Electronics Information – Test 4

Decibels can be used for determining voltage gains and losses as well as power gains and losses. The formula for this function is:
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:
10^20/20 = 10¹ = 10
Output voltage level is:
input level * gain ratio = 2 V * 10 = 20 V

There are two possible reasons for the no-voltage reading across R4.
• An open circuit between the source and R4
• A short across R4 or a parallel branch—R5

Different types of diodes perform different functions, one of which is rectifier whichconverts AC to pulsating DC

The decibel (db) is the unit used to measure levels in telecommunications. It is based onthe base 10 logarithm of a ratio (Reference Sheet). The ratio can be the ratio of input to output for indicating relative gains and losses in a system. The ratio can also be the ratio of a level relative to an absolute value. This question considers both types of decibels.To determine the power level at the input to the receiver, convert the 1 mW (milliwatt) to dBm (decibels relative to a 1 mW reference) as follows:

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 totalloss of the link is:

5 dB/1000ft * 6 = 30 dB loss.

The level into the receiver = 0 dBm – 30 dB = -30 dBm.

c Transmission lines are typically rated for the amount of capacitance per foot. Given that there are 5,280 feet in one mile, a 1-mile transmission line rated at 25 pF/ft has a total capacitance of:
5,280 * 25 pF = 132,000 pF.
Moving the decimal point three places to the left will provide the answer in nF (nanoFarads), 132 nF.

Inductive reactance and frequency have a direct relationship. As frequency increases, inductive reactance increases. Therefore, in the circuit shown in Figure 1C, the higher frequencies coming from the source will be attenuated more by the inductor than the lower frequencies will. Attenuating higher frequencies while passing lower frequencies is the function of a low pass filter.

A solid state device used to protect equipment against high voltage transients (spikes) must have a very fast turn-on time. The component most commonly used for this purpose is the metal oxide varistor (MOV).

The BTU is equivalent to 1.055 × 103 joules

An electric current is said to exist when there is a net flow of electric charge through a region. Electric charge is carried by charged particles, so an electric current is a flow of charged particles.

The physical band located on a diode indicates the cathode (negative) end of the diode. A diode typically will drop approximately 0.6 volts across it when negative voltage is applied to the cathode and positive voltage is applied to the anode. This is referred to as forward bias.
When measuring voltage across a diode with a multi-meter, the red lead is connected to the anode and the black lead is connected to the cathode. In this configuration, 0.6 volts will be seen when the diode is forward biased. If the test leads are reversed and the diode is forward biased, a reading of –0.6 volts will be seen.

Current and voltage will be out of phase in a reactive circuit. If the circuit is inductive, the voltage will lead the current. If the circuit is capacitive, the current will lead the voltage. Across purely resistive elements in a reactive circuit, the current and voltage will still be in phase. Since the circuit in the question is capacitive, the current will lead the voltage across the capacitive element. However, there will be no phase difference between I and E across the resistor.

The resistance of a conductor is inversely proportional to the amount of cross-sectional area. The greater the cross-sectional area, the more freely electrons are allowed to move, lowering resistance to flow.

The American Wire Gauge (AWG) numbers are based on the number of times a large diameter wire of specific starting size must be pulled through a successive number of reducing dies in order to reach the end wire size. Therefore, high AWG numbers refer to large wires which must be drawn through many reducing dies in order to reach the final diameter, and low AWG numbers refer to small wires which require fewer reductions.
An AWG 22 wire has a greater diameter than an AWG 26 wire. Greater diameter means greater cross-sectional area which means lower resistance, so resistance will increase as wire gauge increases.

Since this circuit was simplified in Question 13, the total resistance (25 Ω) is already known. Use Ohm's law (Reference Sheet) to solve for I which gives the total current flowing through the circuit. Since the resistors are in series, the same amount of current flows through each resistor.

The next step is to solve for E across R1, using Ohm’s law.
I = E/R = 48/25 = 1.92 A
E = IR = 1.92 * 10 = 19.2 V

The circuit in the illustration consists of a combination of series and parallel paths. In order to find the total resistance, the circuit must be simplified. Simplification is the combination of series or parallel branches into a single value of resistance. The process is performed
until there is only one value left.

To combine series values, just add them together. To combine parallel branches in which the values in each branch are the same (as in the illustration), divide the value of one branch
by the number of branches. Simplifying the illustration:

a) Combine parallel branches R4 and R5: 10 Ω/2 = 5 Ω.
b) Combine Step a in series with R3: 5 Ω + 5 Ω = 10 Ω.
c) Combine Step b in parallel with R2: 10 Ω/ 2 = 5 Ω.
d) Combine Step c in series with R1 and R6: 5 Ω + 10 Ω + 10 Ω = 25 Ω.

By knowing any two values of the Voltage, Current or Resistance, Ohm's law can be used to find the third missing value.