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Subelement G5
Electrical Principles
Section G5C
Resistors, capacitors, and inductors in series and parallel; transformers
What causes a voltage to appear across the secondary winding of a transformer when an AC voltage source is connected across its primary winding?
  • Capacitive coupling
  • Displacement current coupling
  • Correct Answer
    Mutual inductance
  • Mutual capacitance

Mutual Inductance is what causes a voltage to appear across the secondary winding of a transformer when an AC voltage source is connected across its primary winding. A transformer is made up of two wire windings (coils) on either side of a common core. When a current flows through the first or primary winding, it causes a magnetic field to be generated in the core. As the magnetic field fluctuates in strength and polarity with the input AC cycle, it induces a current to be generated in the secondary winding.

For more info see Wikipedia: Transformer

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Tags: arrl chapter 4 arrl module 12

What happens if a signal is applied to the secondary winding of a 4:1 voltage step-down transformer instead of the primary winding?
  • Correct Answer
    The output voltage is multiplied by 4
  • The output voltage is divided by 4
  • Additional resistance must be added in series with the primary to prevent overload
  • Additional resistance must be added in parallel with the secondary to prevent overload

The secondary voltage becomes 4 times the primary voltage

In a normal transformer, the first number in the ratio refers to the amount of windings in the primary, and the second number the secondary, in proportion to each other.

The clue to look out for is that this is a "step down" transformer; therefore, reversing the transformer will be a step up transformer. A step up transformer with 1:4 ratio (reversed from 4:1) steps up the voltage by 4 times.

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Which of the following components increases the total resistance of a resistor?
  • A parallel resistor
  • Correct Answer
    A series resistor
  • A series capacitor
  • A parallel capacitor

A resistor in series should be added to an existing resistor to increase the resistance. The total resistance of resistors in series is the sum of the individual resistances. So adding a resistor in series will add to the total resistance.

For more info see Wikipedia: Resistors, Series and parallel circuits

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What is the total resistance of three 100 ohm resistors in parallel?
  • 0.30 ohms
  • 0.33 ohms
  • Correct Answer
    33.3 ohms
  • 300 ohms

When the resistors all have the same value, you can divide the total resistance by the number of resistors:

\[\frac{100\ \Omega}{ 3 } = 33.3\ \Omega\]


For resistors in parallel regardless of the resistances, take the reciprocal of the sum of reciprocals:

\[\frac{ 1}{ R_{Total}} = \frac{ 1}{R_1} + \frac{ 1}{R_2} + \ldots + \frac{ 1}{R_n}\]

We can re-write that as:

\[R_{Total} = \frac {1}{ \frac{ 1}{R_1} + \frac{ 1}{R_2} + \ldots + \frac{ 1}{R_n}}\]

For an intuitive explanation, check out the Khan Academy video about resistors in parallel.

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If three equal value resistors in series produce 450 ohms, what is the value of each resistor?
  • 1500 ohms
  • 90 ohms
  • Correct Answer
    150 ohms
  • 175 ohms

To get the total equivalent resistance of resistors in series, sum the values.

Rtotal = R1 + R2 + R3 + … + Rn

The question states that the resistors are of equal value, so you can divide the total equivalent resistance by the number of resistors:

  • 450 Ω / 3 = 150 Ω

For an intuitive explanation of this concept see the Khan Academy video on resistors in series.

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What is the RMS voltage across a 500-turn secondary winding in a transformer if the 2250-turn primary is connected to 120 VAC?
  • 2370 volts
  • 540 volts
  • Correct Answer
    26.7 volts
  • 5.9 volts

The voltage ratio is equal to the turn ratio in the primary and secondary windings.

\[{E_s \over E_p} = {N_s \over N_p}\]

A smidgen of algebraic transformation yields the following formula:

\[E_s=E_p \times \frac{N_s}{N_p}\]

To find the voltage induced in the secondary winding (\(E_s\)), take the voltage of the primary winding (\(E_p\)) and multiply by the turn ratio of the secondary to the primary (\(\frac{N_s}{N_p}\)).

So solving for this problem:

\[E_s = 120 \text{V} \times \frac{500}{2250} = 26.7 \text{V}\]

Easy hint: If ratio of 2,250 and 500 = 4.5. Then, divide 120V by same ratio of 4.5 = 26.7V.

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Tags: arrl chapter 4 arrl module 12

What is the turns ratio of a transformer used to match an audio amplifier having 600 ohm output impedance to a speaker having 4 ohm impedance?
  • Correct Answer
    12.2 to 1
  • 24.4 to 1
  • 150 to 1
  • 300 to 1

The impedance ratio of the primary (\(P\)) to secondary (\(S\)) coil is equal to the turns ratio squared:

\[\frac{ Z_P }{ Z_S } = \bigg ( \frac{ N_P }{ N_S } \bigg )^2\]

To find the turns ratio from impedances, take the square root of the impedance ratio

\[\frac{ N_P }{ N_S } = \sqrt{ \frac{ Z_P }{ Z_S } }\]

