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Unformatted text preview: A LTERNATING C URRENT Problems 1, 3, 4, 5, 11, 18, 27, 34, 39, 42, 55. 31.1. IDENTIFY: The maximum current is the current amplitude, and it must not ever exceed 1.50 A. SET UP: I rms = I / 2 . I is the current amplitude, the maximum value of the current. EXECUTE: I = 1 . 50A gives I rms = 1 . 50 A 2 = 1 . 06 A . EVALUATE: The current amplitude is larger than the root-mean-square current. 31.3. IDENTIFY and SET UP: Apply Eq.(31.5). EXECUTE: (a) (b) Since the voltage is sinusoidal, the average is zero. EVALUATE: The voltage amplitude is larger than 31.4. IDENTIFY: We want the phase angle for the source voltage relative to the current, and we want the capacitance if we know the current amplitude. SET UP: X C = V I and 1 2 C X fC π = . EXECUTE: (a) o 90 φ = − . The source voltage lags the current by 90°. (b) 60.0 V 11.3 5.30 A C V X I = = = Ω . Solving 1 2 C X fC π = for C gives 4 1 1 1.76 10 F 2 2 (80.0 Hz)(11.3 ) C C fX π π − = = = × Ω ....
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This note was uploaded on 12/18/2011 for the course PHYS 2B 2b taught by Professor Hirsch during the Spring '10 term at UCSD.
- Spring '10