A capacitor ol capacitance C = 8.5 is initially uncharged. It is connected in se

Question

A capacitor ol capacitance C = 8.5 is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 5.5 kQ, and a battery which provides a potential difference of Part (a) Immediately after the switch is closed, what is the voltage drop Vq across the capacitor? Part (b) Immediately after the switch is closed, what is the voltage drop Vr across the resistor in terms of VB? No Answer Given Submission Detail Totals Part (c) Immediately after the switch is closed, what is the current I through the resistor in terms of Vq and R? No Answer Given Submission Detail Part (c) Immediately after the switch is closed, what is the current I through the resistor in terms of V# and Rl No Answer Given Submission Detail Part (d) Find an expression for the time t1/2 after the switch is closed when the current in the resistor equals half its maximum value. No Answer Given Submission Detail Part (e) Find an expression for the charge Q on the capacitor when the current in the resistor equals one half its maximum value. No Answer Given Submission Detail Part (c) Immediately after the switch is closed, what is the current I through the resistor in terms of Vq and R? No Answer Given Submission Detail Part (c) Immediately after the switch is closed, what is the current I through the resistor in terms of V# and Rl No Answer Given Submission Detail Part (d) Find an expression for the time t1/2 after the switch is closed when the current in the resistor equals half its maximum value. No Answer Given Submission Detail Part (e) Find an expression for the charge Q on the capacitor when the current in the resistor equals one half its maximum value. No Answer Given Submission Detail

Explanation / Answer

a) capacitor does not allow sudden change of voltage.

hence vc at t=0-   =vc at t=0+ =0 v

b) just after switch close capacitor act as a short circuit hence

VR = VB

c) just after switch close capacitor act as a short circuit hence

I= VB/R

d)the expression for current by using formula

i(t)=Ifinal+(Iinitial-Ifinal)e^(-t/RC)

i(t)=0+(55/5.5-0)e^(-t/.04675) mA

2.5=5e^(-t/.04675)

t=.0324sec

e)by using q=cv

v is viltage across capacitor when current flow in resistance is 2.5 mA

hence v=55-2.5mA*5.5k ohm

v=41.25

q=8.5*41.25

q=350.625 micro culom

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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