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Unformatted text preview: Problem 1.9 Determine the net charge ∆Q that ﬂowed through a resistor over the speciﬁed time interval for each of the following currents: (a) i(t ) = 0.36 A, from t = 0 to t = 3 s (b) i(t ) = [40t + 8] mA, from t = 1 s to t = 12 s (c) i(t ) = 5 sin(4π t ) nA, from t = 0 to t = 0.05 s (d) i(t ) = 12e−0.3t mA, from t = 0 to t = ∞ Solution: (a) ∆Q(0, 3) = (b) ∆Q(1, 12) =
12 12 3 3 i dt =
0 0 0.36 dt = 0.36t 3 = 1.08 0 (C). i dt =
1 1 (40t + 8) dt × 10−3
12 = (c) ∆Q(0, 0.05) =
0.05 40t 2 + 8t 2 × 10−3 = 2.948
1 (C). 0.05 i dt =
0 0 5 sin 4π t dt × 10−9
0.05 = −5 cos 4π t 4π × 10−9
0 = (−0.32 + 0.40) × 10−9 = 80 (d) ∆Q(0, ∞) =
∞ ∞ (pC). i dt =
0 0 12e−0.3t dt × 10−3 = −12e−0.3t 0.3 ∞ × 10−3 = 40
0 (mC). ...
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This note was uploaded on 04/19/2010 for the course EE 220 taught by Professor Rawat during the Spring '09 term at Nevada.
 Spring '09
 Rawat

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