All practice questions

Question 961Question →

A benchtop DC power supply delivers $45\text{ V}$ across a fixed resistor in a test circuit. A series ammeter measures a steady current flow of $5\text{ A}$ through the component. What is the electrical resistance of the resistor?

225\ \Omega

9\ \Omega

40\ \Omega

90\ \Omega

Show answer & explanation

Answer: $9\ \Omega$

Answer

9\ \Omega

According to Ohm's Law, electrical resistance is equal to the voltage across a component divided by the current passing through it (

R = V / I

). Substituting

45\text{ V}

and

5\text{ A}

yields

R = 45 / 5 = 9\ \Omega

.

Step-by-Step Solution

1

Identify the given circuit parameters and standard Ohm's Law formula

Voltage

V = 45\text{ V}

, Current

I = 5\text{ A}

, Ohm's Law:

V = I \times R

Ohm's Law defines the relationship between potential difference, current flow, and resistance in a DC circuit.

2

Rearrange Ohm's Law to solve for resistance (

R

)

R = \frac{V}{I}

Dividing both sides of

V = I \times R

by current isolates the unknown resistance.

3

Substitute the given values into the formula and evaluate

R = \frac{45\text{ V}}{5\text{ A}} = 9\ \Omega

Dividing 45 by 5 yields the exact resistance value in ohms.

Key Concept

Ohm's Law Resistance Calculation (

R = V / I

)

Estimated Time:45s

Question 962Question →

An industrial control panel utilizes a direct-current bus to power three parallel branches. Branch 1 contains a $15\ \Omega$ resistor, Branch 2 contains a $30\ \Omega$ resistor, and Branch 3 contains a $10\ \Omega$ resistor. If an ammeter connected in Branch 2 measures a current of $3\text{ A}$ , what is the total current delivered by the power supply to the entire circuit?

4.5\text{ A}

6.0\text{ A}

9.0\text{ A}

18.0\text{ A}

Show answer & explanation

Answer: $18.0\text{ A}$

Answer

18.0\text{ A}

The supply voltage across the parallel circuit is constant. Using Branch 2,

V = 3\text{ A} \times 30\ \Omega = 90\text{ V}

. The remaining branch currents are

I_1 = 90\text{ V} / 15\ \Omega = 6\text{ A}

and

I_3 = 90\text{ V} / 10\ \Omega = 9\text{ A}

. Summing all three branch currents yields

I_{\text{total}} = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

.

Step-by-Step Solution

1

Calculate the common supply voltage across the parallel circuit using Ohm's Law on Branch 2.

V = I_2 \times R_2 = 3\text{ A} \times 30\ \Omega = 90\text{ V}

In a parallel circuit, the voltage drop across every individual branch is equal to the total supply voltage.

2

Determine the current flowing through Branch 1 and Branch 3 using the common voltage of

90\text{ V}

.

I_1 = \frac{90\text{ V}}{15\ \Omega} = 6\text{ A}

and

I_3 = \frac{90\text{ V}}{10\ \Omega} = 9\text{ A}

Current in each individual branch is determined by dividing the constant branch voltage by that branch's resistance.

3

Apply Kirchhoff's Current Law to sum the currents of all parallel branches.

I_{\text{total}} = I_1 + I_2 + I_3 = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

The total current leaving the power supply equals the sum of the individual branch currents in a parallel circuit.

Key Concept

In a parallel circuit, voltage remains constant across all branches (

V_{\text{total}} = V_1 = V_2 = V_3

), and total circuit current is the sum of all individual branch currents (

I_{\text{total}} = I_1 + I_2 + I_3

).

Alternative Method

Calculate total equivalent resistance first:

\frac{1}{R_{\text{eq}}} = \frac{1}{15\ \Omega} + \frac{1}{30\ \Omega} + \frac{1}{10\ \Omega} = \frac{2 + 1 + 3}{30} = \frac{6}{30} = \frac{1}{5\ \Omega}

, giving

R_{\text{eq}} = 5\ \Omega

. With

V = 90\text{ V}

, total current is

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{90\text{ V}}{5\ \Omega} = 18\text{ A}

.

