All practice questions

Question 101Question →

An audio receiver circuit contains three capacitors arranged in a sub-network. Two $8\text{ }\mu\text{F}$ capacitors are connected in series with each other, and this series combination is connected in parallel with a third capacitor rated at $16\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this sub-network in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

20 μF

The equivalent capacitance of two

8\text{ }\mu\text{F}

capacitors in series is

4\text{ }\mu\text{F}

because series capacitors follow the formula

\frac{1}{C_s} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}

, resulting in

C_s = 4\text{ }\mu\text{F}

. When this

4\text{ }\mu\text{F}

equivalent section is wired in parallel with the

16\text{ }\mu\text{F}

capacitor, their values add directly (

4\text{ }\mu\text{F} + 16\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

).

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two series-connected capacitors (

C_1 = 8\text{ }\mu\text{F}

and

C_2 = 8\text{ }\mu\text{F}

).

C_{\text{series}} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}\text{ }\mu\text{F}^{-1}

, which yields

C_{\text{series}} = 4\text{ }\mu\text{F}

.

2

Add the series branch equivalent capacitance (

4\text{ }\mu\text{F}

) to the parallel branch capacitor (

C_3 = 16\text{ }\mu\text{F}

).

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

Capacitors connected in parallel add directly:

C_{\text{total}} = C_{\text{series}} + C_3 = 4\text{ }\mu\text{F} + 16\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

.

Key Concept

Series and Parallel Capacitance Calculation

Estimated Time:1m 0s

Question 102Question →

A battery pack is constructed using six $1.5\text{ V}$ cells connected in series. If five of the cells are connected in series-aiding polarity and one cell is accidentally connected in reverse (series-opposing polarity), what is the total electromotive force (EMF) across the output terminals in volts?

Show answer & explanation

Answer: 6

Answer

The net electromotive force across the terminals is

6.0\text{ V}

.

Electromotive force (EMF) values in a series circuit combine algebraically according to polarity. The five aiding cells provide

+7.5\text{ V}

(

5 \times 1.5\text{ V}

), while the single reversed cell opposes the chain with

-1.5\text{ V}

. The resulting net output voltage is

7.5\text{ V} - 1.5\text{ V} = 6.0\text{ V}

.

Step-by-Step Solution

1

Calculate the combined electromotive force of the cells connected in series-aiding polarity.

5 \times 1.5\text{ V} = 7.5\text{ V}

Cells oriented in the same aiding polarity add their potential differences directly.

2

Subtract the potential difference of the reversed (series-opposing) cell from the aiding sum.

7.5\text{ V} - 1.5\text{ V} = 6.0\text{ V}

A reversed cell creates an electromotive force in the opposite direction, opposing the aiding cells and subtracting from the net terminal voltage.

Key Concept

Series cell electromotive force addition and opposing polarity subtraction

Estimated Time:1m 0s

Question 103Question →

A military field radio power pack contains four $3.6\text{ V}$ direct-current battery cells connected in series. During maintenance, three cells are placed in series-aiding polarity, but one cell is accidentally reversed into a series-opposing polarity. The power pack is also connected in series with an auxiliary charging module that supplies an aiding electromotive force (EMF) of $850\text{ mV}$ and a protective component that produces an opposing voltage drop of $650\text{ mV}$ . What is the net electromotive force, in volts ( $\text{V}$ ), delivered across the radio terminals?

Show answer & explanation

Answer: 7.4

Answer

The net electromotive force delivered across the radio terminals is

7.4\text{ V}

.

In a series circuit, total electromotive force is the algebraic sum of all individual aiding and opposing potentials. The three aiding

3.6\text{ V}

cells contribute

+10.8\text{ V}

, while the reversed cell subtracts

-3.6\text{ V}

, leaving

+7.2\text{ V}

. Converting the auxiliary units (

850\text{ mV} = +0.85\text{ V}

and

650\text{ mV} = -0.65\text{ V}

) yields a net auxiliary contribution of

+0.20\text{ V}

. Adding these gives a total net terminal EMF of

7.2\text{ V} + 0.20\text{ V} = 7.4\text{ V}

.

