All practice questions

Question 721Question →

Which of the following statements correctly identifies the primary function and operating principle of a capacitor in an electronic circuit?

It stores energy in a magnetic field to oppose sudden changes in electrical current.

It converts alternating current into direct current by permitting electron flow in only one direction.

It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

It allows continuous direct current flow across its internal dielectric layer to dissipate energy as heat.

Show answer & explanation

Answer: It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

Answer

It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

A capacitor is designed with two conductive plates separated by an insulating dielectric material. When a voltage is applied, opposite charges accumulate on the plates, establishing an electrostatic field that stores electrical energy. Because charge accumulation requires finite time, a capacitor naturally opposes any sudden change in voltage across its terminals.

Step-by-Step Solution

1

Identify the fundamental physical construction and energy storage mechanism of a capacitor.

A capacitor consists of two conducting plates separated by an insulating dielectric, which accumulates opposite charges to establish an electrostatic field.

Capacitors store potential electrical energy in an electrostatic field rather than a magnetic field.

2

Determine how a capacitor interacts with circuit voltage and current over time.

Because electrical charge requires finite time to build up or dissipate across the plates (

q = C \cdot V

), the component naturally resists instantaneous changes in voltage.

This property makes capacitors essential for filtering voltage ripples in power supplies and setting time constants in electronic circuits.

Key Concept

Capacitor Energy Storage and Voltage Regulation

Estimated Time:45s

Question 722Question →

A fleet mechanic inspects a cargo van that maintains correct factory ride height when loaded, yet suffers from rapid tire bounce and uncontrolled wheel hop after driving over uneven pavement. Which suspension component is failing to perform its primary function?

Leaf spring assembly

Upper spring shackle

Hydraulic shock absorber

Rubber jounce bumper

Show answer & explanation

Answer: Hydraulic shock absorber

Answer

The hydraulic shock absorber is failing to perform its primary function of damping axle oscillations.

The hydraulic shock absorber is designed to control and dampen spring oscillation and axle hop by forcing fluid through calibrated internal valves. Because the vehicle does not sag under load, the springs are functioning, but the inability to stop repeated bouncing demonstrates a loss of hydraulic damping.

Step-by-Step Solution

1

Evaluate the static vs. dynamic symptoms described in the scenario.

The vehicle maintains proper static ride height under load, confirming that the load-supporting components are structurally sound.

Suspension springs (such as leaf or coil springs) are responsible for supporting vehicle weight and establishing ride height.

2

Differentiate between weight support and motion damping to identify the faulty part.

Excessive tire bounce and wheel hop indicate a lack of kinetic energy dissipation after encountering bumps.

Shock absorbers convert the kinetic energy of oscillating springs into thermal energy through hydraulic fluid restriction, keeping tires planted on the road.

Key Concept

Functional separation between suspension springs (weight support and ride height) and shock absorbers (oscillation damping).

Estimated Time:1m 0s

Question 723Question →

An automotive starter circuit uses an electromechanical relay between the ignition switch and the starter motor. What is the primary purpose of using a relay in this circuit?

To allow a low-current switch on the dashboard to safely control the high-current starter motor

To convert direct current from the vehicle battery into alternating current for the starter motor

To permanently break the circuit by melting an internal metal strip during an overcurrent event

To step up the battery voltage to a higher potential to overcome heavy wire resistance

Show answer & explanation

Answer: To allow a low-current switch on the dashboard to safely control the high-current starter motor

Answer

Allowing a low-current switch on the dashboard to safely control the high-current starter motor

An electromechanical relay uses a small control current flowing through an electromagnet coil to close heavy-duty contacts. In an automotive starting system, this enables a driver to use a small, safe ignition switch to engage the hundreds of amperes demanded by the starter motor without running heavy cables through the passenger cabin.

Step-by-Step Solution

1

Identify the basic operating principle of an electromechanical relay

A small control current through a wire coil generates a magnetic field that moves an armature to open or close heavy-duty load contacts.

Relays provide electrical isolation and current amplification between the control circuit and the load circuit.

2

Apply the relay function to the automotive starter scenario

The starter motor requires hundreds of amperes of current, which would overheat standard dashboard toggle switches and wiring if routed directly.