Estimated Time:1m 30s

Question 963Question →

While troubleshooting a direct-current ( $12\text{ V}$ ) automotive auxiliary circuit, a technician observes that the branch fuse blows immediately every time the power switch is closed. Which of the following electrical faults is the most probable cause of this issue?

An open circuit in the wiring leading to the load

A short circuit directly to ground before the load

High contact resistance due to a corroded connector terminal

Reversed polarity connection at the power source

Show answer & explanation

Answer: A short circuit directly to ground before the load

Answer

A short circuit directly to ground before the load is the most probable cause because it creates a near-zero resistance path that draws excessive current, exceeding the fuse rating.

A short circuit directly to ground creates a zero- or near-zero-resistance bypass around the intended load. According to Ohm's law (

I = \frac{V}{R}

), as resistance approaches zero, the circuit current rapidly exceeds the protective fuse's current rating, instantly melting the fuse link when the switch is closed.

Step-by-Step Solution

1

Analyze the observed symptom and the operational role of a fuse.

A fuse is an overcurrent protection device designed to melt and open when current exceeds its specified ampere rating.

Understanding why fuses blow establishes that the root cause must be excessive current draw.

2

Apply Ohm's law (

I = \frac{V}{R}

) to evaluate how each potential fault affects circuit current.

To produce massive current (

I

) from a constant

12\text{ V}

source, the total resistance (

R

) must drop close to

0\,\Omega

.

Ohm's law dictates an inverse relationship between resistance and current.

3

Identify the circuit condition that causes near-zero resistance.

A short circuit bypasses the normal load resistance directly to ground, causing an instantaneous current spike that blows the fuse.

Open circuits and high-resistance connections reduce or eliminate current flow rather than increasing it.

Key Concept

A short circuit provides an unintended low-resistance bypass path that causes current to surge above safe levels, blowing protective fuses.

Estimated Time:45s

Question 964Question →

A step-up transformer utilized in an industrial ignition system has $80\text{ turns}$ on its primary coil and $1{}200\text{ turns}$ on its secondary coil. When the primary coil is energized by a $16\text{ V AC}$ source, the secondary circuit delivers a current of $40\text{ mA}$ to its load. Assuming an ideal transformer, what is the current drawn by the primary coil?

2.67\text{ mA}

600\text{ mA}

0.6\text{ mA}

40\text{ mA}

Show answer & explanation

Answer: $600\text{ mA}$

Answer

600\text{ mA}

In an ideal transformer, power is conserved (

P_p = P_s

), which means voltage and current vary inversely with the turns ratio. For this transformer, the secondary-to-primary turns ratio is

\frac{1{}200}{80} = 15

. While voltage steps up by a factor of 15, current steps down by a factor of 15. Therefore, the primary current must be 15 times the secondary current:

I_p = 40\text{ mA} \times 15 = 600\text{ mA}

(or

0.6\text{ A}

).

Step-by-Step Solution

1

Calculate the turns ratio of the transformer

Turns ratio

\frac{N_s}{N_p} = \frac{1{}200}{80} = 15

The turns ratio establishes how voltage and current scale between the primary and secondary windings.

2

Apply the inverse current transformation formula for ideal transformers

I_p = I_s \times \left(\frac{N_s}{N_p}\right) = 40\text{ mA} \times 15 = 600\text{ mA}

Because transformers conserve power (

V_p I_p = V_s I_s

), a step-up in voltage results in a proportional step-down in current, meaning the primary current must be greater than the secondary current by the turns ratio.

Key Concept

Transformer Current and Turns Ratio Relationship

Estimated Time:1m 0s

Question 965Question →

An oscilloscope displays a periodic alternating voltage waveform on its screen. One complete cycle of the waveform occupies $4\text{ horizontal divisions}$ across the graticule. If the oscilloscope's horizontal time base (Time/Div) control is set to $15\text{ ms/div}$ , what is the period of the waveform in milliseconds?

Show answer & explanation

Answer: 60

Answer

60\text{ ms}

The period (

T

) of a signal displayed on an oscilloscope is found by multiplying the number of horizontal divisions for one complete cycle by the Time/Div control setting:

4\text{ divisions} \times 15\text{ ms/div} = 60\text{ ms}

.

