You are Here: Home >< A-levels

# easy question, reply needed urgently watch

1. here's the ms and the question, so i need to know how to draw the graph for the last part, i.e. shape and axis intercepts
thanks
2. (Original post by gorilla_baby)
here's the ms and the question, so i need to know how to draw the graph for the last part, i.e. shape and axis intercepts
thanks
I don't see them...
3. oops, sorry
here they are, in the same doc
Attached Files
4. A circuit is set up as shown in the diagram.doc (36.0 KB, 111 views)
5. (Original post by Muppety_Kid)
I don't see them...
i posted them above
6. (Original post by gorilla_baby)
here's the ms and the question, so i need to know how to draw the graph for the last part, i.e. shape and axis intercepts
thanks
I think you should probably post this in the physics section here

7. You need to find the "half life" of the voltage , i.e the time taken for the voltage to halve , and use the formula t(1/2) = ln2.T to find time constant T.

The second part is straightforward but you have to first find the resistance given the initial amperage using Ohm's Law.
8. ^^ For that part, can't you find T by taking 1/e of original value? i.e. (9V * 1/e) = time constant, cause it is an exponential graph. The NAS book does that for I = Io e^ (-t / RC) ... but since V is directly proportional to I, we can take time constant using a V -t graph too, right? I think the MS did include it.

My method is basically as I described above - find (1/e) * 9V, find the time taken for the voltage to fall from 9V to that value. That's the time constant.

Next part : When Io = 0.19 * 10^-3 A. Io = (Vbattery)/Resistance, so you can find the resistance. And since t = RC (time constant = resistance * capacitance), C = t/R.

3rd part : As the potential difference across the resistor is decreasing, the potential difference across the capacitor is increasing. So its increasing but in a curve-way and it gets less steep (as the MS says). Not sure why it has to end at 7.5V / 15 s though!
9. (Original post by rohitkhannak)

3rd part : As the potential difference across the resistor is decreasing, the potential difference across the capacitor is increasing. So its increasing but in a curve-way and it gets less steep (as the MS says). Not sure why it has to end at 7.5V / 15 s though!
yea, that's what i thought, and i also dont get the 7.5V/15s thing??!!

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 8, 2008
Today on TSR

### Cambridge interviews

Find out which colleges are sending invitations

### University open days

• University of East Anglia
Fri, 23 Nov '18
• Norwich University of the Arts
Fri, 23 Nov '18
• Edge Hill University
Sat, 24 Nov '18
Poll

## All the essentials

### Student life: what to expect

What it's really like going to uni

### Essay expert

Learn to write like a pro with our ultimate essay guide.

### Create a study plan

Get your head around what you need to do and when with the study planner tool.

### Resources by subject

Everything from mind maps to class notes.

### Study tips from A* students

Students who got top grades in their A-levels share their secrets