Placing the relay near the battery and starter allows short, heavy-gauge cables for the high-current path while using lightweight wiring for the ignition switch.

Key Concept

Electromechanical Relay Operation and High-Current Load Switching

Estimated Time:40s

Question 724Question →

Complete the statement describing the dual-action operating mechanisms inside a standard industrial circuit breaker.

Fill in the blanks below

In a thermal-magnetic circuit breaker, sustained moderate overloads cause a strip to heat and deflect over time, whereas severe short-circuit currents create an intense magnetic field in an coil to trip the mechanism instantaneously.

Show answer & explanation

Answer

Blank 1: bimetallic (or bimetal); Blank 2: electromagnetic (or magnetic)

Thermal-magnetic circuit breakers provide coordinated protection using two distinct physical principles: a bimetallic strip that flexes under prolonged thermal stress to prevent wire insulation degradation, and an electromagnetic solenoid that provides virtually instantaneous tripping when massive fault currents occur.

Step-by-Step Solution

1

Identify the thermal overload component

Bimetallic strip

Prolonged overcurrent conditions produce resistive heat proportional to

I^2R \cdot t

. Two bonded metals with different coefficients of thermal expansion expand at unequal rates, causing the strip to bend and unlatch the breaker contacts after a calibrated time delay.

2

Identify the short-circuit protection component

Electromagnetic coil (solenoid)

High-magnitude fault currents (such as direct phase-to-ground or phase-to-phase shorts) generate an immediate, powerful magnetic field inside a wire coil/solenoid, pulling an internal armature to trip the breaker instantly before conductors reach destructive temperatures.

Key Concept

Dual-trip circuit breakers utilize a slow-responding bimetallic element for thermal overcurrent protection and a rapid-responding electromagnetic solenoid for short-circuit protection.

Estimated Time:1m 15s

Question 725Question →

An automotive technician tests a $12\text{ V}$ direct current (DC) circuit powering an auxiliary cooling fan. If the multimeter measures a current of $3\text{ A}$ flowing through the fan motor, what is the electrical resistance of the motor?

36\text{ }\Omega

0.25\text{ }\Omega

4\text{ }\Omega

9\text{ }\Omega

Show answer & explanation

Answer: $4\text{ }\Omega$

Answer

4\text{ }\Omega

According to Ohm's law, resistance equals voltage divided by current (

R = \frac{V}{I}

). Dividing

12\text{ V}

by

3\text{ A}

yields

4\text{ }\Omega

.

Step-by-Step Solution

1

Identify the known values and target quantity from Ohm's law.

Voltage (

V

) =

12\text{ V}

, Current (

I

) =

3\text{ A}

, Resistance (

R

) = unknown.

Setting up the given parameters ensures the proper mathematical formula is selected.

2

Rearrange Ohm's law (

V = I \times R

) to solve for resistance (

R

).

R = \frac{V}{I}

Dividing both sides by current isolates resistance as the subject.

3

Substitute the measured values into the formula and calculate.

R = \frac{12\text{ V}}{3\text{ A}} = 4\text{ }\Omega

Carrying out the division gives the total electrical resistance of the fan motor.

Key Concept

Ohm's Law:

R = \frac{V}{I}

Estimated Time:45s

Question 726Question →

An avionics technician reads a specification sheet indicating that an auxiliary battery provides an electromotive force (EMF) of $28\text{ V}$ . What does this $28\text{ V}$ EMF rating fundamentally represent?

The battery supplies

28\text{ joules}

of electrical energy for every

1\text{ coulomb}

of charge moved.

The battery maintains a continuous current flow of

28\text{ coulombs}

of charge per second.

The battery exerts an internal opposition to electron flow equal to

28\text{ ohms}

.

The battery produces alternating polarity reversals at a rate of

28\text{ cycles}

per second.

Show answer & explanation

Answer: The battery supplies $28\text{ joules}$ of electrical energy for every $1\text{ coulomb}$ of charge moved.

Answer

The battery supplies

28\text{ joules}

of electrical energy for every

1\text{ coulomb}

of charge moved.

