CIE iGCSE Co-ordinated Sciences-P4.5.6 The transformer- Study Notes- New Syllabus
CIE iGCSE Co-ordinated Sciences-P4.5.6 The transformer – Study Notes
CIE iGCSE Co-ordinated Sciences-P4.5.6 The transformer – Study Notes -CIE iGCSE Co-ordinated Sciences – per latest Syllabus.
Key Concepts:
Supplement
- Describe the construction of a basic transformer with a soft-iron core, as used for voltage transformations
- Use the terms primary, secondary, step-up and step-down
- Recall and use the equation \(\mathrm{\frac{V_p}{V_s} = \frac{N_p}{N_s}}\) where p and s refer to primary and secondary
- Recall and use the equation for 100% efficiency in a transformer \(\mathrm{V_p I_p = V_s I_s} \) where p and s refer to primary and secondary
- Describe the use of transformers in high- voltage transmission of electricity
- Recall and use the equation \(\mathrm{P = I^2 R}\) to explain why power losses in cables are smaller when the voltage is greater
CIE iGCSE Co-Ordinated Sciences-Concise Summary Notes- All Topics
Transformers
A transformer is a device that changes the voltage of an alternating current (a.c.) supply using electromagnetic induction.
Construction of a Basic Transformer
Consists of a soft iron core (easily magnetised and demagnetised, strengthens magnetic field).
- Two separate coils of insulated wire are wound on the core:
- Primary coil: Connected to the input a.c. supply.
- Secondary coil: Connected to the output circuit.
- There is no electrical connection between the coils → energy is transferred by the changing magnetic field in the core.
Important Terms Used
- Step-up transformer: More turns on the secondary coil → increases voltage, decreases current.
- Step-down transformer: Fewer turns on the secondary coil → decreases voltage, increases current.
- Primary coil (p): The coil connected to the input a.c. supply.
- Secondary coil (s): The coil connected to the output circuit.
Transformer Equation
The ratio of voltages is equal to the ratio of turns on the coils:
\(\mathrm{\frac{V_p}{V_s} = \frac{N_p}{N_s}}\)
- \(\mathrm{V_p}\) = voltage across primary coil
- \(\mathrm{V_s}\) = voltage across secondary coil
- \(\mathrm{N_p}\) = number of turns on primary coil
- \(\mathrm{N_s}\) = number of turns on secondary coil
Key Notes:
- A.c. supply is essential → a changing magnetic field is needed to induce e.m.f. in the secondary coil.
- A d.c. supply does not work in a transformer (no changing flux).
| Transformer Type | Coil Turns | Effect |
|---|---|---|
| Step-up | \(\mathrm{N_s > N_p}\) | Increases voltage, decreases current |
| Step-down | \(\mathrm{N_s < N_p}\) | Decreases voltage, increases current |
Example :
A transformer has 500 turns on its primary coil and 100 turns on its secondary coil. The input voltage is 230 V. What is the output voltage?
▶️ Answer/Explanation
Step 1: Use formula: \(\mathrm{\frac{V_p}{V_s} = \frac{N_p}{N_s}}\).
Step 2: Rearrange: \(\mathrm{V_s = V_p \times \frac{N_s}{N_p}}\).
Step 3: Substitute values: \(\mathrm{V_s = 230 \times \frac{100}{500} = 46 \, V}\).
Final Answer: The transformer is a step-down transformer with an output of 46 V.
Transformer Efficiency
An ideal transformer is assumed to have 100% efficiency, meaning no energy is lost as heat, sound, or eddy currents. In this case:
\(\mathrm{V_p I_p = V_s I_s} \)
- \(\mathrm{V_p}\) = primary voltage
- \(\mathrm{I_p}\) = primary current
- \(\mathrm{V_s}\) = secondary voltage
- \(\mathrm{I_s}\) = secondary current
Meaning of the Equation:
- The input power to the primary coil is equal to the output power from the secondary coil.
- If the transformer steps up the voltage (\(\mathrm{V_s > V_p}\)), the current decreases (\(\mathrm{I_s < I_p}\)).
- If the transformer steps down the voltage (\(\mathrm{V_s < V_p}\)), the current increases (\(\mathrm{I_s > I_p}\)).