Step-by-Step Solution

1

Determine the net voltage contribution from the battery cells

Net battery EMF =

7.2\text{ V}

When cells are connected in series, aiding voltages add while opposing voltages subtract. With three cells aiding and one cell opposing:

V_{\text{cells}} = (3.6\text{ V} + 3.6\text{ V} + 3.6\text{ V}) - 3.6\text{ V} = 7.2\text{ V}

.

2

Convert auxiliary voltage values from millivolts to volts

Auxiliary aiding EMF =

0.85\text{ V}

, opposing drop =

0.65\text{ V}

Since

1\text{ V} = 1000\text{ mV}

, divide millivolts by

1000

:

850\text{ mV} = 0.85\text{ V}

and

650\text{ mV} = 0.65\text{ V}

.

3

Calculate the net contribution from the auxiliary components

Net auxiliary voltage =

+0.20\text{ V}

Subtract the opposing voltage drop from the aiding auxiliary EMF:

0.85\text{ V} - 0.65\text{ V} = +0.20\text{ V}

.

4

Combine the net battery EMF and net auxiliary voltage

Total EMF =

7.4\text{ V}

Add the net auxiliary contribution to the net battery bank voltage:

7.2\text{ V} + 0.20\text{ V} = 7.4\text{ V}

.

Key Concept

Series Electromotive Force (EMF) Addition, Opposing Polarity, and Metric Prefix Conversion

Estimated Time:1m 30s

Question 104Question →

An avionics communication filter assembly contains a capacitor network designed for signal conditioning. The network consists of two $40\text{ }\mu\text{F}$ capacitors connected in series. This series combination is connected in parallel with an $80\text{ }\mu\text{F}$ capacitor. Finally, this entire parallel arrangement is connected in series with a single $25\text{ }\mu\text{F}$ capacitor. What is the total equivalent capacitance of the network, in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

The total equivalent capacitance of the network is

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

.

The correct equivalent capacitance is

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

. In capacitor networks, series components combine reciprocally (

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

) and parallel components add directly (

C_p = C_1 + C_2

). Resolving the initial series branch yields

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

, which in parallel with

80\text{ }\mu\text{F}

produces

100\text{ }\mu\text{F}

. Finally, combining

100\text{ }\mu\text{F}

in series with

25\text{ }\mu\text{F}

results in

\frac{100 \times 25}{100 + 25} = 20\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the initial series branch consisting of two

40\text{ }\mu\text{F}

capacitors.

C_{s1} = 20\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{s1}} = \frac{1}{40\text{ }\mu\text{F}} + \frac{1}{40\text{ }\mu\text{F}} = \frac{2}{40\text{ }\mu\text{F}} = \frac{1}{20\text{ }\mu\text{F}}

.

2

Calculate the equivalent capacitance of the parallel combination formed by

C_{s1}

and the

80\text{ }\mu\text{F}

capacitor.

C_p = 100\text{ }\mu\text{F}

Capacitors in parallel add directly:

C_p = C_{s1} + C_3 = 20\text{ }\mu\text{F} + 80\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

.

3

Calculate the total circuit capacitance by placing the parallel equivalent

C_p

in series with the

25\text{ }\mu\text{F}

capacitor.

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

Applying the series formula for the final combination:

\frac{1}{C_{\text{total}}} = \frac{1}{C_p} + \frac{1}{C_4} = \frac{1}{100\text{ }\mu\text{F}} + \frac{1}{25\text{ }\mu\text{F}} = \frac{1+4}{100\text{ }\mu\text{F}} = \frac{5}{100\text{ }\mu\text{F}} = \frac{1}{20\text{ }\mu\text{F}}

.

Key Concept

Calculation of total equivalent capacitance in compound series-parallel networks using reciprocal series addition and direct parallel addition rules.