Electromotive force (EMF), measured in volts, represents the electrical potential energy imparted to charge carriers by a source. By definition, one volt equals one joule of energy per one coulomb of charge (

1\text{ V} = 1\text{ J/C}

). Therefore, a

28\text{ V}

rating indicates that the battery supplies

28\text{ joules}

of energy for each coulomb of charge transferred.

Step-by-Step Solution

1

Recall the fundamental physical definition of electromotive force (EMF) and electric potential difference.

Voltage (

V

) is defined as electrical work or energy (

W

) per unit charge (

Q

), expressed by the formula

V = \frac{W}{Q}

.

Electromotive force represents the electric pressure or energy supplied to charge carriers by a power source.

2

Apply the standard units of measurement to the voltage formula.

1\text{ volt} = 1\text{ joule per coulomb}

(

1\text{ V} = 1\text{ J/C}

).

Joules measure energy or work done, while coulombs measure quantity of electric charge.

3

Evaluate a potential difference of

28\text{ V}

using these units.

A rating of

28\text{ V}

means

28\text{ joules}

of electrical energy are supplied to each coulomb of charge.

Direct multiplication:

28\text{ V} = 28\text{ J/C}

.

Key Concept

Voltage and Electromotive Force (EMF) as electrical potential energy per unit charge (

1\text{ V} = 1\text{ J/C}

)

Estimated Time:1m 0s

Question 727Question →

An industrial electronics technician is servicing a timing circuit on an automated assembly line. The circuit network contains two $12\text{ }\mu\text{F}$ capacitors connected in series, and this combination is connected in parallel with a third capacitor rated at $14\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this circuit network in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

20\text{ }\mu\text{F}

For the series branch, two identical

12\text{ }\mu\text{F}

capacitors yield an equivalent capacitance of

6\text{ }\mu\text{F}

. When placed in parallel with the

14\text{ }\mu\text{F}

capacitor, the capacitances add directly (

6\text{ }\mu\text{F} + 14\text{ }\mu\text{F}

), giving a total equivalent capacitance of

20\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = 6\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

, or

C_s = \frac{C}{n} = \frac{12}{2} = 6\text{ }\mu\text{F}

.

2

Add the capacitance of the parallel branch to the series equivalent.

C_{\text{total}} = 20\text{ }\mu\text{F}

Capacitors in parallel add directly together:

C_{\text{total}} = C_s + C_3 = 6\text{ }\mu\text{F} + 14\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

.

Key Concept

Series and parallel capacitor combinations and equivalent capacitance calculations

Estimated Time:1m 0s

Question 728Question →

A power adapter uses a step-down transformer with $400\text{ turns}$ on the primary coil and $20\text{ turns}$ on the secondary coil. If an input voltage of $120\text{ V AC}$ is applied across the primary coil, what is the secondary output voltage in volts?

Show answer & explanation

Answer: 6

Answer

The secondary output voltage is

6\text{ V}

.

In a transformer, the ratio of secondary voltage to primary voltage equals the ratio of secondary turns to primary turns (

V_s / V_p = N_s / N_p

). With

400

primary turns and

20

secondary turns, the turns ratio is

20 / 400 = 1/20

. Multiplying the primary voltage (

120\text{ V}

) by

1/20

yields a secondary output voltage of

6\text{ V}

.

Step-by-Step Solution

1

Identify the given transformer parameters from the problem statement.

V_p = 120\text{ V}

,

N_p = 400\text{ turns}

, and

N_s = 20\text{ turns}

.

These parameters are required to determine the relationship between primary and secondary voltages.

2

Set up the transformer turns ratio equation relating primary and secondary voltages to coil turns.

\frac{V_s}{V_p} = \frac{N_s}{N_p}

The voltage induced across a transformer winding is directly proportional to the number of turns in that winding.

3

Substitute the known values into the equation and compute the secondary voltage

V_s

.

V_s = 120\text{ V} \times \left(\frac{20}{400}\right) = 120\text{ V} \times 0.05 = 6\text{ V}

Direct algebraic calculation yields the step-down output voltage.

