FCC Exam Question: 3-15B6
What is the time constant of a circuit having two 220-microfarad capacitors and two 1-megohm resistors all in series?
Explanation: The time constant ($\tau$) of an RC circuit is found by multiplying the total resistance (R) by the total capacitance (C): $\tau = R \times C$. First, calculate the total resistance. When resistors are in series, their values add: Total R = 1 M$\Omega$ + 1 M$\Omega$ = 2 M$\Omega$ (or $2 \times 10^6$ ohms). Next, calculate the total capacitance. When capacitors are in series, their combined capacitance is less than the smallest individual capacitance. For two equal capacitors in series, the total capacitance is half of one capacitor's value: Total C = 220 $\mu$F / 2 = 110 $\mu$F (or $110 \times 10^{-6}$ Farads). Finally, calculate the time constant: $\tau = (2 \times 10^6 \text{ ohms}) \times (110 \times 10^{-6} \text{ Farads}) = 220$ seconds. Therefore, 220 seconds is the correct time constant.
3-58H2
3-79L6
3-72K6
3-74L4
3-11B1
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Includes Elements 1, 3, 6, 7R, 8, and 9.