ANSWER KEY: EC EXERCISE 7

1. The amount of useful output power provided by a device is 1.5 W. It is powered by a 48-V supply with 100 mA of current. How much power is wasted in heat?
P = IE = 0.1 * 48 = 4.8 W
4.8 W – 1.5 W = 3.3 W of wasted power


2. What is the peak voltage of a sine wave that measures 220 VAC rms?
Vpeak = rms/0.707 = 220/0.707 = 311 VAC


3. A 1-mW signal is attenuated at the rate of 5 dB/1,000ft. What is the power level into a receiver that is 6,000 feet from the signal source?
10 * log(value/reference value) = 10 * log(1mW/1mW) = 0 dBm
Since the output level equals the reference level, the value of 1 mW is 0 dBm. The total
loss of the link is:
5 dB/1000ft * 6 = 30 dB loss.
The level into the receiver = 0 dBm – 30 dB = -30 dBm.


4 . A power amplifier has a gain of 20 dB and an input level of 2 volts. Assuming that the input and output impedances are the same, what is the voltage level at the amplifier output?
dB = 20 * log(voltage ratio)
In order to determine the output voltage, solve for the voltage ratio and then multiply the
input level by the ratio to find the output level. Solving for the gain ratio:
1020/20 = 101= 10
Output voltage level is:
input level * gain ratio = 2 V * 10 = 20 V


5. A precision current shunt is measuring 100 millivolts across it. This indicates a current of 25 A. What is the actual resistance of this shunt?
R = E/I = .1 volts/25 amps = 0.004 Ω.


6. A power system is providing 200 VAC at 25 A. The phase angle between current and voltage is 25°. What is the true power used by the system?
Determine the apparent power: P = IE = 25 * 200 = 5000 VA (volt-amps)1
Determine the PF: PF = cos(25°) = .9063

Multiply the PF times the Papp: PTrue = PApp * PF = 5000 * .9063 = 4531 Watts

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