Key Concept

Transformer Turns Ratio and Voltage Transformation

Estimated Time:45s

Question 729Question →

An auxiliary automotive circuit connects a $12\ \text{V}$ direct current (DC) power source in series with two resistors valued at $2\ \Omega$ and $4\ \Omega$ . What is the total current flowing through the circuit?

2\ \text{A}

9\ \text{A}

0.5\ \text{A}

72\ \text{A}

Show answer & explanation

Answer: $2\ \text{A}$

Answer

The total current flowing through the circuit is

2\ \text{A}

.

In a series circuit, total resistance is the direct sum of the individual resistors:

R_T = 2\ \Omega + 4\ \Omega = 6\ \Omega

. Using Ohm's law (

I = V / R_T

), dividing the source voltage of

12\ \text{V}

by the total resistance of

6\ \Omega

yields a total current of

2\ \text{A}

.

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_T

) of the series circuit.

R_T = R_1 + R_2 = 2\ \Omega + 4\ \Omega = 6\ \Omega

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

2

Calculate the total circuit current (

I

) using Ohm's law.

I = \frac{V_T}{R_T} = \frac{12\ \text{V}}{6\ \Omega} = 2\ \text{A}

Ohm's law states that current equals voltage divided by total resistance.

Key Concept

Series Circuit Resistance and Ohm's Law

Estimated Time:45s

Question 730Question →

A $36\text{ V}$ DC power supply is connected across a series circuit containing a fixed $16\ \Omega$ resistor and a variable rheostat. If an in-line ammeter measures a circuit current of $1.5\text{ A}$ , what is the resistance setting of the rheostat?

24\ \Omega

8\ \Omega

54\ \Omega

9.6\ \Omega

Show answer & explanation

Answer: $8\ \Omega$

Answer

8\ \Omega

In a series circuit, the current is identical at all points. Applying Ohm's Law gives the total resistance:

R_T = \frac{V_T}{I} = \frac{36\text{ V}}{1.5\text{ A}} = 24\ \Omega

. Since resistances in series add together linearly (

R_T = R_{\text{fixed}} + R_{\text{rheostat}}

), the rheostat resistance is

24\ \Omega - 16\ \Omega = 8\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) using Ohm's Law.

R_T = \frac{V_T}{I} = \frac{36\text{ V}}{1.5\text{ A}} = 24\ \Omega

The current in a series circuit is uniform throughout the entire loop, allowing Ohm's Law to determine total loop resistance from total voltage and current.

2

Subtract the fixed resistance from the total series resistance to solve for the unknown rheostat resistance (

R_{\text{rheostat}}

).

R_{\text{rheostat}} = R_T - R_{\text{fixed}} = 24\ \Omega - 16\ \Omega = 8\ \Omega

In a series circuit, total resistance equals the sum of all individual resistances (

R_T = R_1 + R_2

).

Key Concept

Series Total Resistance and Ohm's Law

Estimated Time:1m 0s

Question 731Question →

In a closed direct-current (DC) circuit operating under steady-state conditions, continuous electrical current requires a complete conducting loop through both the external load and the power source. Which of the following correctly describes the direction of physical electron movement throughout the complete circuit?

Electrons flow from the positive terminal to the negative terminal through the external load, and from the negative terminal to the positive terminal inside the power source.

Electrons flow from the negative terminal to the positive terminal through the external load, and are driven from the positive terminal to the negative terminal inside the power source.

Electrons circulate in a single uninterrupted direction from the negative terminal to the positive terminal through both the external conductor and inside the power source.

Electrons and conventional current both travel from the positive terminal to the negative terminal through the external load, driven directly by electromotive force.

Show answer & explanation

Answer: Electrons flow from the negative terminal to the positive terminal through the external load, and are driven from the positive terminal to the negative terminal inside the power source.

Answer

Electrons flow from the negative terminal to the positive terminal through the external load, and are driven from the positive terminal to the negative terminal inside the power source.

In any direct-current circuit, free electrons in the external conductors drift from the negative terminal toward the positive terminal due to electrostatic forces. To sustain continuous current, the power source utilizes electromotive force to pump electrons internally from the positive terminal back to the negative terminal, completing the closed cycle.

