8463/1H June 2020

Coach Name: Sir Muhammad Abdullah Shah

2.​​ Figure 3​​ shows an LED torch.

i-​​ The torch contains one LED, one switch and three cells.

Which diagram shows the correct circuit for the torch?

[1 mark]

Tick () one box.

ANSWER

We need to tick that circuit in which the current comes out of the positive terminal of the battery, flows to switch and than through the diode in forward direction. Keeping this in mind, option A is incorrect as one of the cell is wrong way around;​​ as a result, the potential difference of this cell​​ cancels out the potential difference of one of the other cells. Therefore, the total potential difference across the entire circuit is equivalent to p.d. of 1 cell only. Whereas in case of option B, the potential difference is reversed as the cells are connected in other way around, there is no current at all because diode has very high resistance in reverse direction.

 

ii-​​ Write down the equation which links charge flow (Q), current (I) and time (t).​​ 

[1 mark]

ANSWER

Q=I ×t

iii-​​ The torch worked for 14 400 seconds before the cells needed replacing.​​ 

The current in the LED was 50 mA.​​ 

Calculate the total charge flow through the cells.​​ 

[3 marks]

SOLUTION​​ 

Q=I ×t

Don’t forget to change milli ampere (mA) to ampere​​ (A). Take the help from the following table.​​ 

Q=50×10-3 A ×144000

Q=720 C

iv-​​ When replaced, the cells were put into the torch the wrong way around. Explain why the torch did not work.​​ 

[2 marks]

ANSWER

A diode is a component that allows current to go in only one direction. When​​ the potential difference is ‘reversed’, there is no current at all. The diode has a very high resistance in the reverse direction.

v-​​ Write down the equation which links efficiency, total power input and useful power output.​​ 

[1 mark]

ANSWER

efficiency= useful power output total power input

vi-​​ The​​ total power input to the LED was 0.24 W. The efficiency of the LED was 0.75 Calculate the useful power output of the LED.​​ 

[3 marks]

SOLUTION​​ 

efficiency=useful power outputtotal power input

useful power output=efficiency  ×power inout

=0.75 ×0.24

useful power output=0.18 W