JAMB - Physics (1987 - No. 33)
Two capacitance of 6\(\mu\)F and 8\(\mu\)F are connected in series. What additional capacitance must be connected in series with this combination to give a total of 3\(\mu\)F
3\(\mu\)F
16\(\mu\)F
24\(\mu\)F
30\(\mu\)F
Объяснение
3 = \(\frac{6 \times 8 \times c}{(6 \times 8) + (6 \times c) + (8 \times c}\)
therefore, C = 24\(\mu\)F
therefore, C = 24\(\mu\)F