(16%) Problem 2: An AC power supply is connected across a capacitor with capacit
ID: 1770670 • Letter: #
Question
(16%) Problem 2: An AC power supply is connected across a capacitor with capacitance C. When the power supply is switched on, at time 0, it immediately malfunctions and produces a time-dependent voltage v(t) - Ar2, where A is constant Refer to the table in the figure. The capacitor is initially uncharged Using a multimeter, you collect data for the voltage across the power supply with respect to time. The data is shown in the table below time (sec) voltage (V) 0.260 1.040 2.340 4.160 6.500 9.360 12.74 16.64 21.06 26.00 10 ©theexpertta.com 25% Part (a) Enter an expression for the magnitude of the time-dependent current through the capacitor, in terms oft, A, and C. Grade Summary Deductions Potential i(t)- I 0% 100% Submissions Attempts remaining: 5 % per attempt) detailed view 123 0 BACKSPACE DELCLEAR Submit Hints: 0% deduction per hint. Hints remaining: Feedback: 0%-deduction per feedback. 1 D 25% Part (b) From the data in the table determine the value of A, in volts per second squared D 25% Part (c) If the capacitance is 1.6 F, find the magnitude of the current through the capacitor, in microamperes, at time t = 7 s là 25% Part (d) The capacitor is designed to fail open at a voltage of 550 V. How long, in seconds, after the power supply is switched on will the capacitor fail?Explanation / Answer
a)
We know that,
I = Q/t
Q = CV
I = CV/t ; V(t)= A t^2
I = C A t^2/t = C A t
Hence, I(t) = C A t
b)
v(t) = A t^2
A = v(t)/t^2
A = 0.26/1^2 = 0.26
A = 1.04/2^2 = 0.26
A = 2.34/3^2 = 0.26
Hence, A = 0.26
c)
C = 1.1 uF ; t = 1s
I = C A t
I = 1.1 x 10^-6 x 0.26 x 1 = 0.286 x 10^-6 A
Hence, I = 0.286 micro Amperes
d)
We have,
v(t) = A t^2
t = sqrt(V/A) = sqrt (850/0.26) = 57.18 s
Hence, t = 57.18 sec
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