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ENGINEERING, PPT 2018 HSC 22d

To operate successfully, a laser diode requires 2.3 V while drawing a current of 25 mA.
 

For the circuit shown, calculate the required resistor value `R` needed to produce the correct operating voltage.   (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

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`268\ text{Ohms}`

Show Worked Solution

`V= V_(\text{total})-V_(\text{led})= 9-2.3= 6.7\ text{V}`

`R= V/I= 6.7\(25 xx 10^(-3))= 268\ text{Ohms}`


♦ Mean mark 43%.

Filed Under: Electricity/Electronics Tagged With: Band 5, smc-3720-20-Circuit diagrams, smc-3720-24-P=VI / V=IR calcs

ENGINEERING, PPT 2019 HSC 23b

An electric scooter is powered by a 12 volt rechargeable battery with a capacity of 18 Ah.

Calculate the energy stored in the battery. Use 1 Wh = 3600 J.   (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

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`777.6\ text{kJ}`

Show Worked Solution

Power = volts × current

12 V × 18 A × 1 h = 216 W in 1 h

`:.` Power stored = 216 Wh

  
1 W = 1 J/s

1 Wh = 3600 J  (1 h = 60 × 60 seconds)

`:.` Energy stored  = 216 Wh × 3600 J  
   = 777 600 J  
   = 777.6 kJ  
♦ Mean mark 46%.

Filed Under: Electricity/Electronics Tagged With: Band 5, smc-3720-24-P=VI / V=IR calcs, smc-3720-40-Electrical motors

ENGINEERING, PPT 2022 HSC 16 MC

The diagram shows a circuit. The diode causes a voltage drop of 0.7 volts.
 

What is the current flowing in this circuit?

  1. 0.35 A
  2. 0.38 A
  3. 0.40 A
  4. 0.51 A
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`A`

Show Worked Solution
`V` `= IR`  
`I` `= V/R= (12-0.7)/32= 11.3/32= 0.35`  

 
`=>A`


Mean mark 56%.

Filed Under: Electricity/Electronics Tagged With: Band 4, smc-3720-20-Circuit diagrams, smc-3720-24-P=VI / V=IR calcs

ENGINEERING, PPT 2021 HSC 22c

An electric skateboard battery is rated at 0.078 watts at an operating voltage of 48 volts.

Calculate the capacity of the battery in amp hours, if it fully discharges in one hour of use. The power (`P`) is given by the formula  `P=V I`.    (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

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`text{Capacity} = 5.85\ text{Ah}`

Show Worked Solution

`text{Power = 0.078 W,  Volts = 48 V}`

`P` `=VI`
`I` `=P/V=0.078/48`

 
`:.\ text{Capacity}=(0.078×3600)/48=5.85\ text{Ah}`


♦ Mean mark 50%.

Filed Under: Electricity/Electronics Tagged With: Band 5, smc-3720-10-Electrical systems, smc-3720-24-P=VI / V=IR calcs

ENGINEERING, PPT 2021 HSC 18 MC

The diagram shows a simplified circuit of an electric scooter.
 

What would be the reading of the ammeter in this circuit?

  1. 3.0 A
  2. 6.0 A
  3. 12.0 A
  4. 26.9 A
Show Answers Only

`C`

Show Worked Solution

Find total resistance (`R_t`):

`1/R_t` `=1/R_1+1/R_2`  
`1/R_t` `= 1/(1+2)+1/3=2/3`  
`R_t` `=3/2 = 1.5\ Ω`  

 
Find current (`I`):

`I=V/R_t=18/1.5=12\ text{A}`

`=>C`

Mean mark 52%.

Filed Under: Electricity/Electronics Tagged With: Band 5, smc-3720-20-Circuit diagrams, smc-3720-24-P=VI / V=IR calcs, smc-3720-40-Electrical motors

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