Alternative Method

Use the product-over-sum shortcut for each two-component series reduction:

C_{s1} = \frac{40 \times 40}{40 + 40} = \frac{1600}{80} = 20\text{ }\mu\text{F}

. Next, add the parallel component directly:

C_p = 20 + 80 = 100\text{ }\mu\text{F}

. Finally, apply product-over-sum for the last series stage:

C_{\text{total}} = \frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}

.

Estimated Time:1m 30s

Question 105Question →

In a series-connected aircraft instrument lighting circuit, three resistive elements have values of $15\ \Omega$ , $35\ \Omega$ , and $50\ \Omega$ . What is the total equivalent resistance of the circuit in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 100

Answer

100 ohms (

\Omega

)

In a series circuit, total resistance is calculated by adding the individual resistance values together (

R_T = R_1 + R_2 + R_3

). Summing

15\ \Omega

,

35\ \Omega

, and

50\ \Omega

gives exactly

100\ \Omega

.

Step-by-Step Solution

1

Identify the formula for total resistance in a series circuit

R_T = R_1 + R_2 + R_3

In a series circuit, there is only one continuous path for current flow, making total opposition to current equal to the sum of all individual resistances.

2

Substitute the individual component values and calculate the sum

R_T = 15\ \Omega + 35\ \Omega + 50\ \Omega = 100\ \Omega

Adding the three series resistances yields the overall equivalent resistance.

Key Concept

Total resistance in a series circuit is equal to the direct sum of all individual resistances.

Estimated Time:45s

Question 106Question →

An automotive auxiliary circuit contains three capacitors connected in parallel with capacitance values of $12\text{ }\mu\text{F}$ , $18\text{ }\mu\text{F}$ , and $30\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this parallel combination in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 60

Answer

The total equivalent capacitance of the parallel circuit is

60\text{ }\mu\text{F}

.

In parallel circuits, capacitor values add together directly (

C_{\text{total}} = C_1 + C_2 + C_3

). Adding

12\text{ }\mu\text{F} + 18\text{ }\mu\text{F} + 30\text{ }\mu\text{F}

yields an equivalent total capacitance of

60\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the circuit configuration and the appropriate formula for parallel capacitors.

C_{\text{total}} = C_1 + C_2 + C_3

Connecting capacitors in parallel effectively combines their plate surface areas, meaning the total capacitance is found by direct summation.

2

Substitute the individual capacitor values into the formula and solve.

C_{\text{total}} = 12\text{ }\mu\text{F} + 18\text{ }\mu\text{F} + 30\text{ }\mu\text{F} = 60\text{ }\mu\text{F}

Performing the addition gives the total equivalent capacitance in microfarads.

Key Concept

Total capacitance in a parallel circuit is the sum of the individual capacitances.

Question 107Question →

An electronics technician measures a potential difference of $3.5\text{ kV}$ across the secondary winding of an ignition transformer. What is this electromotive force expressed in volts ( $\text{V}$ )?

Show answer & explanation

Answer: 3500

Answer

3,500 V

The prefix 'kilo-' (abbreviated as 'k') indicates a factor of

1{}000

. Therefore, an electromotive force of

3.5\text{ kV}

is equivalent to

3.5 \times 1{}000 = 3{}500\text{ V}

.

Step-by-Step Solution

1

Identify the standard metric prefix multiplier for kilo (k)

The prefix kilo represents a multiplier of

10^3

, or

1{}000

.

Electromotive force expressed with standard engineering prefixes must be converted using the appropriate power of ten.

2

Multiply the numerical value in kilovolts by 1,000

3.5 \times 1{}000 = 3{}500\text{ V}

To convert from a larger prefixed unit (kilovolts) to the base electrical unit (volts), multiply by

1{}000

.

Key Concept

Metric prefix conversion of electrical potential difference and electromotive force (EMF)

Estimated Time:30s

Question 108Question →

A pulse-generator circuit contains two $50\text{ }\mu\text{F}$ capacitors connected in parallel. This parallel pair is then connected in series with a third capacitor rated at $25\text{ }\mu\text{F}$ . What is the total equivalent capacitance of the circuit in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

The total equivalent capacitance of the circuit is

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

.

