ass HW-ETA-11 Begin Date: 11/162017 12:00:00 AM Due Date: 11/22/2017 11:59:00 PM

Question

ass HW-ETA-11 Begin Date: 11/162017 12:00:00 AM Due Date: 11/22/2017 11:59:00 PM End Date: 11/22/2017 11:5900 PM (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 power supply with respect to time The data is shown in the table below malfunctions and produces a time-dependent voltage v) A, 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 time (sec) voltage (V) 1 0.260 2 1.040 3 2.340 4.160 6.500 6 9.360 7 12.74 16.64 9 21.06 26.00 10 Ctheexpertta.com 25% Part (a) Enter an expression for the magnitude of the time-dependent current through the capacitor, in terms of, 25% Part (c) 25% Part (d) The capacit r is designed to fail open at a voltage of 850 V. 2s% Part (b) From the data in the table determine the value of A. in volts per second squared and C If the capacitance is 1.1 F, find the magnitude of the current th rough the capacitor, m microamp eres, at time , How long, in seconds, after the power supply is switched on will the capacitor fail? EAS Grade Summary Deductions 0% 100% Potential sino cotangasi cos) acos inbo Sabmissions Ateinpes rensaining D per attempt) atan acotan de,alledi." arch

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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