Step-by-Step Solution

1

Analyze external electron movement

External drift is from the negative terminal to the positive terminal.

Free electrons in metallic conductors carry a negative electric charge and are repelled by the negative battery terminal and attracted toward the positive terminal.

2

Analyze closed-loop continuity inside the source

Internal drift is from the positive terminal to the negative terminal.

To maintain a continuous current without charge buildup, the source's electromotive force (chemical or mechanical energy) does work against the internal electric field to transport electrons from the positive terminal back to the negative terminal.

3

Synthesize full loop current dynamics

External flow is negative-to-positive; internal transport is positive-to-negative.

A complete electrical circuit forms a closed loop, ensuring conservation of electric charge across all nodes.

Key Concept

Closed-Loop Electron Drift and Charge Conservation in DC Circuits

Estimated Time:45s

Question 732Question →

Electric utility companies distribute power across regional power grids using high-voltage alternating current (AC) rather than direct current (DC). Which of the following best explains why alternating current is preferred for long-distance power distribution?

Alternating current maintains a continuous, unidirectional electron flow that completely eliminates heat dissipation in transmission cables.

Alternating current allows power transmission using voltage alone without requiring any electrical current flow through the conductors.

Alternating current voltage levels can be efficiently stepped up and down using transformers to minimize transmission power losses.

Alternating current can be directly stored in standard chemical battery banks at substations without needing rectification.

Show answer & explanation

Answer: Alternating current voltage levels can be efficiently stepped up and down using transformers to minimize transmission power losses.

Answer

Alternating current voltage levels can be efficiently stepped up and down using transformers to minimize transmission power losses.

Alternating current produces a continuously alternating magnetic field that allows transformers to efficiently step up voltage to very high levels for long-distance transmission, which lowers current and minimizes resistive heat losses (

I^2R

). Transformers can then step the voltage down to standard consumer levels.

Step-by-Step Solution

1

Identify the primary electrical requirement for long-distance power distribution.

Long-distance lines suffer power loss proportional to the square of current (

P_{\text{loss}} = I^2R

). Reducing current by raising voltage reduces energy loss.

Transmission cables have inherent resistance over miles of wire, creating significant power dissipation if current is high.

2

Evaluate the operational principle of transformers with alternating current versus direct current.

Transformers rely on mutual electromagnetic induction, which requires a constantly changing magnetic field generated by alternating current.

Steady direct current does not produce a continuous rate of change in magnetic flux, preventing transformers from stepping direct current voltages up or down.

3

Connect transformer functionality to the preference for alternating current in power distribution grids.

Alternating current can be stepped up to hundreds of kilovolts for efficient transmission and stepped down to safe voltages for end users.

This ease of voltage transformation makes alternating current the standard choice for long-distance power grids.

Key Concept

Transformer compatibility and transmission efficiency differences between alternating current and direct current.

Estimated Time:45s

Question 733Question →

A step-up transformer in a specialized power supply is connected to a $120\text{ V AC}$ source on its primary side. The secondary circuit delivers a current of $24\text{ mA}$ across a $25\text{ k}\Omega$ resistive load. If the primary winding has $150\text{ turns}$ , what is the total number of turns on the secondary winding?

30\text{ turns}

75\text{ turns}

750\text{ turns}

18\text{ turns}

Show answer & explanation

Answer: $750\text{ turns}$

Answer

750\text{ turns}

The correct answer is

750\text{ turns}

. First, determine the secondary voltage using Ohm's Law:

V_s = I_s \times R_L = (0.024\text{ A}) \times (25{}000\,\Omega) = 600\text{ V}

. Then, apply the transformer ratio formula

\frac{N_s}{N_p} = \frac{V_s}{V_p}

, yielding

N_s = 150 \times \left(\frac{600}{120}\right) = 750\text{ turns}

.

Step-by-Step Solution

1

Calculate the secondary voltage (

V_s

) across the load resistor using Ohm's Law (

V = I \times R

).

V_s = (24\text{ mA}) \times (25\text{ k}\Omega) = (0.024\text{ A}) \times (25{}000\,\Omega) = 600\text{ V}

Determining the voltage across the secondary winding is necessary to establish the transformer's voltage transformation ratio.

