In the circuit of the figure e = 4.50 kV, C = 8.80 mu F_1 = R_2 = R_3 = 1.05 M o

Question

In the circuit of the figure e = 4.50 kV, C = 8.80 mu F_1 = R_2 = R_3 = 1.05 M ohm. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are (a) current i_1 in resistor 1, (b) current i_2 in resistor 2, and (c) current i_3 in resistor 3? At t = infinity (that is, after many time constants), what are (d) i_1 (e) i_2, and (f) i_3? What is the potential difference V_2 across resistor 2 at (g) t = 0 and (h)t = infinity? (a) Number I Units (b) Number I Units (c) Number I Units (d) Number Units (e) Number I Units (f) Number Units (g) Number I Units (h) Number I Units

Explanation / Answer

When completely discharged, the capacitor will behave as a perfectly conducting wire. So, we can assume it to be a normal wire.

Now, going ahead,

Let's assume R=1.05 M ohms, for the time being.

parallel combo of R2 and R3 will be R/2.

in series, net resistance= R+R/2= 3R/2.

Current in circuit= E2/3R

Across R1= E2/3R = 2.857 milliC

across R2= acorss R3= E/3R = 1.428 millC (Half of that across R1, since they are in parallel)

This should solve your parts a), b) and c).

Potential diff. acroos R2= E/4 (g)

at infinte time, capacitor will block the current and behave completely as an insulator. So, just ignore that wire.

Current through R3 will naturally be zero, in this case.

Current via R1 and R2 will be same, equal to E/(R1+R2) = E/2R(since now, they will be in series.)

This answers your d), e) and f) parts.

Potential diff. across R2 will now be half of the volatage, i.e. E/2 (h)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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