The Student Room Group

Slowing Down A Bomb

https://isaacphysics.org/questions/slowing_down_bomb_sym?board=3e48c048-992e-4b3d-85f5-878cf66b6120&stage=a_level

For part b of this question, I managed to get the right answer but im quite confused on how the logic for this question works out.
if someone could explain where my thought process is right/wrong it would be greatly appreciated.

I basically first treated both capacitors as one equivalent, so C_eq = 2C
Then pd of the equivalent capacitor V = Q_0/2C = V_0/2
and because of kirchoffs 2nd, each capacitor has V_0/2
so each capacitor has charge Q_0/2
but im pretty sure this happens in an instant? And I dont get how charge of Q_0/2 can travel instantly to the other capacitor- how can it discharge instantly?

Anyways, then thinking back to both capacitors C_eq
alpha*V_0 = V_0/2*e^(-t/R*C_eq)
-ln(2alpha) = t/(2RC)
which got me the right answer.

but if you think about each individual capacitor then you use C instead of C_eq, which gets you a different answer for t?
Reply 1
Original post by mosaurlodon
https://isaacphysics.org/questions/slowing_down_bomb_sym?board=3e48c048-992e-4b3d-85f5-878cf66b6120&stage=a_level
For part b of this question, I managed to get the right answer but im quite confused on how the logic for this question works out.
if someone could explain where my thought process is right/wrong it would be greatly appreciated.
I basically first treated both capacitors as one equivalent, so C_eq = 2C
Then pd of the equivalent capacitor V = Q_0/2C = V_0/2
and because of kirchoffs 2nd, each capacitor has V_0/2
so each capacitor has charge Q_0/2
but im pretty sure this happens in an instant? And I dont get how charge of Q_0/2 can travel instantly to the other capacitor- how can it discharge instantly?
Anyways, then thinking back to both capacitors C_eq
alpha*V_0 = V_0/2*e^(-t/R*C_eq)
-ln(2alpha) = t/(2RC)
which got me the right answer.
but if you think about each individual capacitor then you use C instead of C_eq, which gets you a different answer for t?

In an ideal circuit youd instantaneously have an infinite current in the parallel part which would divide the charge across the two parallel capacitors and similar for the voltage across them. Then they would discharge in the series circuit as normal with an equivalent capacitance 2C. You cant just consider one of the paths in the parallel part, when considering the overall series circuit.

The first part is related to the "famous" two capacitor paradox
https://en.wikipedia.org/wiki/Two_capacitor_paradox
(edited 4 weeks ago)
I see so you have to make both capacitors instantly the same state (discharging) in order to do the maths with the equivalent capacitor, and this happens because theres infinite current as the capacitors have 0 resistance, so the charge is transported instantly.

sorry to bring another q to this but part b of this q is very similar: https://isaacphysics.org/questions/capacitive_touchscreen?board=a2c9be40-53c0-4275-b108-564270adc01f&stage=a_level

In this q you effectively add a capacitor of Delta_C parallel to the capacitor of C, so the equivalent capacitance C_eq = C + Delta_C.
But in this q, you have the pd of the equivalent capacitor as V_0 = Q_0/(C+Delta_C)
so 1/2 V_0 = V_0 *e^(-t/RC_eq)
but you only get the correct answer if you treat C_eq = Delta_C? Instead of C_eq = C + Delta_C, which doesnt really make sense to me.
Reply 3
Original post by mosaurlodon
I see so you have to make both capacitors instantly the same state (discharging) in order to do the maths with the equivalent capacitor, and this happens because theres infinite current as the capacitors have 0 resistance, so the charge is transported instantly.
sorry to bring another q to this but part b of this q is very similar: https://isaacphysics.org/questions/capacitive_touchscreen?board=a2c9be40-53c0-4275-b108-564270adc01f&stage=a_level
In this q you effectively add a capacitor of Delta_C parallel to the capacitor of C, so the equivalent capacitance C_eq = C + Delta_C.
But in this q, you have the pd of the equivalent capacitor as V_0 = Q_0/(C+Delta_C)
so 1/2 V_0 = V_0 *e^(-t/RC_eq)
but you only get the correct answer if you treat C_eq = Delta_C? Instead of C_eq = C + Delta_C, which doesnt really make sense to me.

