SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

PHYSICS, M6 2019 VCE 5-6* MC

A 40 V AC generator and an ideal transformer are used to supply power. The diagram below shows the generator and the transformer supplying 240 V to a resistor with a resistance of 1200 \( \Omega \).
 

Question 5

Which of the following correctly identifies the parts labelled \(\text{X}\) and \(\text{Y}\), and the function of the transformer?

  Part \(\text{X}\) Part \(\text{Y}\) Function of transformer
A. primary coil secondary coil step-down
B. primary coil secondary coil step-up
C. secondary coil primary coil step-down
D.  secondary coil primary coil step-up

 
Question 6

Which one of the following is closest to the current in the primary circuit?

  1. \(0.04\ \text{A}\)
  2. \(0.20\ \text{A}\)
  3. \(1.20\ \text{A}\)
  4. \(1.50\ \text{A}\)
Show Answers Only

\(\text{Question 5:}\ B\)

\(\text{Question 6:}\ C\)

Show Worked Solution

\(\text{Question 5}\)

→ The primary coil is connected to the power supply (i.e. Part \(\text{X}\)).

→ For a step-up transformer, there are more coils in the secondary coil then there are in the primary coil.

\(\Rightarrow B\)
 

\(\text{Question 6}\)

\(I_s=\dfrac{V_s}{r_s}=\dfrac{240}{1200}=0.2\ \text{A}\)

\(\dfrac{I_s}{I_p}\) \(=\dfrac{N_p}{N_s}\)  
\(I_p\) \(=\dfrac{N_s \times I_s}{N_p}\)  
  \(=\dfrac{6000 \times 0.2}{1000}\)  
  \(=1.2\ \text{A}\)  

 

\(\Rightarrow C\)

Mean mark Q6 57%.

PHYSICS, M6 2020 VCE 7 MC

An ideal transformer has an input DC voltage of 240 V, 2000 turns in the primary coil and 80 turns in the secondary coil.

The output voltage is closest to

  1. 0 V
  2. 9.6 V
  3. 6.0 × 10\(^{3}\) V
  4. 3.8 × 10\(^{7}\) V
Show Answers Only

\(A\)

Show Worked Solution

→ Transformers require an AC current to provide the change in flux that generates an EMF in the secondary coil from the primary coil.

→ The question refers to a DC current. There will therefore be no change in flux in the transformer and no output voltage.

\(\Rightarrow A\)

♦♦♦ Mean mark 16%.
COMMENT: Attention! DC current will not work on a transformer.

PHYSICS, M6 2021 VCE 7*

A mobile phone charger uses a step-down transformer to transform 240 V AC mains voltage to 5.0 V. The mobile phone draws a current of 3.0 A while charging. Assume that the transformer is ideal and that all readings are RMS.

Calculate the current drawn from the mains during charging?   (2 marks)

Show Answers Only

\(0.06\ \text{A}\)

Show Worked Solution
\(\text{Power}_{\text{in}}\) \(=\ \text{Power}_{\text{out}}\)  
\(V_p I_p\) \(=V_s I_s\)  
\(I_p\) \(=\dfrac{V_s I_s}{V_p}\)  
  \(=\dfrac{5 \times 3}{240}\)  
  \(=0.06\ \text{A}\)  

PHYSICS, M6 2023 VCE 4*

A transformer is used to provide a low-voltage supply for six outdoor garden globes. The circuit is shown in Figure 3. Assume there is no power loss in the connecting wires. 
 

The input of the transformer is connected to a power supply that provides an AC voltage of 240 V. The globes in the circuit are designed to operate with an AC voltage of 12 V. Each globe is designed to operate with a power of 20 W.

  1. Assuming that the transformer is ideal, calculate the ratio of primary turns to secondary turns of the transformer.  (1 mark)

The globes are turned on.

  1. Calculate the current in the primary coil of the transformer.  (2 marks)
  1. Explain why the input current to the primary coil of the transformer must be AC rather than constant DC for the globes to shine.  (2 marks)

Show Answers Only

a.    \(20:1\)

b.    \(0.5\ \text{A}\)

c.    → A change in flux from the primary coil induces an EMF in the secondary coil.

→ Only an AC current will provide the change in flux required for the transformer to work. 

→ DC currents provide a constant flux and therefore will not work on a transformer.

Show Worked Solution

a.    \(\dfrac{N_p}{N_s}=\dfrac{240}{12}=20:1\)
 

b.    The total power drawn on the secondary side \(=6 \times 20=120\ \text{W}\)

Total power on the primary side is \(120\ \text{W}\).

