Turn on thread page Beta
    • Thread Starter
    Offline

    20
    ReputationRep:
    nI was given a graph of q/t which was an exponential decay I think.
    The question said what was the current at
    (i) t=0s
    and
    (ii) t=15s


    for (i) I just put 0 as I have no idea how to start

    for (ii) I worked out time of half-life(20s) and then subbed into the equation given to get 1.41x10-5A
    the charge at 15s on the graph is 240x10-6 C not sure how to use it
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    (Original post by bruh2132)
    nI was given a graph of q/t which was an exponential decay I think.
    The question said what was the current at
    (i) t=0s
    and
    (ii) t=15s


    for (i) I just put 0 as I have no idea how to start

    for (ii) I worked out time of half-life(20s) and then subbed into the equation given to get 1.41x10-5A
    the charge at 15s on the graph is 240x10-6 C not sure how to use it
    The clue is in the question:

    q/t = instantaneous current. i.e. remember that current is defined as the rate at which charge flows.

    The current at any point on the curve is therefore given by the gradient of q/t.

    For i), draw a tangent to the curve at t=0 and continue the line, noting where it crosses the time axis. The current at t=0 is then simply:

    i_{(t=0)} = \frac{q_{(t=0)}}{t_{(q=0)}}

    Do the same for t=15 seconds.

    At t=o, the point at where the initial discharge gradient crosses the time axis, is known as the time-constant of the discharge as long as it's an exponential decay.
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by uberteknik)
    The clue is in the question:

    q/t = instantaneous current. i.e. remember that current is defined as the rate at which charge flows.

    The current at any point on the curve is therefore given by the gradient of q/t.

    For i), draw a tangent to the curve at t=0 and continue the line, noting where it crosses the time axis. The current at t=0 is then simply:

    i_{(t=0)} = \frac{q_{(t=0)}}{t_{(q=0)}}

    Do the same for t=15 seconds.

    At t=o, the point at where the initial discharge gradient crosses the time axis, is known as the time-constant of the discharge as long as it's an exponential decay.
    Thanks a lot, I wasn't expecting to have to draw stuff, was expecting something more theoretical, thanks a lot though!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 8, 2017

1,015

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.