Not worked it through/checked, but they ask for the change in time and as time is proportional to RC, the change in time will be proportional to Delta_C
So just to check, in the original q, they say that t=0 when the capacitor is added in parallel so Delta_t + t = Delta_t so the change in time is still just Delta_t .
But I dont really get why in the second q, we cant also just say t=0 when the "capacitor is added in parallel" - why is the change in time != time calculated using the method used in the original q, if the start time = 0?
(edited 3 weeks ago)
Reply 5
Original post by mosaurlodon
So just to check, in the original q, they say that t=0 when the capacitor is added in parallel so Delta_t + t = Delta_t so the change in time is still just Delta_t .
But I dont really get why in the second q, we cant also just say t=0 when the "capacitor is added in parallel" - why is the change in time != time calculated using the method used in the original q, if the start time = 0?

Im not sure what youre unsure about. In the first question they ask for the time that a bomb takes to go off when you have two capacitors in parallel, so thats the usual total capacitance in the exponential decay. Its not asking for the change in time compared to just one capacitor. However, the second question asks for the change in time.
I tried to draw diagrams for each question to see if i was missing something obvious
so bomb question:
electric.png

and touchscreen q:
electric 2.png


Lets say t=0, when the capacitors are added in both scenarios....

For the bomb question, the time is just how long it takes from the capacitor being added till pd is reached of alpha*V_0 of BOTH capacitors.

But I still dont really get what "change in time" means - so for the touchscreen q, the 2 capacitors both reach the same voltage at the same time, so it cant be a change in time between them right?

The only thing I can think of for "change in time" is the time between when the capacitor is added in parallel and the pd of V/2 for BOTH capacitors, but the answer only makes you use ONE of them?

But then I cant really think why they would use the phrasing "change in time", when it really just means time until a certain pd, like the bomb q.
(edited 3 weeks ago)
Reply 7
Original post by mosaurlodon
I tried to draw diagrams for each question to see if i was missing something obvious
so bomb question:electric.pngand touchscreen q:electric 2.pngLets say t=0, when the capacitors are added in both scenarios....
For the bomb question, the time is just how long it takes from the capacitor being added till pd is reached of alpha*V_0 of BOTH capacitors.
But I still dont really get what "change in time" means - so for the touchscreen q, the 2 capacitors both reach the same voltage at the same time, so it cant be a change in time between them right?
The only thing I can think of for "change in time" is the time between when the capacitor is added in parallel and the pd of V/2 for BOTH capacitors, but the answer only makes you use ONE of them?
But then I cant really think why they would use the phrasing "change in time", when it really just means time until a certain pd, like the bomb q.

For the touchscreen one, its simply comparing (subtract one from the other) the charging times for

Original system with C

New system with C + DeltaC

so as time is proportional to the (total) capacitance, the change (difference) in time is proportional to the change in capacitance DetaC?
(edited 3 weeks ago)
Ohhhhhhh I think it just clicked.

So usually the starting voltage for the 1 capacitor would be 12V, but with 2 capacitors its 11.6V as in my diagram.
So to get to the same end pd of 6V by discharging, the 2 capacitors will do it slightly quicker.

By working out the exact time difference for a finger you can compare, so any time below that time difference has too low of a capacitance and any time above has too high of a capacitance.

If the time difference is roughly the same, the touchscreen knows there must have been a finger.

The way I did it confused me since I didnt really understand how my answer was a "change" and how Delta_C was proportional to the difference but actually doing the difference calculation makes so much more sense, even though it really is just the same.

Thank you very much :smile:

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