In the circuit at the right C = 180 microF, R = 1.70 kohms and E = 50.0V. How lo

Question

In the circuit at the right C = 180 microF, R = 1.70 kohms and E = 50.0V. How long
after switch S is closed will the charge on the capacitor rise to 78% of the maximum
charge that will be stored in it?
Answer: (a) 140 ms. (b) 246 ms. (c) 260 ms. (d) 412 ms. (e) 463 ms. (f) ________ S

Circuit first is Epsilon (battery) then next to it is the capacitor, next is the resistor on top of rectangular circuit then there is a open switch on tghe bottom directly down from the resistor.

please help me solve and please show the formula.

Explanation / Answer

time constant, = RC = 1.7*10^3*180*10^-6 = 0.306 s

using the equation,

charge on the capacitor, Q = Qo*(1- e^(-t/))    where Qo = V*E = 180*10^-6*50 = 0.009 C

so, for 78% charge ie 0.78*Qo, Q = 0.78*Qo = Qo*(1- e^(-t/0.009))

so, t = 136 ms   <---------------answer

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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