Step-by-Step Solution

1

Identify the number of horizontal graticule divisions occupied by one full waveform cycle.

4\text{ divisions}

On an oscilloscope display, the horizontal axis represents time, and one full cycle corresponds to the period (

T

).

2

Multiply the division count by the horizontal scale factor (

15\text{ ms/div}

).

4\text{ div} \times 15\text{ ms/div} = 60\text{ ms}

Multiplying the span in divisions by the time allocated per division gives the total duration of one cycle.

Key Concept

Calculating waveform period from oscilloscope horizontal division settings

Estimated Time:45s

Question 966Question →

An electronics technician needs to verify the insulation integrity of high-voltage motor windings to ensure there is no current leakage to ground. Which specialized test instrument is designed specifically to measure these extremely high resistance values?

Signal generator

Megohmmeter

Logic probe

Frequency counter

Show answer & explanation

Answer: Megohmmeter

Answer

Megohmmeter

A megohmmeter (commonly referred to as a megger) applies a high DC test voltage across an insulator or motor winding to measure resistance in the megohm range, making it the proper instrument for evaluating insulation integrity.

Step-by-Step Solution

1

Identify the diagnostic requirement in the scenario.

The task requires measuring extremely high resistance values and checking insulation integrity against breakdown under high voltage.

Standard multimeters operate with low test voltages (

3\text{ to }9\text{ V}

) and cannot expose insulation breakdown that occurs under high operational potentials.

2

Select the specialized test instrument designed for high-voltage resistance and insulation testing.

A megohmmeter (megger) applies a high DC test voltage (

250\text{ to }1000\text{ V}

) to measure resistance in the megohm or gigohm range.

The megohmmeter is specifically built for insulation testing and detecting minute leakage currents.

Key Concept

Specialized Testers - Megohmmeter (Megger) Insulation Testing

Estimated Time:45s

Question 967Question →

A portable electric heater is connected to a $120\text{ V}$ direct-current power supply. If the heating element has an internal resistance of $24\ \Omega$ , what is the current flowing through the circuit?

0.2\text{ A}

5\text{ A}

50\text{ A}

2{}880\text{ A}

Show answer & explanation

Answer: $5\text{ A}$

Answer

5\text{ A}

Using Ohm's Law (

I = \frac{V}{R}

), dividing the applied voltage of

120\text{ V}

by the resistance of

24\ \Omega

yields exactly

5\text{ A}

.

Step-by-Step Solution

1

Identify the known values and state Ohm's Law formula for current.

Voltage

V = 120\text{ V}

, Resistance

R = 24\ \Omega

, Formula:

I = \frac{V}{R}

Ohm's Law states that current is directly proportional to voltage and inversely proportional to resistance.

2

Substitute the circuit values into the formula and solve for current (

I

).

I = \frac{120\text{ V}}{24\ \Omega} = 5\text{ A}

Dividing

120

by

24

computes the magnitude of current flow in amperes.

Key Concept

Ohm's Law expresses the fundamental relationship among voltage, current, and resistance (

I = V / R

).

Estimated Time:45s

Question 968Question →

A quartermaster is recording the number of uniform items issued per day over an $8$ -day period. The recorded numbers are $24$ , $19$ , $28$ , $17$ , $33$ , $16$ , $20$ , and $27$ . What is the median number of uniform items issued?

22

23

25

44

Show answer & explanation

Answer: 22

Answer

The median number of uniform items issued is 22.

The correct answer accurately sorts the dataset into numerical order (

16, 17, 19, 20, 24, 27, 28, 33

) and calculates the average of the two middlemost numbers,

20

and

24

, which results in

22

.

Step-by-Step Solution

1

Sort the recorded numbers in ascending numerical order.

The sorted list is:

16, 17, 19, 20, 24, 27, 28, 33

.

The median is the middle value of an ordered dataset, so ordering is always the first step.

2

Identify the middle value(s).

Since there are

8

data points (an even number), there are two middle values: the 4th and 5th numbers, which are

20

and

24

.

An even number of data points does not have a single exact middle number.

3

Calculate the average (mean) of the two middle values.

\frac{20 + 24}{2} = \frac{44}{2} = 22

The median for an even number of data points is found by taking the mean of the two middlemost values.

