Question

Accuracy is important. Do not round intermediates and check the answer range to make sure you are correct.

Values:

[01] 2.08 [02] 68.47 [03] 0.239 [04] 1173.7 [05] 1403. [06] 53.8

THE CORRECT ANSWERS WILL BE IN THE FOLLOWING RANGES:

8-1a. 140, 220 Hz

8-1b. 40:0, 62:0 mA

8-2a. 70:0, 99:0 V

8-2b. 140:0, 160:0 V

8-2c. 60:0, 99:0 V

8-2d. 60:0, 90:0 V

8-3a. 40:0; 99:0 pF

8-3b. 100; 400 mV

8-3c. 1:00; 3:00 mV

8-4a. 100; 150 V

8-4b. 50:0;90:0 mA

8-4c. 500; 990 mA

8-5a. 100; 300 kW

8-5b. 20:0; 60:0 kW

8-5c. 1:00; 1:50 A

8-5d. 10:0; 30:0 W

A [01] ____-mu F capacitor is connected across an alternating voltage with an rms value of 9.28 V. The rms current in the capacitor is 25.2 mA. (a) What is the source frequency? (b) If the capacitor is replaced by an ideal coil with an inductance of 0.167 H, what is the rms current in the coil? In the figure are shown a resistor R = 42.3ohm , an inductor L = 185 mH. and a capacitor C = 65.7 mu F. An ac source with an rms voltage of 115 V and frequency f = [02] Hz is connected between points 1 and 4 in the figure. Calculate the rms voltages between the points (a) 1 and 2, (b) 2 and 3, (c) 3 and 4, (d) 2 and 4. Warning: The sum of the answers to parts (a), (b), and (c) will not equal 115 V. Also, the sum of the answers to parts (b) and (c) will not equal the answer to part (d). The loop rule does not work in the usual way in ac circuits with capacitors and/or inductors. In the figure is shown a circuit with R = 0.214ohm and L = [03] mH. The AC generator produces 25.3 mu V (rms) at 1160.0 kHz. (a) Find the capacitance C such that the magnitude of its reactance is equal to the magnitude of the inductor's reactance. This brings the circuit into resonance with the frequency of the generator, (b) For this value of C, find the rms voltage between points A and B. Hint: Find the impedance of the circuit. Then find the current, (c) Repeat part (b), if the frequency of the generator is changed to [04] kHz without changing the rallies of V, R. L, and C. Warning: When the frequency changes, so does the impedance of the circuit. Note: This is how a radio tunes into a station. The generator represents an antenna which picks up an AC voltage from radio waves (for example, KSL, which broadcasts at 1160 kHz). When you tune the radio, you are adjusting a capacitor so that an LC circuit is in resonance with the frequency of the wave. This causes the signal to be amplified across the capacitor. Other signals with nearby frequencies are not amplified, so you hear only the one radio station. Consider a transformer with 125 turns of wire in the primary winding and[05] turns of wire in the secondary winding. The secondary winding is connected across a 1540-ohm resistor. The primary winding is connected to an AC generator which produces 10.5 volts (rms). (a) Find the voltage (rms) across the secondary winding, (b) Find the current (rms) through the resistor, (c) Find the current (rms) through the primary winding. An AC power generator produces [06] A (rms) at 3630 V (rms). (a) Find the power produced by this generator, (b) If this power is transmitted through a long-distance power line that has a resistance of 11.3 ohm. find the powrer dissipated as heat in the powrer line. If the voltage of the powrer generator is stepped up to 175 kV by an ideal transformer before being transmitted, find (c) the current in the power line and (d) the power dissipated as heat in the power line. Hint: In an ideal transformer, the power output is equal to the powrer input. Note: As you can see, the power dissipated by a power line is greatly reduced if the powrer is transmitted at high voltage.

Answer 1

8.1 Remember that: Vrms = Irmsc Where: 1 Xc = 2n1fC Then: Vrms = Irms (21fc) Solving for f: f-Irms f = 27CVrms Using the giving data: 25.2 x 10-3 f=21(2.08 x 10-6) (9.28) f = 208 Hz If we replace the capacitor by and inductor, then: Vrms = Irms XL = Irms (2nfl) Solving for Irms Vrms Irms 2011 Using the giving data: 9.28 211 * 208 * 0.167 Irms = 0.0425 A = 42.5 mA

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