Ch 35 Physics for Scientists and Engineers

# Ch 35 Physics for Scientists and Engineers - AC CIRCUITS...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
35-2 Chapter 35 Phah at f=60ms 35.5. Solve: Visualize: Please refer to Figure 35.4 for an AC resistor circuit. (a) For a circuit with a single resistor, the peak current is (b) The peak current is the same as in part (a) because the current is independent of frequency. 35.6. frequency w. Visualize: Solve: Model: Current and voltage phasors are vectors that rotate counterclockwise around the origin at angular Please refer to Figure (a) From the figure we note that V, = 10 V and I, = 0.50 A. Using Ohm's law, The frequency is 01 1 28 T 0.04 s f =-=-=-- - 25 Hz (c) The voltage and current are vR = VR COSW = (10 V)cos[2~(25 Hz)t] For both the voltage and the current at t = 15 ms, the phase angle is i, = IR COS@ = (0.50 A)cos[2~(25 Hz)~] ux = 2d25 Hz)(15 ms) = 2N0.375) rad = 135" That is, the current and voltage phasors will make an angle of 135" with the starting t = 0 s position. I S='O" ' ql=;.50A I I v\ 0 I Assess: and voltage are in phase. law applies to both the instantaneous and peak currents and voltages. For a resistor, the current
AC Circuits 35-3 35.7. Visualize: Figure 35.7 shows a simple one-capacitor circuit. Solve: (a) The capacitive reactance at o= 2@= 2n(lOO Hz) = 628.3 rad/s is = 5305 R 1 1 x,=-= WC (628.3 rad / s)(0.30 x 10" F) 10.0 v ,,=E= =1.88~10-~ A=1.88mA X, 5.305~10~ C2 (b) The capacitive reactance at [email protected] kHz) = 628,300 rad/s is 1 1 x, =-= = 5.305 R (6.283 x lo5 rad / s)(0.30 x lo4 F) Assess: increase in w , as observed above. Using reactance is just like using resistance in Ohm's law. Because Xc = w-', decreases' with an 35.8. Solve: (a) For a simple one-capacitor circuit, When the frequency is doubled, the new current is I; = W'CV, = (2W)CVC = 2(WCv,) = 21, = 20.0 mA (b) Likewise, when the voltage is doubled, the current doubles to 20.0 mA. (c) When the frequency is halved and the emf is doubled, the current remains the same at 10.0 mA.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.

### Page1 / 22

Ch 35 Physics for Scientists and Engineers - AC CIRCUITS...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online