Investigating how the charge on a capacitor changes with time. This experiment uses the same circuit as the one with the motor, but instead of the motor we need a device whose resistance will remain constant, so we use a resistor. We also want to be able to measure the current flowing as the capacitor discharges, so we have an ammeter in the circuit Charge the capacitor, and then allow it to discharge through the resistor. As it discharges, take readings of current every 20 seconds and fill in the table below. Then plot a graph of current against time. Time (s) Current (  A) XY ‘Flying’ lead A  F  Teacher Note. Use the large separate  F capacitors, normal resistance boxes and orange digital ammeters with mA shunts

Analysis of the experiment. 1. You have seen this shape of graph before. Which physics topic was it associated with ? Radioactive Decay 2. If the graph has the same shape, it may have the same maths. How could you check this ? See if it has a constant ‘half life’.Using very light pencil lines, do this test. It should work These two aspects of physics do have exactly the same maths, but with capacitors, instead of the fraction ‘half’ or 0.5, we use the fraction 0.37 ! This may seem a bit strange but we will see the reason shortly. 3. Now check to see whether the graph has a constant ‘0.37 life’, and find out what the ‘0.37 life’ is in seconds. You can do this in heavier pencil ! 4. This is where we find out why we use the fraction 0.37 for capacitors. Find the value and the units of the quantity you get when you multiply the capacitance and the resistance in this circuit together. Conclusions. 1. The quantity ‘resistance  capacitance’ is a Time ! It is called the ‘Time Constant’,  (Tau).

2. The Time Constant is the time taken for the discharge current to decrease to 0.37 of its original value. Time Current Current at start The capacitor provides no ‘resistance’ at the start, so: Check this for the values obtained by experiment 3. Charge is always proportional to current (Q = It), so the graph for charge against time is exactly the same. Time Charge Charge at start We know that Q = CV, so: The Time Constant also is the time taken for the charge to decrease to 0.37 of its original value.

Understanding where the figure 0.37 actually comes from Consider the circuit while the capacitor is discharging. R C There will be a voltage ‘V’ across the plates which will depend on the charge on the capacitor.... And a current I will flow which will depend on the resistance in the circuit. Combining these two formulae: As the current flows, the charge will decrease according to the formula The problem is that we can’t really use this formula because current changes as charge changes. To solve this, we use a mathematical trick, which says that if we take a short enough time, the current will remain constant during that time. (- because charge is decreasing) Substituting Derivation not required for exam questions, and to be understood by mathematicians only ! Final result needed by everybody ! or This is called a ‘differential equation’

Differential equations make sense when you put them into words, but you can’t put numbers into them and get answers ! Means : The rate at which charge flows off a capacitor is proportional to the amount of charge on it. This is obvious because if there is more charge on the capacitor, there is more repulsion to push charge off ! To ‘solve’ differential equations, you have to do another mathematical trick called integrating and You need to be able to use both these formulae. The ‘e’ is the exponential function which appears in many branches of physics.

Now, what happens when the time ‘t’ is equal to the time constant RC. The formula becomes Now use your calculator to find the value of e -1 So, at t = RC, Charging Capacitors through a fixed resistor. The charging process is slightly more complicated then the discharge process, because, obviously, the charge on the capacitor increases, but the current flowing in the circuit starts high and then decreases so the graphs are as shown below:

Time Current Current at start The capacitor provides no ‘resistance’ at the start, so: As for discharge Time Charge NB. At any time during the charge or discharge process: Supply Voltage = Voltage across Capacitor + Voltage across resistor.

Experiment: Set up the circuit shown below. Be very careful with the connection of the Earth (black) leads from the oscilloscope. They should both be connected to the Earth (green) terminal of the Signal generator. The output from the signal generator to the resistor should be from the ‘low impedance’ (left hand) yellow terminal. Settings should be as follows: Signal Generator:Frequency 150 Hz Amplitude Fully Clockwise. Wave formSquare Wave Oscilloscope:Time Base 1 mS Channel 1 volts/div 5 Channel 2 volts/div 5

To oscilloscope channel 1 input To oscilloscope channel 2 input Signal Generator 47  22  F

2. Get your teacher to check that you have roughly the correct traces on your oscilloscope. 1. Adjust the ‘Y’ position of the two channels so that the output from the signal generator appears in the top half of the screen, and the output from the resistor appears in the bottom half. 3. Sketch what appears on your oscilloscope screen onto the grid below..