Key Concept

To find the median of a dataset with an even number of values, the data must first be ordered from least to greatest, and then the mean of the two central values must be calculated.

Question 969Question →

In a solid copper conductor connected to a direct-current (DC) power source, the physical drift velocity of individual conduction electrons through the wire occurs at nearly the speed of light.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because free electrons in a metallic conductor undergo frequent collisions with copper atoms, resulting in a net drift velocity of only fractions of a millimeter per second, even though the electrical impulse travels across the circuit near the speed of light.

Step-by-Step Solution

1

Differentiate between the speed of electric field propagation and the drift velocity of charge carriers.

The electric field sets up almost instantaneously across the conductor at near light speed, while charge carriers collide repeatedly with lattice atoms.

Understanding the physical mechanism of electrical current requires separating the transmission of electromagnetic energy from the physical motion of electrons.

2

Evaluate the actual physical movement rate of electrons in typical DC conductor currents.

Free electrons undergo constant random thermal collisions, resulting in a net directional drift velocity of only fractions of a millimeter to a few millimeters per second.

Frequent scattering against stationary copper ions severely impedes direct linear acceleration of individual electrons.

3

Determine the validity of the statement based on the evaluated speeds.

The statement claiming that electron drift velocity occurs at nearly the speed of light is false.

The statement falsely equates signal propagation speed with individual particle drift speed.

Key Concept

Distinction between electron drift velocity and electric field propagation speed in electrical current

Estimated Time:45s

Question 970Question →

A multi-component series circuit has failed and is drawing zero current due to an open fault. Arrange the following troubleshooting steps in the correct chronological sequence to safely and logically locate the open component using a digital multimeter (DMM) while the circuit is energized.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is to first verify the source voltage, anchor the negative reference lead, systematically probe downstream with the positive lead, observe the full source voltage across intact paths, and pinpoint the fault where the voltage abruptly drops to zero.

The voltage drop method is the most effective and professional way to locate an open fault in an energized circuit. In an open circuit, current cannot flow. According to Ohm's Law (

V = I \times R

), if current (

I

) is zero, the voltage drop across any intact resistor is zero. Therefore, measuring anywhere between the positive terminal and the open fault will yield the full source voltage. Measuring anywhere after the open fault will yield zero volts because that section is physically isolated from the positive terminal but electrically connected to the negative reference. The steps outlined provide the logical, chronological, and safe procedure to trace this voltage path and isolate the break.

Step-by-Step Solution

1

Understand the electrical behavior of an open circuit.

Current is 0 Amps, meaning intact resistors will have a voltage drop of 0 Volts (

V = 0 \times R = 0

).

This principle dictates that the full source voltage will be measurable all the way up to the break.

2

Verify the baseline power supply.

The source voltage is confirmed to be present at the main terminals.

Troubleshooting downstream is useless if the main power source is dead or the battery is disconnected.

3

Establish a consistent measurement reference.

The black lead is fixed to ground, ensuring all readings are relative to 0 Volts.

Moving both leads simultaneously can cause confusion; anchoring one lead provides a steady reference point.

4

Trace the voltage potential through the circuit.

The meter reads full source voltage at each test point prior to the open fault.

The circuit wiring is intact up to the fault, extending the positive potential forward.

5

Identify the point of failure.

The fault is located exactly between the last point reading full voltage and the first point reading zero volts.

The open break physically separates the positive potential from the remainder of the circuit, causing the downstream voltage reading to drop to zero relative to ground.

Key Concept

Voltage Drop Troubleshooting Method

Estimated Time:1m 30s

Question 971Question →

Match each electronic component or resistor specification on the left with its correct circuit function or definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potentiometer matches with the three-terminal variable resistor used as an adjustable voltage divider; Rheostat matches with the two-terminal variable resistor used to adjust current flow; Fixed Resistor matches with the component providing a single, constant resistance value; Tolerance Band matches with the color-coded marking indicating allowable percentage variation from nominal resistance.

Each item is correctly matched to its standard electrical engineering and ASVAB definition: potentiometers provide variable three-terminal voltage division, rheostats provide two-terminal variable current control, fixed resistors supply a steady ohmic value, and tolerance bands indicate the precision percentage.

