What is the primary difference between a step-up and a step-down transformer regarding their turns ratio?

In a step-up transformer, the number of turns on the secondary coil ($n_s$) is greater than the primary coil ($n_p$). Conversely, in a step-down transformer, the primary coil has more turns than the secondary coil ($n_p > n_s$).

How does the relationship between voltage and current differ in an ideal transformer?

They are inversely proportional. Because power must be conserved ($V_p I_p = V_s I_s$), an increase in voltage in the secondary coil results in a proportional decrease in current.

Why is high voltage preferred over high current for long-distance power transmission?

Power loss in cables is proportional to the square of the current ($P = I^2 R$). By using high voltage, the current is kept low, which significantly reduces energy lost as heat in the transmission lines.

What common error occurs when setting up the transformer voltage ratio equation?

Students often 'cross-wire' the ratios, such as putting $V_p$ over $V_s$ but $n_s$ over $n_p$. To avoid this, ensure the subscripts match on the top and bottom of both fractions: $\frac{V_p}{V_s} = \frac{n_p}{n_s}$.

Why is it incorrect to assume a transformer can increase the total power output?

This violates the Law of Conservation of Energy. A transformer can only transform energy from one form (high voltage/low current) to another (low voltage/high current); it cannot generate additional energy.

What happens if you try to use a transformer with a Direct Current (DC) source?

The transformer will not work because electromagnetic induction requires a changing magnetic field. DC provides a constant field, so no voltage will be induced in the secondary coil.

Define an 'Ideal Transformer'.

An ideal transformer is a theoretical device that is 100% efficient, meaning all input power is transferred to the output without losses from resistance, magnetic leakage, or eddy currents.

What is the 'Turns Ratio'?

The turns ratio is the ratio of the number of turns in the secondary coil to the number of turns in the primary coil ($n_s/n_p$), which determines the factor by which voltage is stepped up or down.

State the formula for power loss in a wire.

The formula is $P = I^2 R$, where $P$ is power lost in Watts, $I$ is the current in Amperes, and $R$ is the resistance of the wire in Ohms.

What is the mathematical relationship between primary and secondary power in an ideal transformer?

The relationship is $P_p = P_s$, which can be expanded to $V_p I_p = V_s I_s$. This shows that the product of voltage and current is constant across both sides.

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13. Electromagnetic Induction

The Ideal Transformer Equation

Summary

The ideal transformer equation describes the relationship between voltage, current, and the number of turns in a transformer, assuming 100% efficiency and no energy loss. It is the fundamental principle used to step up or step down alternating current (AC) voltages for efficient power transmission and device compatibility.

1. Definition & Core Concepts

An ideal transformer is a theoretical model of a transformer that operates with 100% efficiency, meaning no energy is lost as heat, sound, or magnetic leakage.
The device consists of two coils of wire, the primary coil (input) and the secondary coil (output), wound around a common soft iron core.
The primary function of a transformer is to change the magnitude of an alternating potential difference (voltage) through electromagnetic induction.

Schematic of an ideal transformer showing primary and secondary coils wound around an iron core.

2. The Transformer Voltage Equation

3. Power Conservation & Current

4. Key Distinctions: Step-Up vs. Step-Down

5. Efficiency in Power Transmission

Transformers are critical for the National Grid because they allow electricity to be transmitted at high voltages and low currents.
Power loss in transmission cables is caused by heating due to resistance, calculated as $P_{loss} = I^2 R$ .
By using a step-up transformer to reduce the current ( $I$ ), the power lost as heat in the wires is significantly minimized, making the energy transfer much more efficient.

6. Exam Strategy & Tips