Process of elimination/deduction
Always start with what you know.
(i) There are three components in the circuit - resistor, diode & capacitor.
(ii) There are three terminals (A,B,C)
(iii) There is only one component connected between each terminal.
There are only two possible unique configurations of this circuit.
(i) Either the anode of the diode is connected to the end of the resistor (and the cathode to the end of the capacitor)
(ii) Or the Cathode of the diode is connected to the end of the resistor (and the anode to the end of the capacitor)
Between any TWO TERMINALS you will have a circuit consisting of the 'component' which is connected to the other two component connected in series with each other and parallel to the component.
By using DC to test the components you (should) know that;
(i) the resistor will conduct the same amount of current in both directions AND the capacitor in the parallel path (once charged) will prevent current flowing through the diode regardless of polarity
(ii) the diode will appear as a short circuit in one direction only AND the (charged) capacitor will prevent current flowing in the RC parallel circuit.
(iii) Applying the test across the capacitor only the parallel circuit (resistor/diode) will conduct (- in one direction it will give the same (or similar) reading as the resistor, in the opposite direction it will show an open circuit (no current).)
Its now down to you to work out which terminal is A, which B and which C and which way around is the diode connected. For extra credit you could also calculate the value of the resistor in ohms.
A capacitor is made of two metallic plates separated by insulating material.
Correct.
During the charging of the capacitor electrons flow towards the opposite direction the battery's electric field.
OK.
Shouldn't the electrons flow through the insulator at a very very slow speed so that means some of the charge, which was supposed to be stored, be lost?
In a perfect insulator there will be no charge flow (current) under static voltage conditions. For a less than perfect insulator there will be a leakage current and the capacitor will lose its charge.
Figure 1. Extract from a random electrolytic capacitor series datasheet.
The datasheet shows that for these large Vishay electrolytics that the leakage current, ILS, is measured after the rated voltage, UR, has been applied for 5 minutes. (We can take it from this that there is something that will change a little with "soakage" time.)
If we look at the first entry, a 10,000 μF, 16 V model the IL is listed as 1.2 mA. The charge on the capacitor is given by \$ Q = C V = 0.01 \ \text F \times 16 \ \text V = 0.16 \ \text C\$.
\$ 1.2 \ \text {mA} = 0.0012\ \text {C/s} \$ so the capacitor is leaking at a rate of \$ \frac {0.0012}{0.16} =
0.0075 \ \text {/s} = 0.75\%\text{/s} \$ while fully charged.
Just by using the UR and IL figures we can calculate the equivalent leakage resistance as \$ R_L = \frac {U_R}{I_L} = \frac {16}{1.2m} = 13.3\ \text k\Omega \$.
simulate this circuit – Schematic created using CircuitLab
Figure 2. Equivalent circuit at 16 V.
Best Answer
Don't use 2N2222, use 2N7000 MOSFETs (that motor must be small) and it will be much easier. Try this:
You will have to tweak the value of R5 to fine tune the timing. D2 is there to avoid C1 discharging through anything that isn't R5, and to charge C1 as fast as possible. The time constant between R5 and C1 determines the turn-off time.
This circuit could be improved by doing a harder switching action on M1 to lower switching losses, but if it is not switched on-off very frequently, it will work fine.
Oh, and dont forget about the flywheel diode D3, the motor inductive spike can be very nasty.