Please do a-e. Thanks (17%) Problem 6: A capacitor of capacitance C 3.5 F is ini

Question

Please do a-e. Thanks

(17%) Problem 6: A capacitor of capacitance C 3.5 F is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 13 kQ, and a battery which provides a potential difference of 175 V ©theexpertta.com > 20% Part (a) Immediately after the switch is closed, what is the voltage drop rc, in volts, across the capacitor? Grade Summa Deductions Potential ry 0% 100% 7 8 9 sin0 cotan0asin acos0 atan0 acotan sinhO cosh0 tanh cotanh0 cosO Submissions Attempts remaining: 20 (0%per attempt) detailed view Degrees Radians NO Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. 20% Part (b) Immediately after the switch is closed, what is the voltage drop VR, in volts, across the resistor? 20% Part (c) Immediately after the switch is closed, what is the current I in amperes, through the resistor? 20% Part (d) Find an expression for the time t12 after the switch is closed when the current in the resistor equals half its maximum value - 20% Part (e) What is the charge Q, in microcoulombs, on the capacitor when the current in the resistor equals one half its maximum value

Explanation / Answer

a)

By KVL

VB = VR + VC

VB = IR + Q/C

Qt =CVB[1-e-t/RC]

At t= 0s, Q= 0 C

Hence,

VC = 0 V

b)

VR = VB = 175 V

c)

VR = IR

I= VR /R = 175/13000 = 0.0135 A

d)

It= I0*e-t/RC

It=VB /R*e-t/RC

When It=0.5I0

0.5I0 = I0*e-t/RC

0.5 = e-t/RC

ln0.5 = -t/RC

t=(1/1.4427)*RC

t1/2 = t = (1/1.4427)*(13000*3.5*10^-6) = 31.54ms

e)

Qt =CVB[1-e-t/RC]

Qt = (3.5*10^-6*175)*[1-e-0.031.54/(1300*3.5*10^-6)] = 0.612mC

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