The correct answer is

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

. When two

50\text{ }\mu\text{F}

capacitors are wired in parallel, their total capacitance is

50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

. When this

100\text{ }\mu\text{F}

equivalent capacitance is connected in series with the

25\text{ }\mu\text{F}

capacitor, the product-over-sum formula yields

\frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two parallel capacitors (

C_1

and

C_2

).

C_p = 50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

Capacitors in parallel add their plate surface areas directly, so their individual capacitance values are summed (

C_p = C_1 + C_2

).

2

Calculate the total series capacitance combining the

100\text{ }\mu\text{F}

parallel equivalent with the

25\text{ }\mu\text{F}

third capacitor (

C_3

).

C_{\text{total}} = \frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}

Capacitors in series combine reciprocally according to the product-over-sum rule for two components:

C_{\text{total}} = \frac{C_p \cdot C_3}{C_p + C_3}

.

Key Concept

Capacitors combine opposite to resistors: parallel capacitors add directly (

C_p = C_1 + C_2

), whereas series capacitors combine using reciprocal formulas (

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

).

Alternative Method

Use the standard reciprocal formula for the series stage:

\frac{1}{C_{\text{total}}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{100} + \frac{1}{25} = \frac{1}{100} + \frac{4}{100} = \frac{5}{100} = \frac{1}{20}

, hence

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

.

Estimated Time:1m 0s

Question 109Question →

An avionics technician constructs a filtering network using three capacitors. Two capacitors, rated at $0.05\text{ }\mu\text{F}$ and $50\text{ nF}$ , are connected in series with each other. This series combination is then connected in parallel with a third capacitor rated at $15\text{ nF}$ . What is the total equivalent capacitance of the network in nanofarads ( $\text{nF}$ )?

Show answer & explanation

Answer: 40

Answer

The total equivalent capacitance of the network is

40\text{ nF}

.

First, convert

0.05\text{ }\mu\text{F}

to nanofarads:

0.05\text{ }\mu\text{F} \times 1000\text{ nF}/\mu\text{F} = 50\text{ nF}

. Next, the two series capacitors (

50\text{ nF}

and

50\text{ nF}

) combine via the reciprocal formula to give

25\text{ nF}

. Finally, adding the parallel capacitor (

15\text{ nF}

) yields

25\text{ nF} + 15\text{ nF} = 40\text{ nF}

.

Step-by-Step Solution

1

Convert the microfarad rating to nanofarads to ensure uniform units across all calculations.

0.05\text{ }\mu\text{F} = 0.05 \times 10^3\text{ nF} = 50\text{ nF}

Capacitor combination formulas require all terms to share the same metric prefix.

2

Calculate the equivalent capacitance of the two capacitors connected in series (

C_1 = 50\text{ nF}

and

C_2 = 50\text{ nF}

).

C_{\text{series}} = \frac{1}{\frac{1}{50\text{ nF}} + \frac{1}{50\text{ nF}}} = 25\text{ nF}

Capacitors in series combine reciprocally, meaning two identical series capacitors yield half of an individual capacitor's value.

3

Add the parallel branch capacitance (

C_3 = 15\text{ nF}

) to the series combination equivalent.

C_{\text{total}} = C_{\text{series}} + C_3 = 25\text{ nF} + 15\text{ nF} = 40\text{ nF}

Capacitors in parallel add directly because their plate areas effectively combine to store more charge at a given voltage.

Key Concept

Series and parallel equivalent capacitance rules with metric prefix conversion

Alternative Method

Convert all components into microfarads first:

C_1 = 0.05\text{ }\mu\text{F}

,

C_2 = 0.05\text{ }\mu\text{F}

, and

C_3 = 0.015\text{ }\mu\text{F}

. The series pair gives

\frac{0.05}{2} = 0.025\text{ }\mu\text{F}

. Adding the parallel capacitor gives

0.025\text{ }\mu\text{F} + 0.015\text{ }\mu\text{F} = 0.040\text{ }\mu\text{F}

. Finally, converting

0.040\text{ }\mu\text{F}

back to nanofarads gives

0.040 \times 1000 = 40\text{ nF}

.

