For RC circuit shown here, Kirchhoff\'s voltage loop rule would result in V_0 -

Question

For RC circuit shown here, Kirchhoff's voltage loop rule would result in V_0 - IR - Q/C = 0. If the capacitor initially has a charge Q_0.on it a. time 0 (it is partially pm-charged), the charge on it at later times is given by Q = V_o C (1 - e^1/RC) + Q_o e^1/Rc. This expression is a combination of the capacitor both charging and discharging simultaneously. Find the current in the circuit by differentiating the above expression for charge. SHOW YOUR WORK IN ORDER TO GET CREDIT - "BARE" ANSWERS WILL NOT GET CREDIT. Show that with the expressions for charge as given in the statement of the problem, and the current that you found in part i), that Kirchhoffs loop equation [V_o - IR -Q/C = 0] is satisfied.

Explanation / Answer

i) Q=VoC[1-e-t/RC]

dQ/dt=d/dt(VoC)-d/dt(VoCe-t/RC)

dQ/dt=0-VoC(-1/RC)e-t/RC

ii) Vo-IR-Q/C=0

Vo-(Vo/R(e-t/RC))-VoC/C[1-e-t/RC]=0

Vo-Vo=0

i=Vo/R(e-t/RC)

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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