Step-by-Step Solution

1

Identify the distinction between the two variable resistor types based on terminal count and primary function.

A potentiometer has 3 terminals and functions as a voltage divider; a rheostat has 2 active terminals in series to regulate current.

Variable resistors differ in wiring configuration: voltage division requires three connections across a potential difference, whereas current regulation requires two inline connections.

2

Identify the definition of a fixed resistor.

A fixed resistor provides a constant, unchangeable ohmic value.

Unlike potentiometers or rheostats, fixed resistors have no moving wiper mechanism.

3

Identify the purpose of the resistor tolerance band.

The tolerance band gives the allowable percentage deviation from the nominal value.

Manufacturing variations mean actual resistance lies within a percentage range defined by the 4th band.

Key Concept

Classification, terminal configurations, and functional specifications of fixed and variable resistors.

Estimated Time:40s

Question 972Question →

When trimming excess component leads on a printed circuit board after soldering through-hole parts, standard beveled cutting pliers pinch the wire and transmit mechanical shock into the solder joint, potentially fracturing the pad. Which specialized electrical hand tool features cutting edges that meet flat on the reverse side to produce a clean, flat cut without placing stress on the solder connection?

Standard diagonal cutting pliers

Flush cutters

Multi-gauge wire strippers with bolt cutters

Lineman's side-cutting pliers

Show answer & explanation

Answer: Flush cutters

Answer

Flush cutters

Flush cutters are engineered specifically for electronics fabrication with single-beveled or shear-action blades that are perfectly flat on the back side. When placed against a circuit board, they slice through copper component leads cleanly without the crushing action that transfers damaging kinetic shock into the solder fillet.

Step-by-Step Solution

1

Analyze the mechanical requirements of the task

Identified the need to trim component leads close to a circuit board without transmitting cutting shock to fragile solder pads.

Standard double-bevel cutting edges squeeze metal inward from both sides before severing it, creating an axial shockwave along the lead into the joint.

2

Evaluate the jaw geometry of electrical cutting hand tools

Flush cutters feature a zero-bevel flat outer face that shears through copper leads with a slicing action rather than a pinching action.

The flat back of flush cutters rests parallel to the board surface and produces a smooth cut with minimal residual mechanical strain.

3

Select the appropriate hand tool for delicate electronics work

Flush cutters are the designated tool for post-soldering lead trimming on printed circuit boards.

Other pliers either possess heavy double-beveled jaws or are specialized for insulation removal rather than lead trimming.

Key Concept

Hand tool cutting blade geometry and application in electronics assembly

Estimated Time:1m 15s

Question 973Question →

An aircraft avionics system is powered by a $28\text{ V}$ DC supply connected in series with three components: a $40\ \Omega$ indicator lamp, a $70\ \Omega$ current-limiting resistor, and a $30\ \Omega$ monitoring resistor. What is the voltage drop across the $70\ \Omega$ current-limiting resistor?

28\text{ V}

14\text{ V}

8\text{ V}

5.6\text{ V}

Show answer & explanation

Answer: $14\text{ V}$

Answer

14\text{ V}

The correct answer is

14\text{ V}

. In a series circuit, total resistance equals the sum of the individual resistors:

40\ \Omega + 70\ \Omega + 30\ \Omega = 140\ \Omega

. The circuit current is

I = 28\text{ V} / 140\ \Omega = 0.2\text{ A}

. Multiplying this current by the

70\ \Omega

resistance gives a voltage drop of

0.2\text{ A} \times 70\ \Omega = 14\text{ V}

.

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_{\text{total}}

) of the series circuit.

R_{\text{total}} = 40\ \Omega + 70\ \Omega + 30\ \Omega = 140\ \Omega

In a series circuit, total resistance is the algebraic sum of all individual resistances.

2

Determine the common circuit current (

I

) using Ohm's Law.

I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{28\text{ V}}{140\ \Omega} = 0.2\text{ A}

Current remains constant through every component along a single series path.

3

Calculate the voltage drop across the

70\ \Omega

resistor (

V_{70\Omega}

).

