Kirchhoff\'s Law states that the sum of voltage drops around a circuit (loop) is

Question

Kirchhoff's Law states that the sum of voltage drops around a circuit (loop) is zero. We consider a DC circuit in which a capacitor is charging from a battery via a resistor. For a passive RC circuit, in which a capacitance C Farads, a resistor of R Ohms, a switch and a battery of V0 Volts are all connected in series, we obtain the following first-order differential equation, in which q(t) Coulombs is the charge on the capacitor at time t seconds after the switch is closed. R dq(t)/dt + q(t)/C - V0 = 0 Use either the integrating factor method or separation of variables to find the general solution q(t) of this equation. The switch is closed at t = 0, at which time the charge on the capacitor is zero Coulombs. Find the particular solution satisfying this initial condition. The potential difference (voltage) across the capacitor at time t > 0 is given by VC(t) = q(t)/C. The quantity RC has the dimensions of time and is often called the time constant for the circuit. How many time constants does it take for a capacitor to charge to 90% of the applied voltage, V0? Justify your answer.

Explanation / Answer

a) Rdq(t)/dt +q(t)/C - Vo = 0

==> Rdq(t)/dt =Vo - q(t)/C

==> Rdq(t)/dt = [CVo- q(t)]/C

--> dq(t)/(CVo - q(t)) = dt/RC

==> -ln(CVo - q(t)) = t/RC + k1

==> ln(CVo - q(t)) = -t/RC -k1

==> CVo - q(t) = exp(-t/RC - k1)

==> CVo - q(t) = exp(-t/RC)*exp(-k1)

as k1 is constant so exp(-k1) is also a constant let it be equal to K

==> CVo - q(t) = K*exp(-t/RC)

b) at t= 0 q = 0 i.e q(0) = 0

or, CVo - q(0) = K*exp(-0/RC)

==> CVo = K

so CVo - q(t) = CVo*exp(-t/RC)

==> q(t) = CVo(1-exp(-t/RC)

c) Vc(t) = q(t)/C = CVo(1-exp(-t/RC)/C = Vo(1-exp(-t/RC)

0.9 Vo = Vo(1-exp(-t/RC)

==> 0.9 = 1-exp(-t/RC)

==>exp(-t/RC) = 0.1

==> -t/RC = ln(0.1)

==> -t/RC = -2.3

==> t = 2.3*RC

so time taken = 2.3 times of RC

Related Questions

Navigate

• Browse (All)
• Browse K
• Subjects

---

---

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