An ideal transformer in a power supply has a primary voltage of 230 V and a secondary voltage of 9.0 V. The secondary coil supplies a current of 1.5 A to a ...

(a) Meaning of an ideal transformer [1]

An ideal transformer is one that is 100% efficient. It transfers all the electrical power from the primary coil to the secondary coil with no energy losses whatsoever. [1]

(b) Power output [2]

Power is calculated using \( P = V \times I \):

\( P = V_s \times I_s = 9.0 \times 1.5 \) [1]

P = 13.5 W [1]

(c) Relationship between power input and power output [1]

For an ideal transformer:

Power input = Power output

\( V_p I_p = V_s I_s \) [1]

Since there are no energy losses, every joule of electrical energy entering the primary coil exits the secondary coil.

(d) Current in the primary coil [2]

Since the transformer is ideal, power input = power output = 13.5 W.

\( I_p = \frac{P}{V_p} = \frac{13.5}{230} \) [1]

I_p = 0.059 A (or 59 mA) [1]

Notice how the primary current is much smaller than the secondary current. A step-down transformer reduces voltage but increases current on the secondary side, while the primary side draws a small current at high voltage.

(e) Why the real primary current would be slightly larger [2]

In practice, transformers are not 100% efficient. Energy is lost as heat due to the resistance of the copper coils (\( I^2 R \) losses) and eddy currents induced in the iron core. [1] Because some energy is wasted, the input power must be greater than the output power to compensate. Since \( P = VI \) and the primary voltage is fixed at 230 V, a larger power input requires a larger primary current. [1]