V_{70\Omega} = I \times R = 0.2\text{ A} \times 70\ \Omega = 14\text{ V}

Applying Ohm's Law (

V = I \times R

) to the specific component yields its proportional voltage drop.

Key Concept

Series Circuit Resistance and Voltage Division

Alternative Method

Use the voltage divider formula directly:

V_{70\Omega} = 28\text{ V} \times \left(\frac{70\ \Omega}{40\ \Omega + 70\ \Omega + 30\ \Omega}\right) = 28\text{ V} \times \frac{70}{140} = 28\text{ V} \times 0.5 = 14\text{ V}

.

Estimated Time:1m 0s

Question 974Question →

An automotive technician tests an unpowered single-pole single-throw (SPST) toggle switch that has been removed from a vehicle's dashboard. With the digital multimeter set to the resistance ( $\Omega$ ) mode and the switch set to the OFF position, what should the meter display across the switch terminals if the switch is functioning normally?

An infinite resistance or "OL" (overload) reading, indicating an open circuit

A reading of approximately

0\text{ }\Omega

with a continuous beep, indicating a complete path for current

A constant reading of

12\text{ V}

, representing the supply voltage held across the open contacts

A reading of several kilo-ohms (

\text{k}\Omega

) that gradually drops to zero once the circuit is energized

Show answer & explanation

Answer: An infinite resistance or "OL" (overload) reading, indicating an open circuit

Answer

An infinite resistance or "OL" (overload) reading, indicating an open circuit

When an SPST toggle switch is placed in the OFF position, its internal mechanical contacts are physically separated. This creates an open circuit with infinite electrical resistance. A digital multimeter set to resistance (

\Omega

) or continuity mode will display this condition as "OL" (overload / open loop) or infinite resistance (

\infty

), confirming that no current can flow across the open terminals.

Step-by-Step Solution

1

Identify the physical state of the switch contacts in the OFF position

The internal mechanical contacts are physically separated (open)

An SPST switch in the OFF position interrupts the current path by creating an air gap between terminals.

2

Determine how a digital multimeter interprets an open path in resistance (

\Omega

) mode

The meter measures infinite resistance, displayed as "OL" or "

\infty

"

An open circuit provides no path for the multimeter's internal sensing current to return.

Key Concept

Interpreting open circuit resistance readings on a digital multimeter

Estimated Time:1m 0s

Question 975Question →

An electronics technician is designing a control panel that uses a single DC supply line to illuminate either a green 'Standby' lamp or a red 'Run' lamp, ensuring that only one lamp can receive power at any given time. Which type of switch configuration is required to route power between these two alternate loads?

Double-pole, single-throw (DPST) switch

Single-pole, double-throw (SPDT) switch

Single-pole, single-throw (SPST) switch

Thermal-magnetic circuit breaker

Show answer & explanation

Answer: Single-pole, double-throw (SPDT) switch

Answer

Single-pole, double-throw (SPDT) switch

A single-pole, double-throw (SPDT) switch consists of a single common movable contact (one pole) that can connect to either of two stationary output contacts (two throws). Moving the switch actuator connects the single supply line to either the Standby lamp or the Run lamp, preventing both from being energized simultaneously.

Step-by-Step Solution

1

Identify the number of input circuits (poles) required by the scenario.

One incoming DC supply line means 1 pole (Single-Pole, SP).

The 'pole' count defines how many independent input circuits or movable contact arms are operated by the switch mechanism.

2

Identify the number of output positions (throws) needed for switching between loads.

Two distinct output paths (Standby lamp and Run lamp) means 2 throws (Double-Throw, DT).

The 'throw' count defines how many individual output positions or contacts each pole can connect to.

3

Combine pole and throw nomenclature to select the appropriate component.

Single-Pole, Double-Throw (SPDT).

An SPDT switch directs current from one common terminal to either of two branch terminals in a mutually exclusive manner.

Key Concept

Switch pole and throw configurations (SPST, SPDT, DPST, DPDT)

Estimated Time:1m 0s

Question 976Question →

In a standard electronic schematic diagram, what does the triangular arrowhead of a semiconductor diode symbol indicate?