Estimated Time:1m 30s

Question 110Question →

A marine communication device is connected to a $12\text{ V}$ direct current (DC) power source. If the circuit has a total resistance of $4\text{ }\Omega$ , what is the current flowing through the circuit in amperes ( $\text{A}$ )?

Show answer & explanation

Answer: 3

Answer

The current flowing through the circuit is

3\text{ A}

.

According to Ohm's law (

I = \frac{V}{R}

), the current in a direct current circuit is calculated by dividing the total voltage by the total circuit resistance. With a voltage of

12\text{ V}

across a resistance of

4\text{ }\Omega

, the current is

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

.

Step-by-Step Solution

1

Identify the given circuit parameters.

Voltage

V = 12\text{ V}

and resistance

R = 4\text{ }\Omega

.

Extracting the known quantities is necessary before applying the formula.

2

Select the appropriate form of Ohm's law to solve for current.

I = \frac{V}{R}

Ohm's law defines current as the ratio of electromotive force (voltage) to resistance.

3

Perform the division.

I = \frac{12}{4} = 3\text{ A}

Dividing

12\text{ V}

by

4\text{ }\Omega

computes the total current.

Key Concept

Ohm's law mathematical relationship where current is directly proportional to voltage and inversely proportional to resistance (

I = V / R

).

Estimated Time:45s

Question 111Question →

A series-connected thermopile module contains $14$ thermocouple junctions. If each junction generates an electromotive force (EMF) of $350\text{ mV}$ when heated, what is the total output potential difference of the module in volts ( $\text{V}$ )?

Show answer & explanation

Answer: 4.9

Answer

The total output potential difference is

4.9\text{ V}

.

Because all

14

thermocouple junctions are connected in series aiding, their individual voltages sum directly:

14 \times 350\text{ mV} = 4{}900\text{ mV}

. Converting millivolts to volts by dividing by

1{}000

yields exactly

4.9\text{ V}

.

Step-by-Step Solution

1

Calculate the total series electromotive force in millivolts

E_{\text{total}} = 14 \times 350\text{ mV} = 4{}900\text{ mV}

When voltage sources or EMF-generating elements are connected in a series-aiding string, their individual electromotive forces add directly.

2

Convert the total potential difference from millivolts to standard volts

V = \frac{4{}900\text{ mV}}{1{}000\text{ mV/V}} = 4.9\text{ V}

The metric prefix 'milli-' indicates one thousandth (

10^{-3}

) of a base unit, requiring division by

1{}000

to convert millivolts to volts.

Key Concept

Series electromotive forces (EMFs) add algebraically in a series-aiding configuration, and millivolts convert to standard volts by a factor of

10^{-3}

(

1{}000\text{ mV} = 1\text{ V}

).

Estimated Time:1m 0s

Question 112Question →

An electronics system contains a pulse-forming network where an electromotive force (EMF) source must do $0.24\text{ J}$ of work to move a charge of $2000\text{ \mu C}$ through the load. The power supply for this system is built by connecting identical $24\text{ V}$ battery modules in series. Assuming no internal resistance, how many $24\text{ V}$ battery modules must be connected in series to provide the required EMF?

Show answer & explanation

Answer: 5

Answer

5

The total EMF required is found by dividing the work done (

0.24\text{ J}

) by the charge in coulombs (

0.002\text{ C}

), which equals

120\text{ V}

. Since each battery module provides

24\text{ V}

and they are connected in series, the total EMF is divided by the module voltage.

120\text{ V}

divided by

24\text{ V}

per module equals exactly

5

modules.

Step-by-Step Solution

1

Recall the physical definition of electromotive force.

