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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