rosemariechua
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#1
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#1
https://pastpapers.co/cie/view.php?i..._s18_qp_12.pdf


Hi can someone explain Q15,28,32,34,36 please? Thank you so much!
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Eimmanuel
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#2
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#2
(Original post by rosemariechua)
https://pastpapers.co/cie/view.php?i..._s18_qp_12.pdf


Hi can someone explain Q15,28,32,34,36 please? Thank you so much!
For Q15:
Name:  force_01.JPG
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Resolve the 4 N force that is slanted and this is shown as red line arrows in the diagram.

When the respective red line arrows are translated horizontally and vertically to overlap the vertical 3 N force and horizontal 4 N force as shown as yellow line arrows in the diagram.

The non-overlap part in the horizontal and vertical direction is the components of the resultant force.
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Eimmanuel
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(Original post by rosemariechua)
https://pastpapers.co/cie/view.php?i..._s18_qp_12.pdf


Hi can someone explain Q15,28,32,34,36 please? Thank you so much!
For Q28:

Note X and Y are in phase (make sure you know why). So Y will also travel downwards but the amplitude of Y should be less than 2 m.
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Eimmanuel
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(Original post by rosemariechua)
https://pastpapers.co/cie/view.php?i..._s18_qp_12.pdf


Hi can someone explain Q15,28,32,34,36 please? Thank you so much!
For Q32:

Since the wires are in series with the kettle, you can form an equation using V = IRtotal as the first equation.
The power dissipated in the circuit is P = I2Rtotal and you use this equation to form the second equation. You have 2 unknowns with 2 equations, so you can solve the problem.
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Eimmanuel
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(Original post by rosemariechua)
https://pastpapers.co/cie/view.php?i..._s18_qp_12.pdf


Hi can someone explain Q15,28,32,34,36 please? Thank you so much!
Q34

Name:  circuit_05.jpg
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I attach the above diagram from the following link.

https://www.swtc.edu/Ag_Power/electr...l_circuits.htm

You can work out the current I1 and I2 or see the link.

Next work out the ratio of the resistances with respect to the ratio of current in each resistor.

You can then use similar analysis to attempt Q34.
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rosemariechua
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(Original post by Eimmanuel)
For Q32:

Since the wires are in series with the kettle, you can form an equation using V = IRtotal as the first equation.
The power dissipated in the circuit is P = I2Rtotal and you use this equation to form the second equation. You have 2 unknowns with 2 equations, so you can solve
Hmmmmm... well the thing is I know what todo but IDK which value to sub in .....
What I did was I find the I=10A ... after that step then I m confuse.....
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Eimmanuel
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(Original post by rosemariechua)
Hmmmmm... well the thing is I know what todo but IDK which value to sub in .....
What I did was I find the I=10A ... after that step then I m confuse.....
The current in the kettle is 10 A.
Can you show how do you get the current as 10 A?

You can use Kirchhoff’s voltage rule to write an equation of the potential drop through the circuit to deduce the p.d. across the kettle.
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Eimmanuel
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#8
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(Original post by rosemariechua)
Hmmmmm... well the thing is I know what todo but IDK which value to sub in .....
What I did was I find the I=10A ... after that step then I m confuse.....
(Original post by Eimmanuel)
The current in the kettle is 10 A.
Can you show how do you get the current as 10 A?

You can use Kirchhoff’s voltage rule to write an equation of the potential drop through the circuit to deduce the p.d. across the kettle.
One more thing, you should note that

Rtotal = 1.0 Ω + R
where R is the resistance of kettle.
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rosemariechua
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#9
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(Original post by Eimmanuel)
The current in the kettle is 10 A.
Can you show how do you get the current as 10 A?

You can use Kirchhoff’s voltage rule to write an equation of the potential drop through the circuit to deduce the p.d. across the kettle.
pwer=2.4kW
voltagr=240
so P=VI
2.4k/240=10A
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Eimmanuel
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(Original post by rosemariechua)
pwer=2.4kW
voltagr=240
so P=VI
2.4k/240=10A
Ok, have a look at post #8 or the quote below.

(Original post by Eimmanuel)
One more thing, you should note that


Rtotal = 1.0 Ω + R

where R is the resistance of kettle.

When you multiply the current to the equation of total resistance, what do you get?

How do you interpret IR? and IRtotal?
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rosemariechua
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#11
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(Original post by Eimmanuel)
Ok, have a look at post #8 or the quote below.




When you multiply the current to the equation of total resistance, what do you get?

How do you interpret IR? and IRtotal?
alrite i got it!!! so just find the powerwhich is...
P=I^2R
=10^2(0.50+0.50)
=100W
So 2400W-100W=2300W
to find V
P=VI
2300W=V(10)
V=230V
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Eimmanuel
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(Original post by rosemariechua)
alrite i got it!!! so just find the powerwhich is...
P=I^2R
=10^2(0.50+0.50)
=100W
So 2400W-100W=2300W
to find V
P=VI
2300W=V(10)
V=230V

:clap2::congrats:
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Eimmanuel
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#13
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#13
(Original post by rosemariechua)
alrite i got it!!! so just find the powerwhich is...
P=I^2R
=10^2(0.50+0.50)
=100W
So 2400W-100W=2300W
to find V
P=VI
2300W=V(10)
V=230V
By the way, if you need help for the questions given in this thread, let me know.
https://www.thestudentroom.co.uk/sho....php?t=5665260
As it may be long forgotten as no one was answering.
It may be better you ask in the physics forum.
Last edited by Eimmanuel; 3 years ago
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rosemariechua
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(Original post by Eimmanuel)
By the way, if you need help for the questions given in this thread, let me know.
https://www.thestudentroom.co.uk/sho....php?t=5665260
As it may be long forgotten as no one was answering.
It may be better you ask in the physics forum.
An old-fashioned 60W lamp converts 95% of its energy supply into heat. A 4.0W modern lamp
has the same power output of light as the old-fashioned lamp.
What is the efficiency of the modern lamp?
A 5.0% B 6.7% C 75% D 95%


ans c
can explain ?
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Eimmanuel
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#15
(Original post by rosemariechua)
An old-fashioned 60W lamp converts 95% of its energy supply into heat. A 4.0W modern lamp
has the same power output of light as the old-fashioned lamp.
What is the efficiency of the modern lamp?
A 5.0% B 6.7% C 75% D 95%


ans c
can explain ?
For the old-fashioned lamp, how much power is converted into the useful output? Say x useful power.

For the efficiency of the modern lamp,


 e = \dfrac{x}{4.0} \times 100 \%



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rosemariechua
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#16
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#16
(Original post by Eimmanuel)
For the old-fashioned lamp, how much power is converted into the useful output? Say x useful power.

For the efficiency of the modern lamp,


 e = \dfrac{x}{4.0} \times 100 \%



so what is x?
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Eimmanuel
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#17
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#17
(Original post by rosemariechua)
An old-fashioned 60W lamp converts 95% of its energy supply into heat. A 4.0W modern lamp
has the same power output of light as the old-fashioned lamp.
What is the efficiency of the modern lamp?
A 5.0% B 6.7% C 75% D 95%


ans c
can explain ?
(Original post by rosemariechua)
so what is x?
If 95% of the energy is dissipated as heat, what is the % of useful output?
Last edited by Eimmanuel; 3 years ago
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lillyalexagupp
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#18
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#18
efficiency of 60W lamp is 5% as 95% is heat energy. So only 3W is power output. Therefore power output of 4W lamp is the same 3W as stated by question. 3/4 X100= 75%
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