Practical Note:
- Real transformers are not 100% efficient due to energy losses (heating in coils, eddy currents in the core).
- However, good transformers in power transmission can be over 95% efficient.
| Condition | Effect |
|---|---|
| Step-up Transformer | \(\mathrm{V_s > V_p}\), so \(\mathrm{I_s < I_p}\) |
| Step-down Transformer | \(\mathrm{V_s < V_p}\), so \(\mathrm{I_s > I_p}\) |
Example :
A transformer reduces the voltage from 230 V to 23 V. If the current in the secondary is 4 A, what is the current in the primary (assuming 100% efficiency)?
▶️ Answer/Explanation
Step 1: Use equation: \(\mathrm{V_p I_p = V_s I_s}\).
Step 2: Rearrange: \(\mathrm{I_p = \frac{V_s I_s}{V_p}}\).
Step 3: Substitute: \(\mathrm{I_p = \frac{23 \times 4}{230} = \frac{92}{230} = 0.4 \, A}\).
Final Answer: The primary current is 0.4 A.
Use of Transformers in High-Voltage Transmission
Transformers are used in the National Grid to transmit electricity efficiently over long distances. The aim is to reduce energy losses in the transmission cables.
Why High Voltage is Used:
- When electricity is transmitted, energy is lost as heat in the cables due to the heating effect of current (\(\mathrm{P = I^2 R}\)).
- To reduce these losses, the current must be made as small as possible.
- For the same power (\(\mathrm{P = V \times I}\)), increasing the voltage allows the current to be reduced.
- Therefore, electricity is transmitted at very high voltages (up to 400,000 V) and low current.
Role of Transformers in the National Grid:
- Step-up transformers: Increase the voltage and reduce the current at the power station before transmission → minimizes energy loss in cables.
- Step-down transformers: Reduce the voltage to safer levels for domestic and industrial use:
- ~230 V for homes.
- Higher voltages (e.g. 11,000 V) for some industrial applications.
Advantages:
- Greatly reduces energy wasted as heat in transmission lines.
- Makes long-distance supply of electricity efficient and economical.
Example :
Why is it more efficient to transmit electricity at 275,000 V rather than at 230 V?
▶️ Answer/Explanation
Step 1: For the same power, higher voltage means lower current (\(\mathrm{P = V \times I}\)).
Step 2: Lower current reduces energy loss in cables (\(\mathrm{P_{loss} = I^2 R}\)).
Final Answer: Transmitting at high voltage greatly reduces heating losses in cables, making the system efficient.
Power Losses in Transmission Cables
Equation:
$\mathrm{P_{loss} = I^2 R}$
- \(\mathrm{P_{loss}}\) = power lost as heat in the cables (W)
- \(\mathrm{I}\) = current in the cables (A)
- \(\mathrm{R}\) = resistance of the cables (Ω)
Explanation:
- The resistance of long transmission cables causes electrical energy to be wasted as heat.
- The wasted power is proportional to the square of the current (\(\mathrm{I^2}\)).
- For the same useful power transmitted (\(\mathrm{P = V \times I}\)): – If voltage is low → current is high → large power loss. – If voltage is high → current is small → small power loss.
- This is why the National Grid uses very high voltages (up to 400,000 V) to keep current low and reduce losses.
| Condition | Current | Power Loss (\(I^2R\)) |
|---|---|---|
| Low Voltage Transmission | High Current | Large power losses |
| High Voltage Transmission | Low Current | Small power losses |
Example :
A power station supplies 200 kW through cables of resistance 2 Ω. Compare the power loss if the transmission voltage is (a) 1000 V, (b) 10,000 V.
▶️ Answer/Explanation
Case (a):
\(\mathrm{P = VI \Rightarrow I = P/V = 200,000/1000 = 200 \, A}\).
Power loss = \(\mathrm{I^2R = 200^2 \times 2 = 80,000 \, W = 80 \, kW}\).
Case (b):
\(\mathrm{I = 200,000/10,000 = 20 \, A}\).
Power loss = \(\mathrm{20^2 \times 2 = 800 \, W}\).
Final Answer: At 1000 V, 80 kW is lost (40% of useful power). At 10,000 V, only 0.8 kW is lost (<1%). High voltage greatly reduces power loss.