Confusing Physics question Watch

OJ Emporium
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The question: https://gyazo.com/7e5e59c048341c772bfd944ccdcffa76

The previous sub-question from this was figuring out the resistance from the time constant, which was easy and I got 2702 Ohms. The voltage of that circuit was 9.8 V. The question itself: https://gyazo.com/9f1582f2a8537fc6288ad7a4b8b2da10

Not sure what to do in this question
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Joinedup
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(Original post by OJ Emporium)
The question: https://gyazo.com/7e5e59c048341c772bfd944ccdcffa76

The previous sub-question from this was figuring out the resistance from the time constant, which was easy and I got 2702 Ohms. The voltage of that circuit was 9.8 V. The question itself: https://gyazo.com/9f1582f2a8537fc6288ad7a4b8b2da10

Not sure what to do in this question
The time constant is 1 sec so between t=0 and t=1s the PD will rise from zero to 1-e-1 (approx 63%) of the supply voltage... and between t=1s and t=2s it'll rise by a further 63% of the difference between the PD at t=1s and the supply voltage.

That's how I'd deal with questions about a low number of whole time constants

The sketch will have the shape shown here http://iamtechnical.com/capacitor-charging-graph
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OJ Emporium
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(Original post by Joinedup)
The time constant is 1 sec so between t=0 and t=1s the PD will rise from zero to 1-e-1 (approx 63%) of the supply voltage... and between t=1s and t=2s it'll rise by a further 63% of the difference between the PD at t=1s and the supply voltage.

That's how I'd deal with questions about a low number of whole time constants

The sketch will have the shape shown here http://iamtechnical.com/capacitor-charging-graph
Wait I'm confused, so do I find like 63% of 9.8 V for example and add on to it??
I saw the charging graph too so i just assumed it increases by 63%
Last edited by OJ Emporium; 1 month ago
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Joinedup
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first look at the change between t=0 and t=1

difference between 0V and 9.8V is 9.8V
so between t=0 and t=1 the ΔV=9.8*0.63
since the starting value of V (at t=0) was 0 the value at t=1 is
0+(9.8*0.63) V
so at t=1 V=6.2V

next look at the change between t=1 and t=2

difference between 6.2V and 9.8V is 3.6V
so between t=1 and t=2 the ΔV=3.6*0.63
since the starting value was 6.2V at (t=1) the value at t=2 is
6.2+(3.6*0.63) V
V=8.5V

---
alternate method
The more 'classy' way of doing it is
V=Vbatt(1-e-t/RC)... but I don't think you get given this on your formula book
I think of t/RC as being 'number of time constants'

fwiw 1-e-1 is 0.63 which is of course where the 63% per time constant comes from.
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