\(I=\dfrac{P}{V}=\dfrac{120}{240}=0.5\ \text{A}\)

♦ Mean mark (b) 43%.

c.    → A change in flux from the primary coil induces an EMF in the secondary coil.

→ Only an AC current will provide the change in flux required for the transformer to work. 

→ DC currents provide a constant flux and therefore will not work on a transformer.

♦ Mean mark (c) 42%.

PHYSICS, M6 2023 HSC 28

An ideal transformer is connected to a 240 V AC supply. It has 300 turns on the primary coil and 50 turns on the secondary coil.

It is connected in the circuit with two identical light globes, \(X\) and \(Y\), as shown.
 

  1. Calculate the voltage across light globe \(X\) when the switch is open. (2 marks)
  1. Explain why, after the switch has been closed, the current in the primary coil is different from when the switch is open.  (3 marks)

Show Answers Only

a.     40 V

b.    Closed switch  \(\Rightarrow \)  globe \(X\) and \(Y\) are connected in a parallel circuit.

→ Once the switch is closed, the total resistance through the circuit is less than when only light globe \(X\) is in the circuit. 

→ Since  \(V=IR\ \ \Rightarrow\ \ R \propto \dfrac{1}{I} \). Therefore, a decrease in resistance through the circuit leads to an increase in the current.

→ Transformer is ideal, so Power in = Power out ( \(V_{p}I_{p}=V_{s}I_{s}\) ). Hence, an increase in \(I_s\) corresponds to an increase in \(I_p\).

→ Therefore, the current in the primary coil is greater when the switch is closed.

Show Worked Solution

a.    \( \dfrac{V_p}{V_s}\) \(= \dfrac{N_p}{N_s}\)  
  \( V_s\) \(=V_p \times \dfrac{N_s}{N_p}\)  
    \(=240 \times \dfrac{50}{300}\)  
    \(=40\ \text{V} \)  

 
→ The voltage across light globe \(X\) is 40 V.
 

b.    Closed switch  \(\Rightarrow \)  globe \(X\) and \(Y\) are connected in a parallel circuit.

→ Once the switch is closed, the total resistance through the circuit is less than when only light globe \(X\) is in the circuit. 

→ Since  \(V=IR\ \ \Rightarrow\ \ R \propto \dfrac{1}{I} \). Therefore, a decrease in resistance through the circuit leads to an increase in the current.

→ Transformer is ideal, so Power in = Power out ( \(V_{p}I_{p}=V_{s}I_{s}\) ). Hence, an increase in \(I_s\) corresponds to an increase in \(I_p\).

→ Therefore, the current in the primary coil is greater when the switch is closed.

♦ Mean mark (b) 39%.

PHYSICS, M6 EQ-Bank 23

The diagram shows a circuit containing two ideal transformers connected with an ammeter. The current through the load is 5.0 A.
 

What is the reading on the ammeter?   (3 marks)

Show Answers Only

`1.0  text{A}`

Show Worked Solution

→ The total power (`P=VI`) of the circuit is constant. This can be calculated from the data given regarding the load.

`P` `=VI`  
  `=24 xx5.0`  
  `=120\ text{W}`  

 
→ Calculating the voltage in the secondary coil of transformer 1:

`(V_(p))/(V_(s))` `=(N_(p))/(N_(s))`  
`V_(s)` `=(V_(p)N_(s))/(N_(p))`  
  `=(240 xx500)/(1000)`  
  `=120\ text{V}`  

 
→ Calculating the reading on the ammeter:

`P` `=VI`  
`120` `=120 xxI`  
`I` `=1.0  text{A}`  

PHYSICS, M6 2015 HSC 25b

The diagram shows a label on a transformer used in an appliance.

     
Explain why the information provided on the label is not correct. Support your answer with calculations.   (3 marks)

Show Answers Only

Calculating the input power:

`P` `=VI`  
  `=240 xx5`  
  `=1200  text{W}`  

 
Calculating the output power:

`P` `=VI`  
  `=2000 xx1`  
  `=2000  text{W}`  

 
→ The output power is greater than the input power.

→ The label is incorrect as this is inconsistent with the law of conservation of energy.

Show Worked Solution

Calculating the input power:

`P` `=VI`  
  `=240 xx5`  
  `=1200  text{W}`  

 
Calculating the output power:

`P` `=VI`  
  `=2000 xx1`  
  `=2000  text{W}`  

 
→ The output power is greater than the input power.

→ The label is incorrect as this is inconsistent with the law of conservation of energy.


♦ Mean mark 45%.

PHYSICS, M6 2015 HSC 18 MC

The diagram shows an ideal transformer.
 

When the switch is closed, the pointer on the galvanometer deflects.

