Question 1:
A step-down transformer is connected to an alternating current power supply. Explain the working principle of the transformer.
Answer:
The alternating voltage produces an alternating current in the primary coil.
The alternating current produces a changing magnetic field.
The magnetic field is linked to the secondary coil through the soft iron core.
The changing magnetic field induces an alternating voltage in the secondary coil.
The number of turns of the secondary coil is less than the number of turns of the primary coil.
The voltage across the secondary coil is lower than the voltage across the primary coil.
Therefore, the transformer steps down voltage.
A step-down transformer is connected to an alternating current power supply. Explain the working principle of the transformer.
Answer:
The alternating voltage produces an alternating current in the primary coil.
The alternating current produces a changing magnetic field.
The magnetic field is linked to the secondary coil through the soft iron core.
The changing magnetic field induces an alternating voltage in the secondary coil.
The number of turns of the secondary coil is less than the number of turns of the primary coil.
The voltage across the secondary coil is lower than the voltage across the primary coil.
Therefore, the transformer steps down voltage.
Question 2:
A pupil collects the following information on a transformer:
(a) Calculate the efficiency of the transformer.
(b) Explain two factors that cause the transformer to be non-ideal.
Answer:
(a)
$$ \begin{aligned} &\text { Efficiency, } \eta =\frac{\text { Output power }}{\text { Input power }} \times 100 \% \\ & =\frac{V_{\mathrm{S}} I_{\mathrm{S}}}{V_{\mathrm{P}} I_{\mathrm{P}}} \times 100 \% \\ & =\frac{6 \times 4.80}{120 \times 0.25} \times 100 \% \\ & =96.00 \% \end{aligned} $$
(b) The resistance of the coil causes heating of the coil when a current flows through it. This results in energy loss in the form of heat.
Hysteresis caused by the magnetisation and demagnetisation of the soft iron core results in energy loss in the form of heat.
A pupil collects the following information on a transformer:
(a) Calculate the efficiency of the transformer.
(b) Explain two factors that cause the transformer to be non-ideal.
Answer:
(a)
$$ \begin{aligned} &\text { Efficiency, } \eta =\frac{\text { Output power }}{\text { Input power }} \times 100 \% \\ & =\frac{V_{\mathrm{S}} I_{\mathrm{S}}}{V_{\mathrm{P}} I_{\mathrm{P}}} \times 100 \% \\ & =\frac{6 \times 4.80}{120 \times 0.25} \times 100 \% \\ & =96.00 \% \end{aligned} $$
(b) The resistance of the coil causes heating of the coil when a current flows through it. This results in energy loss in the form of heat.
Hysteresis caused by the magnetisation and demagnetisation of the soft iron core results in energy loss in the form of heat.
Question 3:
Explain how an induction cooker can heat up food in a steel pot.
Answer:
The alternating current in the coil produces a changing magnetic field.
The changing magnetic field induces eddy currents in the base of the pot.
The eddy currents flow along a path of low resistance and heats up the base of the pot.
Explain how an induction cooker can heat up food in a steel pot.
Answer:
The alternating current in the coil produces a changing magnetic field.
The changing magnetic field induces eddy currents in the base of the pot.
The eddy currents flow along a path of low resistance and heats up the base of the pot.
Question 4:
Transformers are used in the electrical energy transmission and distribution system.
State the type of transformer used:
(a) before transmission of electrical energy
(b) at the distribution substation
Answer:
(a) Step-up transformer
(b) Step-down transformer
Transformers are used in the electrical energy transmission and distribution system.
State the type of transformer used:
(a) before transmission of electrical energy
(b) at the distribution substation
Answer:
(a) Step-up transformer
(b) Step-down transformer