E = \frac{W}{Q}

EMF is defined as the work done by a source per unit of charge it moves.

2

Convert the given charge into standard SI units (Coulombs).

2000\text{ \mu C} = 2000 \times 10^{-6}\text{ C} = 0.002\text{ C}

Standard units (Joules and Coulombs) are required to calculate voltage in Volts.

3

Calculate the total required EMF.

E = \frac{0.24\text{ J}}{0.002\text{ C}} = 120\text{ V}

Substituting the work and charge values yields the total potential difference required by the network.

4

Calculate the number of battery modules needed.

120\text{ V} / 24\text{ V} = 5

In a series connection, the total voltage is the sum of the individual voltages, requiring division.

Key Concept

Electromotive Force (EMF) as Work per Unit Charge and Series Circuit Addition

Question 113Question →

During the maintenance of an aerospace navigation system, an avionics technician isolates a filtering network to verify its total capacitance. The network is configured as follows: a $20 \mu\text{F}$ capacitor and a $5 \mu\text{F}$ capacitor are wired in series. This series combination is placed in parallel across a $36 \mu\text{F}$ capacitor. Finally, this entire grouped section is wired in series with a $120 \mu\text{F}$ capacitor. What is the total equivalent capacitance of this filtering network? Enter your answer as a number.

Show answer & explanation

Answer: 30

Answer

30

To find the total equivalent capacitance, you must simplify the circuit step-by-step applying the specific rules for capacitors, which are the exact reverse of resistors. First, the

20 \mu\text{F}

and

5 \mu\text{F}

series capacitors are combined using the product-over-sum method to yield

4 \mu\text{F}

. Next, this

4 \mu\text{F}

equivalent is added directly to the

36 \mu\text{F}

parallel capacitor, resulting in

40 \mu\text{F}

for the entire parallel section. Finally, this

40 \mu\text{F}

equivalent block is in series with the final

120 \mu\text{F}

capacitor. Using the product-over-sum method once more (

\frac{40 \times 120}{160}

), the final total equivalent capacitance equals

30 \mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the

20 \mu\text{F}

and

5 \mu\text{F}

series pair.

4 \mu\text{F}

Capacitors in series are calculated using the reciprocal formula or product-over-sum rule:

\frac{20 \times 5}{20 + 5} = 4

.

2

Calculate the equivalent capacitance of the parallel section.

40 \mu\text{F}

The

4 \mu\text{F}

series equivalent is in parallel with the

36 \mu\text{F}

capacitor. Capacitors in parallel simply add together:

4 + 36 = 40

.

3

Calculate the final total capacitance of the entire network.

30 \mu\text{F}

The

40 \mu\text{F}

equivalent section is in series with the

120 \mu\text{F}

capacitor. Using the product-over-sum rule again:

\frac{40 \times 120}{40 + 120} = 30

.

Key Concept

Calculating equivalent capacitance in compound series-parallel networks

Estimated Time:3m 0s

Question 114Question →

An industrial control panel supplies a constant $24\text{ V}$ DC to a sensor circuit. During normal operation, the sensor draws a current of $16\text{ mA}$ . Following a component degradation, the measured current rises to $64\text{ mA}$ at the same operating voltage. By how many ohms ( $\Omega$ ) has the circuit's total resistance decreased?

Show answer & explanation

Answer: 1125

Answer

The circuit's total resistance decreased by

1125\text{ }\Omega

.

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

R = \frac{V}{I}

). Converting the currents to amperes gives

0.016\text{ A}

and

0.064\text{ A}

. The initial resistance is

R_1 = \frac{24\text{ V}}{0.016\text{ A}} = 1500\text{ }\Omega

, and the final resistance is

R_2 = \frac{24\text{ V}}{0.064\text{ A}} = 375\text{ }\Omega

. The decrease in resistance is

1500\text{ }\Omega - 375\text{ }\Omega = 1125\text{ }\Omega

.

