A single-loop circuit consists of a 7.1 resistor, 12.1 H inductor, and a 3.3 F c

Question

A single-loop circuit consists of a 7.1 resistor, 12.1 H inductor, and a 3.3 F capacitor. Initially the capacitor has a charge of 6.1 C and the current is zero. Calculate the charge on the capacitor N complete cycles later for (a) N = 5, (b) N =10, and (c) N = 100.

Chapter 31, Problem 024 A single-loop circuit consists of a 7.1 2 resistor, 12 Hinductor, and a 3.3 UF capacitor. Initially the capacitor has a charge of 6.1 uC and the current is zero. Calculate the charge on the capacitor N complete cycles later for (a) N 5, (b) N 10, and (c) N 100 (a) Number Units (b) Number Units (c) Number Units

Explanation / Answer

let charge on the capacitor at any time t is q and current in the circuit is i.

writing KVL around the loop:

R*i+L*(di/dt)+(1/c)*q=0

using i=dq/dt,

R*(dq/dt)+L*(d^q/dt^2)+(1/C)*q=0

taking laplace transform on both sides:

s*R*Q-R*q0+s^2*L*Q-s*L*q0+(1/C)*Q=0

where Q=laplace transform of q

==>Q*(s^2*L+s*R+(1/C))=q0*(R+s*L)

==>Q=q0*(R+s*L)/(s^2*L+s*R+(1/C))

taking inverse laplace transform,

q=q0*exp(-R*t/(2*L))*((R/w*L)*sin(w*t)+cos(w*t))

where w=sqrt((1/LC)-(R^2/4L^2))

using L=12.1 , C=3.3*10^(-6), R=7.1

we get w=158.252 rad/s

time period=2*pi/w=0.0397 seconds

part a:

charge after N=5 cycles:

time=5*time period=0.1985 seconds

charge=q0*exp(-R*t/(2*L))*((R/w*L)*sin(w*t)+cos(w*t))

with t=0.1985 seconds

charge=5.7548 uC

part b:

at N=10 cycles, time=10*time period=0.397 seconds

charge=q0*exp(-R*t/(2*L))*((R/w*L)*sin(w*t)+cos(w*t))

at t=0.397 seconds

charge=5.4291 uC

part c:

at N=100 cycles, time =100*0.0397=3.97 seconds

charge=1.8996 uC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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