:
\( P_{loss} = I^2 R \)
- Higher current → greater losses (since loss depends on \( I^2 \)).
Using High Voltage for Transmission
To reduce current, electricity is transmitted at very high voltages (e.g., \( 132 \, kV \), \( 400 \, kV \)).
- If transmitted power is constant:
\( P = VI \implies I = \dfrac{P}{V} \)
- Increasing voltage (\( V \)) decreases current (\( I \)), which reduces \( I^2 R \) losses.
- Transformers are used:
- Step-up transformers increase voltage for transmission → reduce losses.
- Step-down transformers decrease voltage before distribution to homes (for safety).
Real-Life Application
- National power grids transmit electricity at very high voltages to minimize energy losses.
- Voltage is stepped down before reaching households (e.g., 230 V in many countries).
Example:
A power station generates \( 100 \, MW \) of power at \( 10,000 \, V \). It is transmitted through cables with total resistance of \( 5 \, \Omega \). Calculate the power lost in the cables.
▶️ Answer/Explanation
Step 1: Current in the transmission line: \( I = \dfrac{P}{V} = \dfrac{100 \times 10^6}{10,000} = 10,000 \, A \).
Step 2: Power loss in cables: \( P_{loss} = I^2 R = (10,000)^2 \times 5 = 5 \times 10^8 \, W = 500 \, MW \).
Final Answer: The power lost is \(\boxed{500 \, MW}\), which is even greater than the power transmitted! This shows why higher voltages are needed.
Example:
If the same \( 100 \, MW \) of power is transmitted at \( 200,000 \, V \), with the same cable resistance (\( 5 \, \Omega \)), calculate the new power loss.
▶️ Answer/Explanation
Step 1: Current: \( I = \dfrac{100 \times 10^6}{200,000} = 500 \, A \).
Step 2: Power loss: \( P_{loss} = I^2 R = (500)^2 \times 5 = 1.25 \times 10^6 \, W = 1.25 \, MW \).
Final Answer: The power loss is only \(\boxed{1.25 \, MW}\), much smaller compared to 500 MW earlier.
Example:
Explain why step-down transformers are used before electricity enters homes.
▶️ Answer/Explanation
Step 1: High voltages are dangerous for appliances and people.
Step 2: Household appliances are designed for safe voltages (e.g., 230 V or 120 V).
Step 3: Step-down transformers reduce the voltage from transmission level to safe household levels.
Final Answer: Step-down transformers make electricity safe for domestic and industrial use by lowering the voltage.