Step-by-Step Solution

1

Convert current quantities from milliamperes (mA) to amperes (A)

I_1 = 0.016\text{ A}

and

I_2 = 0.064\text{ A}

Ohm's law requires current to be expressed in standard SI units (amperes) to yield resistance in ohms.

2

Calculate the initial resistance using Ohm's law (

R = \frac{V}{I}

)

R_1 = \frac{24\text{ V}}{0.016\text{ A}} = 1500\text{ }\Omega

Determines the baseline operating resistance of the sensor module.

3

Calculate the degraded state resistance using Ohm's law (

R = \frac{V}{I}

)

R_2 = \frac{24\text{ V}}{0.064\text{ A}} = 375\text{ }\Omega

Determines the lower operating resistance corresponding to the increased current draw.

4

Calculate the net difference between initial and degraded resistance

\Delta R = 1500\text{ }\Omega - 375\text{ }\Omega = 1125\text{ }\Omega

Finds the exact reduction in total circuit resistance.

Key Concept

Ohm's Law (

R = \frac{V}{I}

) applied to determine resistance changes in DC circuits with metric prefix conversion.

Estimated Time:1m 30s

Question 115Question →

An automotive audio technician installs two filter capacitors connected in parallel across a power line to suppress voltage ripple. If the individual capacitor values are $47\text{ }\mu\text{F}$ and $33\text{ }\mu\text{F}$ , what is the total equivalent capacitance of this parallel combination in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 80

Answer

80 microfarads (

80\text{ }\mu\text{F}

)

When capacitors are wired in parallel, the total plate area increases, allowing more charge to be stored at a given voltage. Therefore, the equivalent capacitance is the simple arithmetic sum of the individual capacitances:

47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the network connection type and select the appropriate rule.

Capacitors are connected in parallel.

Connecting capacitors in parallel increases the total effective plate area, so their capacitances add directly.

2

Calculate the total capacitance using the sum formula

C_{\text{total}} = C_1 + C_2

.

C_{\text{total}} = 47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}

Direct addition of the individual capacitance ratings yields the total circuit capacitance.

Key Concept

Parallel Capacitance Calculation

Estimated Time:45s

Question 116Question →

In an electronic filter circuit, two inductors with values $L_1 = 30\text{ mH}$ and $L_2 = 20\text{ mH}$ are connected in parallel. This parallel pair is connected in series with a third inductor $L_3 = 18\text{ mH}$ and a fourth inductor $L_4 = 10\text{ mH}$ . Assuming negligible mutual inductance between the coils, what is the total equivalent inductance of the circuit in millihenries ( $\text{mH}$ )?

Show answer & explanation

Answer: 40

Answer

40 mH

The total inductance is

40\text{ mH}

. Inductors follow the same combination rules as resistors: the parallel pair (

30\text{ mH}

and

20\text{ mH}

) yields an equivalent inductance of

12\text{ mH}

, which when added to the series inductors (

18\text{ mH}

and

10\text{ mH}

) gives

12 + 18 + 10 = 40\text{ mH}

.

Step-by-Step Solution

1

Calculate the equivalent inductance of the two parallel inductors

L_1

and

L_2

.

L_p = \frac{L_1 \times L_2}{L_1 + L_2} = \frac{30\text{ mH} \times 20\text{ mH}}{30\text{ mH} + 20\text{ mH}} = \frac{600}{50} = 12\text{ mH}

Inductors in parallel combine using reciprocal addition, identically to resistors in parallel.

2

Calculate the total circuit inductance by summing the parallel combination with the remaining series inductors

L_3

and

L_4

.

L_{total} = L_p + L_3 + L_4 = 12\text{ mH} + 18\text{ mH} + 10\text{ mH} = 40\text{ mH}

Inductors in series combine via direct addition because the magnetic fields and induced voltages add along the common current path.

Key Concept

Inductors combine like resistors: series inductances add directly, while parallel inductances combine using reciprocal addition.

Estimated Time:1m 30s

Question 117Question →

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 118Question →

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 119Question →

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 120Question →

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

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes