Capacitance Correction?

Question

The capacitance of a portion of a circuit is to be reduced from 4800 pF to 3000 pF. What capacitance can be added to the circuit to produce this effect without removing existing circuit elements?

Answer 1

If you connect two capacitors C1 and C2 in series the total capacity C (which is less than either C1 or C2) is:
1/C = 1/C1 + 1/C2.
In your case C2 = 4800 pF, and when you connect C1 in series C = 3000 pF. From there you find:
1/C1 = 1/C - 1/C2 = 1/3000 - 1/4800 = 1/8000.
Therefore you need to add C1 = 8000 pF in series with C2.