Choosing a time interval | We have been calculating tables of figures using a step-by-step method. We have to choose the time interval carefully. If we make it too small, then it will take lots of steps to get a decent graph. However, if we make it too big, then the whole system becomes unrealistic. For example, if we were to choose a time interval of 1 second, then this would tell us that 250 mC would leave the capacitor in the first second. However, the capacitor only has 20mC on it to start with. So a step size of 1 second gives a ridiculous and unrealistic result. | | | | Measuring discharge | | | On page 24, we saw what happens when a 2 mF capacitor discharges through a 40 W resistor. It took about 50 ms to reach half of its starting charge. The table below has the same calculation that we used on page 24. However, this time, you can change the values of resistance and capacitance – you may also have to change the value of the time interval (see box). Try different values of R and C. What is the half-life in each case? Is there a pattern between half-life and resistance and half-life and capacitance. (Try doubling, tripling, halving etc R or C whilst keeping the other one fixed; this will help you find patterns.) | | | | | | | Picture 5.15 Table of discharge values highlighting what's left after RC | You can download an Excel spreadsheet with this dynamic models. | | | | | |

| | Results | | | You should have found that: - half-life µ R
- half-life µ C
We can say that Half-life µ R.C Making sense of the effects of R and C | **Increasing resistance** If we double the resistance, then the current will always be half as much. This means that it will take twice as long for the same amount of charge to flow. Therefore, it will take twice as long for half the charge to decay away. So doubling the resistance doubles T_{1/2}. **Increasing the capacitance** If we double the capacitance and have the same starting voltage, then there will be twice as much charge to start off with. It will take twice as long for half of it to flow off the plates. So doubling the capacitance doubles T_{1/2}. | | | | | | |

| Picture 5.14 The charge drops to a half in 69% of RC. It drops to 37% in a time of RC.
| | | Time constant | | | The half life is proportional to the product RC. In fact, the half life equals 69% of RC. We call the product RC the **time constant** for the circuit. It has the symbol t: t = RC And rather than measuring the half life (69% of the time constant), we measure how much the capacitor discharges in a time equal to the time constant. This is 37% of the starting value. I.e. instead of measuring the time to fall to a half (as in radioactivity), we measure the time it takes to fall to 37%. **The charge on a capacitor falls to 37% of its starting value for every time constant (equal to RC).** What is special about 37%? It corresponds to 1/e where e is a natural number like p. The natural number e = 2.718 (to 4 significant figures). | | | | |

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