FCC Exam Question: 6A20
C = capacity in farads. Q = the measure of the quantity of charge of electricity in Coulombs. E = theapplied voltage. So Q = CE:
Explanation: The formula Q = CE defines the fundamental relationship between charge, capacitance, and voltage in a capacitor. Here, 'Q' represents the **quantity of electrical charge** stored in the capacitor, measured in Coulombs. 'C' is the capacitance in Farads, and 'E' is the applied voltage in Volts. Therefore, Q = CE directly **determines the quantity of charge present in a capacitor** when a specific voltage is applied across it. Option B is incorrect because "the Q of a circuit" refers to the **Quality Factor** (a different 'Q'), which is a measure of a circuit's selectivity or resonance sharpness, especially in resonant RLC circuits. This is a distinct concept from the quantity of charge stored in a capacitor.
6A131
6A216
6A290
6A614
6A346
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Includes Elements 1, 3, 6, 7R, 8, and 9.