A 12.6-uF capacitor is connected through a 0.895-M(Ohm) resistor to a constant p

Question

A 12.6-uF capacitor is connected through a 0.895-M(Ohm) resistor to a constant potential difference of 60.0 V. 1) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. Express your answers using two significant figures. Enter your answers numerically separated by commas. 2) Compute the charging currents at the same instants. Express your answers using two significant figures. Enter your answers numerically separated by commas. *Show work and complete entire problem for full-rating.

Explanation / Answer

The charge = Q = CV

C = 12.4 uF

tau = RC = (0.895)(12.4) = 11.0 sec

To find the charge Q(t), we need to find V(t)

using the basic equation:

V(t) = (60 volts)(1 - 1/e to the t/tau)

For example, at t = 5.0 sec

: - t/tau = - 5.0/11.0 = - 0.4545

e to the -0.4545 = 1.72,

and plugging these into the basic equation above:

V = (60.0)(1 - 1/1.72) = 60(0.42) = 25.2 volts

Q = CV = (12.4 uF)(25.2 V) = 312 uCoulombs

The charge on the capacitor at t= 10, 20, 100 etc can be calculated in the same way.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.