2

Apply the transformer turns ratio equation that relates winding turns directly to voltage.

\frac{N_s}{N_p} = \frac{V_s}{V_p}

In an ideal transformer, the ratio of secondary-to-primary turns is directly proportional to the ratio of secondary-to-primary voltage.

3

Substitute the known values (

V_s = 600\text{ V}

,

V_p = 120\text{ V}

, and

N_p = 150\text{ turns}

) to solve for the secondary turns (

N_s

).

N_s = 150 \times \left(\frac{600\text{ V}}{120\text{ V}}\right) = 150 \times 5 = 750\text{ turns}

Multiplying the primary turns by the voltage scaling factor gives the required secondary turns.

Key Concept

Transformer turns ratio and multi-step AC secondary load calculations

Estimated Time:2m 0s

Question 734Question →

In electrical circuits, an electric current of one ampere ( $1\text{ A}$ ) is defined as one coulomb of electrical charge passing through a given point in one second.

True

False

Show answer & explanation

Answer: True

Answer

True

Electric current is the measure of how much electrical charge flows past a specific point per unit of time. Under the International System of Units (SI), a current of one ampere is exactly one coulomb of charge traveling past a reference plane per second.

Step-by-Step Solution

1

Identify the standard formula relating electric current, electrical charge, and time.

The formula is

I = \frac{Q}{t}

, where

I

is current,

Q

is charge in coulombs, and

t

is time in seconds.

Current quantifies the rate of electrical charge movement across a conductor cross-section.

2

Substitute standard SI base units into the rate equation.

1\text{ Ampere} = \frac{1\text{ Coulomb}}{1\text{ Second}}

(

1\text{ A} = 1\text{ C/s}

).

This establishes that one ampere directly equals one coulomb per second.

Key Concept

Definition of electric current as the rate of electrical charge flow in coulombs per second (

1\text{ A} = 1\text{ C/s}

).

Estimated Time:30s

Question 735Question →

A DC circuit consists of a $20\text{ V}$ power source connected in series with a $4\text{ }\Omega$ resistor ( $R_1$ ). After passing through $R_1$ , the circuit splits into two parallel branches containing an $8\text{ }\Omega$ resistor ( $R_2$ ) and a $24\text{ }\Omega$ resistor ( $R_3$ ). The branches then recombine and connect back to the negative terminal of the power source. What is the electrical power dissipated by the $8\text{ }\Omega$ resistor ( $R_2$ )?

Show answer & explanation

Answer: 18

Answer

18 Watts

To find the power dissipated by a single component in a compound circuit, you must isolate the parameters for that specific component. By finding the total resistance (

10\text{ }\Omega

) and total current (

2\text{ A}

), we can determine that the voltage dropping specifically across the parallel section is

12\text{ V}

. Because components in parallel share the same voltage, the

8\text{ }\Omega

resistor receives the full

12\text{ V}

. Using Watt's Law,

P = \frac{V^2}{R}

, we find

\frac{12^2}{8} = 18\text{ W}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the parallel section (

R_p

).

R_p = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\text{ }\Omega

Before analyzing the whole circuit, parallel branches must be simplified into a single equivalent resistance.

2

Calculate the total resistance of the circuit (

R_t

).

R_t = 4\text{ }\Omega + 6\text{ }\Omega = 10\text{ }\Omega

The equivalent parallel resistance is in series with the

4\text{ }\Omega

resistor.

3

Calculate the total current (

I_t

) leaving the power source.

I_t = \frac{20\text{ V}}{10\text{ }\Omega} = 2\text{ A}

Ohm's law (

I = \frac{V}{R}

) relates the total source voltage to the total circuit resistance.

4

Calculate the voltage drop across the parallel section (

V_p

).

V_p = 2\text{ A} \times 6\text{ }\Omega = 12\text{ V}

The total current flows through the equivalent parallel resistance. This gives the voltage available specifically to the parallel branches.

5

Calculate the power dissipated by the

8\text{ }\Omega

resistor (

R_2

).