The path of electron flow through the component

The permitted direction of conventional current flow

The orientation required to block current during forward bias

The maximum voltage level the diode can supply

Show answer & explanation

Answer: The permitted direction of conventional current flow

Answer

The permitted direction of conventional current flow

In standard electronic schematic symbols, the triangular arrowhead points from the anode toward the cathode, designating the direction that conventional electric current (from positive to negative) is allowed to flow during forward bias.

Step-by-Step Solution

1

Analyze the anatomy of a standard semiconductor diode schematic symbol.

The symbol features a triangle pointing directly toward a vertical barrier line.

Recognizing the individual elements of schematic drafting symbols is necessary to read circuit diagrams.

2

Correlate the triangular symbol element with electrical flow conventions.

The triangle represents an arrowhead pointing from the anode (positive terminal) to the cathode (negative terminal), matching conventional current direction.

Standard electronics schematics use conventional current (positive-to-negative flow) to define component polarity arrows.

Key Concept

A semiconductor diode schematic symbol uses a triangular arrowhead pointing from anode to cathode to indicate the permitted direction of conventional current flow.

Estimated Time:45s

Question 977Question →

An industrial control panel is powered by a $36\text{ V}$ DC source. The main circuit splits into three parallel branches. Branch one powers a cooling unit that dissipates $108\text{ W}$ of power. Branch two contains an $18\ \Omega$ limiting resistor. Branch three contains a sensor module with an unknown resistance. If a technician measures the total equivalent resistance of this entire parallel network to be exactly $4\ \Omega$ , what is the resistance of the sensor module?

4\ \Omega

9\ \Omega

12\ \Omega

18\ \Omega

Show answer & explanation

Answer: $9\ \Omega$

Answer

The resistance of the sensor module in the third branch is

9\ \Omega

.

To find the resistance of a single parallel branch, you need the voltage across it and the current through it. The voltage is the source voltage (

36\text{ V}

). The total current is

9\text{ A}

(

36\text{ V} / 4\ \Omega

). Subtracting the current from the first branch (

108\text{ W} / 36\text{ V} = 3\text{ A}

) and the second branch (

36\text{ V} / 18\ \Omega = 2\text{ A}

) leaves

4\text{ A}

flowing through the third branch. Using Ohm's Law (

R = V/I

),

36\text{ V} / 4\text{ A}

equals

9\ \Omega

.

Step-by-Step Solution

1

Calculate the total current (

I_{total}

) drawn by the entire circuit.

I_{total} = \frac{V}{R_{total}} = \frac{36\text{ V}}{4\ \Omega} = 9\text{ A}

Ohm's law applies to the circuit as a whole just as it does to individual components.

2

Calculate the current flowing through Branch 1 (

I_1

).

I_1 = \frac{P_1}{V} = \frac{108\text{ W}}{36\text{ V}} = 3\text{ A}

In a parallel circuit, the voltage across each branch is equal to the source voltage. We use the power formula (

P = V \times I

) rearranged to solve for current.

3

Calculate the current flowing through Branch 2 (

I_2

).

I_2 = \frac{V}{R_2} = \frac{36\text{ V}}{18\ \Omega} = 2\text{ A}

Using Ohm's law (

I = V/R

) with the known voltage and resistance for this specific branch.

4

Determine the current flowing through the unknown Branch 3 (

I_3

).

I_3 = I_{total} - (I_1 + I_2) = 9\text{ A} - (3\text{ A} + 2\text{ A}) = 4\text{ A}

Kirchhoff's Current Law states that the total current entering a parallel network must equal the sum of the currents in individual branches.

5

Calculate the resistance of Branch 3 (

R_3

).

R_3 = \frac{V}{I_3} = \frac{36\text{ V}}{4\text{ A}} = 9\ \Omega

Applying Ohm's law once more to the specific values of the third branch.

Key Concept

Parallel Circuit Current and Power Laws

Question 978Question →

Match each electrical material to its correct classification and physical property.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Silver matches with the conductor having high free electron concentration; Silicon matches with the semiconductor having four valence electrons; Rubber matches with the flexible insulator; Ceramic matches with the rigid, heat-resistant insulator.

Materials are classified by their atomic structure and ability to carry electrical current. Silver contains loosely bound electrons making it an excellent conductor; Silicon contains four valence electrons making it a semiconductor; Rubber has tightly bound electrons making it a flexible insulator; Ceramic has tightly bound ionic and covalent bonds making it a rigid, heat-resistant insulator.