How could the size of the deflection be increased?

  1. Decrease the number of primary coils
  2. Decrease the number of secondary coils
  3. Replace the iron core with a copper core
  4. Place a resistor in series with the galvanometer
Show Answers Only

`B`

Show Worked Solution

In order to increase the deflection of the galvanometer, the current through the secondary coil must increase. 

`(V_(s))/(V_(p))=(n_(s))/(n_(p))`

`V_(s)=(V_(p)n_(s))/(n_(p))\ \ =>\ \ V_(s) prop n_(s)`

→ Decreasing the number of secondary coils will decrease `V_(s)`
 

`V_(s)I_(s)=V_(p)I_(p)\ \ =>\ \ V_(s) prop (1)/(I_(s))`

→ Decreasing the number of secondary coils will increase `I_(s)`

`=>B`


♦ Mean mark 45%.

PHYSICS, M6 2017 HSC 11 MC

An AC source is connected to a transformer having a primary winding of 900 turns. Connected to the secondary winding of 450 turns is a pair of parallel plates 0.01 m apart.
 

The AC input is shown in the graph.
 

What is the maximum field strength (in `text{V m}^(-1)`) produced between the plates?

  1. `1.7`
  2. `6.8`
  3. `1.7 × 10^4`
  4. `6.8 × 10^4`
Show Answers Only

`C`

Show Worked Solution

Maximum voltage in primary coil = 340 `text{V}`  (see graph)

`(V_(p))/(V_(s))` `=(N_(p))/(N_(s))`  
`(340)/(V_(s))` `=(900)/(450)`  
`V_(s)` `=(450 xx 340)/900`  
  `=170\ text{V}`  

 
→ The maximum voltage in the secondary circuit is 170 V.

`E` `=(V)/(d)`  
  `=(170)/(0.01)`  
  `=1.7 xx10^(4)\ text{V m}^(-1)`  

 
`=>C`

PHYSICS M6 2022 HSC 1 MC

An ideal transformer has 20 turns on the primary coil and an input voltage of 100 V.

How many turns are there on the secondary coil if the output voltage is 400 V ?

  1. 4
  2. 5
  3. 80
  4. 400
Show Answers Only

`C`

Show Worked Solution
`(V_(p))/(V_(s))` `=(N_(p))/(N_(s))`  
`(100)/(400)` `=(20)/(N_(s))`  
`N_(s)` `=(20 xx400)/(100)`  
  `=80`  

 
`=>C`

PHYSICS, M6 2019 HSC 24

A step-up transformer is constructed using a solid iron core. The coils are made using copper wires of different thicknesses as shown.
 

The table shows electrical data for this transformer.
 

  1. Explain how the operation of this transformer remains consistent with the law of conservation of energy. Include a relevant calculation in your answer.   (3 marks)
  1. Explain how TWO modifications to this transformer would improve its efficiency.   (4 marks)
Show Answers Only

a.    The energy input into this transformer is 500 J `text{s}^(-1)`

The energy output is given by:

`P` `=V_(s)I_(s)`  
  `=50xx 9`  
  `=450 ` J `text{s}^(-1)`  

 
This is consistent with the law of conservation of energy as 500 J `text{s}^(-1)` of energy is converted into other energy forms such as heat.
   

b.    → Laminating the iron core prevents large eddy currents from being induced in it.

→ This reduces energy loss in the form of heat, increasing efficiency.

→ Increasing the thickness of the wire in the primary coil reduces its resistance, increasing the transformer’s efficiency.

Show Worked Solution

a.   The energy input into this transformer is  500  J `text{s}^(-1)`

The energy output is given by:

`P` `=V_(s)I_(s)`  
  `=50xx 9`  
  `=450 ` J `text{s}^(-1)`  

 
This is consistent with the law of conservation of energy as 500 J `text{s}^(-1)` of energy is converted into other energy forms such as heat.
   

b.     Modification 1:

→ Laminating the iron core prevents large eddy currents from being induced in it.

→ This reduces energy loss in the form of heat, increasing efficiency.

Modification 2:

→ Increasing the thickness of the wire in the primary coil.

→ This reduces its resistance, increasing the transformer’s efficiency.


♦ Mean mark part 48%.

PHYSICS, M6 2021 HSC 7 MC

In a certain ideal transformer, the current in the secondary coil is four times as large as the current in the primary coil.

Which row of the table correctly identifies the type of transformer and the ratio of turns?
 

Show Answers Only

`C`

Show Worked Solution

Larger current in secondary coil → Lower voltage in secondary coil

→ Step down transformer → Primary coil has more turns

`=>C`


♦ Mean mark 42%.