P = \frac{12^2}{8} = \frac{144}{8} = 18\text{ W}

Using Watt's Law (

P = \frac{V^2}{R}

), we apply the voltage specific to the parallel branch (

12\text{ V}

) and the individual resistance (

8\text{ }\Omega

).

Key Concept

Calculating power dissipation in a specific component of a series-parallel circuit requires finding the exact voltage or current acting on that specific component, not the entire circuit.

Question 736Question →

In a bipolar junction transistor (BJT) amplifier circuit operating in its active region, a base current ( $I_B$ ) of $2\text{ mA}$ is applied. If the transistor has a common-emitter DC current gain ( $\beta$ or $h_{FE}$ ) of $50$ , what is the resulting collector current ( $I_C$ ) flowing through the transistor?

25\text{ mA}

52\text{ mA}

100\text{ mA}

0.04\text{ mA}

Show answer & explanation

Answer: $100\text{ mA}$

Answer

The collector current (

I_C

) is

100\text{ mA}

.

In the active region of a bipolar junction transistor, the collector current is directly proportional to the base current by the factor of DC current gain:

I_C = \beta \times I_B

. Multiplying the base current of

2\text{ mA}

by the gain of

50

yields

100\text{ mA}

.

Step-by-Step Solution

1

Identify the given values and formula for transistor DC current gain in the active region.

Given

I_B = 2\text{ mA}

and

\beta = 50

. The governing formula is

I_C = \beta \times I_B

.

Current gain (

\beta

) represents the amplification factor between the controlling base current and the controlled collector current.

2

Substitute the measured base current and gain into the current equation.

I_C = 50 \times 2\text{ mA} = 100\text{ mA}

.

Calculating the product gives the total current flowing through the collector terminal.

Key Concept

Bipolar Junction Transistor DC Current Gain (

\beta

or

h_{FE}

)

Estimated Time:1m 0s

Question 737Question →

A field communications transceiver is powered by a battery string consisting of 7 series-connected cells. Six of the cells each supply an electromotive force (EMF) of $2.4\text{ V}$ in a series-aiding configuration, while one $2.4\text{ V}$ cell is connected in series opposing. What is the net electromotive force, in volts ( $\text{V}$ ), delivered by the battery string?

Show answer & explanation

Answer: 12

Answer

The net electromotive force delivered by the battery string is

12\text{ V}

.

In a series circuit, aiding EMFs add together while opposing EMFs subtract:

V_{\text{net}} = (6 \times 2.4\text{ V}) - (1 \times 2.4\text{ V}) = 14.4\text{ V} - 2.4\text{ V} = 12.0\text{ V}

.

Step-by-Step Solution

1

Calculate the total electromotive force contributed by the series-aiding cells.

6 \times 2.4\text{ V} = 14.4\text{ V}

When cells are connected in series aiding (the positive terminal of one connected to the negative terminal of the next), their voltages add together directly.

2

Subtract the electromotive force of the series-opposing cell from the aiding total.

14.4\text{ V} - 2.4\text{ V} = 12.0\text{ V}

A cell connected in reverse polarity opposes the circuit electromotive force, subtracting its potential difference from the total.

Key Concept

Series electromotive force addition and subtraction: aiding cells add potential (

+V

), whereas opposing cells subtract potential (

-V

).

Alternative Method

Find the effective number of aiding cells:

6 - 1 = 5

net aiding cells. Then calculate net voltage:

5 \times 2.4\text{ V} = 12.0\text{ V}

.

Estimated Time:45s

Question 738Question →

Match each switch configuration and electromechanical relay component with its primary function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

SPST Switch matches with basic single-circuit on-off operation; SPDT Switch matches with directing one common terminal between two output paths; Relay Coil matches with generating a magnetic field to move the armature; Normally Open (NO) Contact matches with maintaining an open circuit until energized.

Each component is correctly matched to its defined operational role: an SPST switch opens or closes one path; an SPDT switch alternates between two paths; a relay coil uses current to create an electromagnetic field; and an NO contact stays open until coil actuation occurs.

Step-by-Step Solution

1

Analyze switch pole and throw designations.

SPST (Single Pole, Single Throw) has one input and one output for simple on/off control. SPDT (Single Pole, Double Throw) routes one input to either of two outputs.