Step-by-Step Solution

1

Identify the conductor among the listed materials.

Silver is a metal with a high density of free valence electrons, allowing electrical current to pass with minimal resistance.

Conductors readily allow electron flow due to loosely bound valence electrons.

2

Identify the semiconductor among the listed materials.

Silicon contains four valence electrons, giving it an intermediate conductivity level between conductors and insulators.

Group IV elements like silicon and germanium serve as the foundational materials for solid-state electronics.

3

Distinguish between the two insulating materials by their mechanical and thermal properties.

Rubber is a flexible insulator commonly used on cables, while ceramic is a rigid, heat-resistant insulator used in high-voltage applications.

Both materials possess tightly bound electrons that resist current flow, but serve distinct structural applications.

Key Concept

Classification of materials into conductors, semiconductors, and insulators based on electron mobility and atomic structure

Estimated Time:45s

Question 979Question →

A four-band carbon-film resistor has the color bands in the following sequence: Blue, Gray, Brown, and Gold. What is the nominal resistance value and tolerance rating of this resistor?

68\ \Omega\text{ with } \pm 5\%\text{ tolerance}

6.8\text{ k}\Omega\text{ with } \pm 5\%\text{ tolerance}

680\ \Omega\text{ with } \pm 5\%\text{ tolerance}

680\ \Omega\text{ with } \pm 10\%\text{ tolerance}

Show answer & explanation

Answer: $680\ \Omega\text{ with } \pm 5\%\text{ tolerance}$

Answer

680\ \Omega\text{ with } \pm 5\%\text{ tolerance}

Decoding the four color bands in order yields: first band (Blue) = 6, second band (Gray) = 8, third band (Brown) =

\times 10^1

(

10

), and fourth band (Gold) =

\pm 5\%

. Calculating

68 \times 10

gives a nominal resistance of

680\ \Omega

with

\pm 5\%

tolerance.

Step-by-Step Solution

1

Identify the significant digits from the first two color bands

Blue corresponds to the digit 6 and Gray corresponds to the digit 8, forming the significant number 68.

In a standard four-band resistor, the first two bands represent the first and second significant digits of the resistance value.

2

Determine the multiplier from the third color band and calculate the nominal resistance

Brown corresponds to a multiplier of

10^1

(

10

). Multiplying

68 \times 10 = 680\ \Omega

.

The third band indicates the power of ten by which the significant digits are multiplied.

3

Identify the tolerance percentage from the fourth color band

Gold corresponds to a tolerance of

\pm 5\%

.

The fourth band indicates the manufacturing tolerance of the resistor.

Key Concept

Four-band resistor color code decoding

Estimated Time:35s

Question 980Question →

A heating resistor with a resistance of $20\text{ \Omega}$ has a steady current of $3\text{ A}$ flowing through it. What is the total electrical power dissipated by the resistor?

60\text{ W}

120\text{ W}

180\text{ W}

18\text{ W}

Show answer & explanation

Answer: $180\text{ W}$

Answer

180\text{ W}

Electrical power dissipated by a resistor when current and resistance are known is determined by the formula

P = I^2 \times R

. Substituting the given values gives

P = (3\text{ A})^2 \times 20\text{ \Omega} = 9 \times 20 = 180\text{ W}

.

Step-by-Step Solution

1

Identify the known circuit parameters and the appropriate power formula.

Known values:

I = 3\text{ A}

,

R = 20\text{ \Omega}

. Applicable formula:

P = I^2 \times R

.

When current and resistance are known, Watt's law in the form

P = I^2 R

gives electrical power directly.

2

Square the current value.

I^2 = (3\text{ A})^2 = 9\text{ A}^2

.

Power dissipation is proportional to the square of the current flowing through a resistor.

3

Multiply the squared current by the resistance.

P = 9 \times 20 = 180\text{ W}

.

Multiplying squared current by resistance yields total dissipated power in watts.

Key Concept

Electrical power dissipation in a resistive component using Joule's/Watt's law (

P = I^2 R

).

Estimated Time:45s

Answer

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