4. Explain, in as much detail as you can, the shape of the output from the resistor.

Investigating how the charge on a capacitor changes with time. This experiment uses the same circuit as the one with the motor, but instead of the motor we need a device whose resistance will remain constant, so we use a resistor. We also want to be able to measure the current flowing as the capacitor discharges, so we have an ammeter in the circuit XY ‘Flying’ lead A Charge the capacitor, and then allow it to discharge through the resistor. As it discharges, take readings of current every 20 seconds and fill in the table below. Then plot a graph of current against time. Time (s) Current (  A)  F 

Analysis of the experiment. 1. You have seen this shape of graph before. Which physics topic was it associated with ? 2. If the graph has the same shape, it may have the same maths. How could you check this ? These two aspects of physics do have exactly the same maths, but with capacitors, instead of the fraction ‘half’ or 0.5, we use the fraction 0.37 ! This may seem a bit strange but we will see the reason shortly. 3. Now check to see whether the graph has a constant ‘0.37 life’, and find out what the ‘0.37 life’ is in seconds. You can do this in heavier pencil ! 4. This is where we find out why we use the fraction 0.37 for capacitors. Find the value and the units of the quantity you get when you multiply the capacitance and the resistance in this circuit together. Conclusions. ‘0.37 life’ = R  C =

2. The Time Constant is the time taken for the discharge current to decrease to 0.37 of its original value. Time Current 3. Charge is always proportional to current (Q = It), so the graph for charge against time is exactly the same. Time Charge

Understanding where the figure 0.37 actually comes from Consider the circuit while the capacitor is discharging. R C There will be a voltage ‘V’ across the plates which will depend on the charge on the capacitor.... And a current I will flow which will depend on the resistance in the circuit. Combining these two formulae: As the current flows, the charge will decrease according to the formula The problem is that we can’t really use this formula because current changes as charge changes. To solve this, we use a mathematical trick, which says that if we take a short enough time, the current will remain constant during that time. (- because charge is decreasing) Substituting Derivation not required for exam questions, and to be understood by mathematicians only ! Final result needed by everybody ! or This is called a ‘differential equation’

Differential equations make sense when you put them into words, but you can’t put numbers into them and get answers ! Means : To ‘solve’ differential equations, you have to do another mathematical trick called integrating and You need to be able to use both these formulae. The ‘e’ is the exponential function which appears in many branches of physics.

Now, what happens when the time ‘t’ is equal to the time constant RC. The formula becomes Now use your calculator to find the value of e -1 Charging Capacitors through a fixed resistor. The charging process is slightly more complicated then the discharge process, because, obviously, the charge on the capacitor increases, but the current flowing in the circuit starts high and then decreases so the graphs are as shown below:

Time Current Current at start The capacitor provides no ‘resistance’ at the start, so: As for discharge Time Charge NB. At any time during the charge or discharge process:

Experiment: Set up the circuit shown below. Be very careful with the connection of the Earth (black) leads from the oscilloscope. They should both be connected to the Earth (green) terminal of the Signal generator. The output from the signal generator to the resistor should be from the ‘low impedance’ (left hand) yellow terminal. Settings should be as follows: Signal Generator:Frequency 150 Hz Amplitude Fully Clockwise. Oscilloscope:Wave FormSquare Wave Time Base 1 mS Channel 1 volts/div 5 Channel 2 volts/div 5

To oscilloscope channel 1 input To oscilloscope channel 2 input Signal Generator 47  22  F

2. Get your teacher to check that you have roughly the correct traces on your oscilloscope. 1. Adjust the ‘Y’ position of the two channels so that the output from the signal generator appears in the top half of the screen, and the output from the capacitor appears in the bottom half. 3. Sketch what appears on your oscilloscope screen onto the grid below..

4. Explain, in as much detail as you can, the shape of the output from the capacitor.

Questions Pg 101