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iGCSE Physics (0625) 4.5.6 The transformer-Exam Style Questions- New Syllabus

iGCSE Physics Exam Style Question Paper 2 - All Chapters

Question

Cables transmit electrical power from the output of the transformer at a power station to the input of another transformer at a substation.
The power at the output of the transformer at the power station is 400 MW.
Which situation delivers the most power to the input of the transformer at the substation?
▶️ Answer/Explanation
Correct Option: C

Detailed solution:

To deliver maximum power, we must minimize power loss ($P_{loss} = I^{2}R$) in the cables. Using a higher potential difference ($V = 400 \text{ kV}$) reduces the current ($I$) since $P = IV$. Lower current significantly reduces $I^{2}R$ losses. Additionally, resistance ($R$) is inversely proportional to cross-sectional area; thus, cables with a large diameter have lower resistance. Combining the highest voltage ($400 \text{ kV}$) and the lowest resistance (large diameter) ensures the least energy is wasted as heat, delivering the most power to the substation.

Question

The primary voltage supplied to a transformer is $3.0\text{ V}$. The secondary voltage is $5.0\text{ V}$ and the secondary current is $2.0\text{ A}$. The transformer is $100\%$ efficient.

What is the primary current in the transformer?

A. $1.2\text{ A}$
B. $2.0\text{ A}$
C. $3.3\text{ A}$
D. $7.5\text{ A}$
▶️ Answer/Explanation
Correct Option: C

Detailed solution:

For a $100\%$ efficient transformer, the input power equals the output power, expressed by the equation $I_p V_p = I_s V_s$.
Given: primary voltage $V_p = 3.0\text{ V}$, secondary voltage $V_s = 5.0\text{ V}$, and secondary current $I_s = 2.0\text{ A}$.
Rearranging the formula to solve for the primary current: $I_p = \frac{I_s V_s}{V_p}$.
Substituting the values: $I_p = \frac{2.0 \times 5.0}{3.0} = \frac{10}{3.0}$.
This calculation yields $I_p \approx 3.33\text{ A}$, which corresponds to option C.
Thus, the primary current required to maintain the power balance is $3.3\text{ A}$.

Question
A simple transformer contains an iron core. What is the function of the iron core in the transformer?
A. to conduct current from the primary coil to the secondary coil
B. to enable the secondary coil to act as an electromagnet
C. to ensure there is a high voltage between the primary coil and the secondary coil
D. to link the magnetic field in the primary coil to the secondary coil
▶️ Answer/Explanation
Correct Option: D

Detailed solution:

A transformer works on the principle of electromagnetic induction, where an alternating current in the primary coil creates a changing magnetic field.
The iron core is a soft magnetic material that provides a high-permeability path for this magnetic flux.
Its primary purpose is to efficiently “link” or guide the magnetic field lines from the primary coil to the secondary coil.
This ensures that the maximum possible magnetic flux passes through the secondary coil to induce an electromotive force ($e.m.f.$).
It does not conduct electricity between the coils, as the primary and secondary circuits are electrically insulated from each other.
Therefore, option D correctly identifies the core’s role in magnetic linkage.

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