Poles define the number of separate circuits controlled, while throws define the number of output positions available per pole.

2

Analyze relay component functions and contact resting states.

The coil provides electromagnetic force to actuate the switch mechanism, while Normally Open contacts only close and allow current flow when this magnetic force is active.

Relay contacts are designated based on their state when the control coil is unpowered (de-energized).

Key Concept

Switch Nomenclature and Electromechanical Relay Operation

Estimated Time:45s

Question 739Question →

An ideal transformer has a primary coil with $600$ turns and a secondary coil with $150$ turns. The primary coil is connected to a $120\text{ V}$ AC source, and the secondary coil is connected to a load consisting of two $15\ \Omega$ resistors wired in parallel. What is the total current flowing through the primary coil?

1\text{ A}

16\text{ A}

4\text{ A}

0.25\text{ A}

Show answer & explanation

Answer: $1\text{ A}$

Answer

The total current flowing through the primary coil is

1\text{ A}

.

The correct answer synthesizes multiple principles. First, the transformer step-down ratio is

600:150

, or

4:1

. This drops the primary

120\text{ V}

to a secondary

30\text{ V}

. Next, the two

15\ \Omega

resistors in parallel create an equivalent resistance of

7.5\ \Omega

. Using Ohm's Law (

I = V/R

), the secondary current is

30\text{ V} / 7.5\ \Omega = 4\text{ A}

. Finally, because a transformer steps current inversely to voltage, a

4:1

voltage step-down means a

1:4

current step-up from primary to secondary. Therefore, the primary current is

4\text{ A} / 4 = 1\text{ A}

.

Step-by-Step Solution

1

Calculate the turns ratio to determine the secondary voltage.

N_p/N_s = 600/150 = 4

. Therefore,

V_s = 120\text{ V} / 4 = 30\text{ V}

.

In a transformer, the voltage ratio is directly proportional to the turns ratio.

2

Determine the equivalent resistance of the secondary load.

For two equal resistors in parallel,

R_{eq} = 15\ \Omega / 2 = 7.5\ \Omega

.

Parallel branches provide multiple paths for current, reducing the total resistance of the circuit.

3

Apply Ohm's law to find the total secondary current.

I_s = V_s / R_{eq} = 30\text{ V} / 7.5\ \Omega = 4\text{ A}

.

The total current drawn by the load depends on the secondary voltage and the equivalent resistance of the load.

4

Calculate the primary current using the inverse turns ratio.

I_p = I_s / (N_p / N_s) = 4\text{ A} / 4 = 1\text{ A}

.

In an ideal transformer, power is conserved (

P_p = P_s

). If voltage steps down by a factor of 4, the current must step up by a factor of 4 from primary to secondary, making the primary current 1/4 of the secondary current.

Key Concept

Transformer turns ratio scaling for voltage and current, combined with parallel circuit analysis.

Question 740Question →

A technician connects a direct current (DC) indicator lamp across a $10\text{ V}$ power supply. A digital multimeter connected in series measures a current of $0.5\text{ A}$ flowing through the circuit. What is the electrical resistance of the indicator lamp in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 20

Answer

The resistance of the indicator lamp is

20\text{ }\Omega

.

According to Ohm's Law, resistance is determined by dividing voltage by current (

R = \frac{V}{I}

). Substituting

10\text{ V}

and

0.5\text{ A}

gives

R = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega

.

Step-by-Step Solution

1

Identify the given circuit parameters from the problem statement.

Voltage

V = 10\text{ V}

, Current

I = 0.5\text{ A}

Applying Ohm's law requires identifying the known electromotive force and circuit current.

2

Select and arrange the Ohm's law equation to solve for resistance (

R

).

R = \frac{V}{I}

Resistance is equal to the voltage divided by the current flowing through the component.

3

Substitute the measured values into the formula and solve.

R = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega

Dividing

10\text{ V}

by

0.5\text{ A}

yields the total resistance of

20\text{ }\Omega

.

Key Concept

Ohm's Law defines the relationship between voltage, current, and resistance in an electrical circuit:

V = I \times R

, which rearranges to

R = \frac{V}{I}

.

Estimated Time:45s

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