For this question:
\(\text{Output impedance}\) (the Primary) = \(600\ \Omega\)
\(\text{Speaker impedance}\) (the Secondary) = \(4\ \Omega\)

So: \begin{align} \text{Turns ratio} &= \sqrt{ \frac{ 600\ \Omega}{ 4\ \Omega } }\\ &= \sqrt{ 150 }\\ &= 12.2\\ \end{align}

Which we can express as:
\[\text{A turns ratio of 12.2 to 1}\]


Alternate explanation: Just find the square root of the ratio. √600÷4

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What is the equivalent capacitance of two 5.0 nanofarad capacitors and one 750 picofarad capacitor connected in parallel?
  • 576.9 nanofarads
  • 1733 picofarads
  • 3583 picofarads
  • Correct Answer
    10.750 nanofarads

When capacitors are in parallel, you can add the individual capacitances together to get the total equivalent capacitance.

Recall that 1000 picofarads are equivalent to 1 nanofarad.

Converting all the values to the same unit yields:

5.0 nF + 5.0 nF + 0.75 nF = 10.75 nF

For an intuitive explanation of this phenomenon check out Khan Academy's video about capacitors in parallel.

Note that capacitors and resistors behave in opposite ways in this regard: you sum up resistors in series and capacitors in parallel.

Silly hint: "750" is in both the question and the answer. "750" is not used in any of the other distractors.

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What is the capacitance of three 100 microfarad capacitors connected in series?
  • 0.30 microfarads
  • 0.33 microfarads
  • Correct Answer
    33.3 microfarads
  • 300 microfarads

The capacitance of three 100 microfarad capacitors connected in series is 33.3 microfarads.

Because each of the capacitors in this problem are of equal value, the total capacitance may be calculated by dividing the common capacitor value (in this case 100 microfarads) by the number, N, of capacitors in the series.

\begin{align} C &= \frac{(Common\ capacitor\ value)}{N}\\ &= \frac{100\ \mu\text{F}}{3}\\ &= 33.3\ \mu\text{F} \end{align}

This is a special case of the generalized equation one can use to calculate the total capacitance of \(N\) capacitors in series:

\[\frac{1}{C_{Total\ Capacitance}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots +\frac{1}{C_N}\]

Which we can also write as:

\[C_{Total\ Capacitance} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots +\frac{1}{C_N}}\]

For more info see Wikipedia: Series and parallel circuits

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What is the inductance of three 10 millihenry inductors connected in parallel?
  • 0.30 henries
  • 3.3 henries
  • Correct Answer
    3.3 millihenries
  • 30 millihenries

(C). The total inductance of three 10 millihenry inductors connected in parallel is 3.3 millihenrys.


Because the inductors are of equal value, the total inductance (L) of these inductors in parallel may be calculated by dividing the common inductance value (in this case 10 millihenries) by the number of inductors (3).

\begin{align} L &= \frac{\text{Common Value}}{N}\\ &= \frac{10\text{ mH}}{3}\\ &= 3.3\text{ mH} \end{align}


Alternate Solution:

To solve the general case of this kind of problem, the total inductance of inductors in parallel is the reciprocal of the sum of the individual inductors' reciprocal values as shown:

\[L = \frac{ 1 }{ \frac{1}{L_1} + \frac{1}{L_2} + \ldots + \frac{1}{L_n}}\]

Plugging in the values:

\begin{align} L &= \frac{ 1 }{ \frac{1}{10\text{ mH}} + \frac{1}{10\text{ mH}} + \frac{1}{10\text{ mH}} }\\ &= \frac{ 1 }{ \frac{3}{10\text{ mH}} }\\ &= \frac{10\text{ mH}}{3}\\ &= 3.3\text{ mH} \end{align}


For more info see Wikipedia: Series and parallel circuits

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What is the inductance of a 20 millihenry inductor connected in series with a 50 millihenry inductor?
  • 0.07 millihenries
  • 14.3 millihenries
  • Correct Answer
    70 millihenries
  • 1000 millihenries

The total inductance of a 20 millihenry inductor in series with a 50 millihenry inductor is 70 millihenrys.

The total inductance of inductors in series is simply the sum of the individual inductances.

L = L1 + L2 + L3 + ...

So for this question:

L = 20 mH + 50 mH = 70 mH

Note: If all of the inductors were equal in value, the total inductance would be the common inductor value times the number of inductors L = (Common Inductor Value) x N.

For more info see Wikipedia: Series and parallel circuits.

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What is the capacitance of a 20 microfarad capacitor connected in series with a 50 microfarad capacitor?
  • 0.07 microfarads
  • Correct Answer
    14.3 microfarads
  • 70 microfarads
  • 1000 microfarads

The total capacitance of a 20 microfarad capacitor in series with a 50 microfarad capacitor is 14.3 microfarads.


The total capacitance of capacitors in series is the reciprocal of the sum of the individual capacitor's reciprocal values as shown:

\[C = \frac{ 1 }{ \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_N}}\]

Solving the problem by this method:

\begin{align} C &= \frac{ 1 }{ \frac{1}{20\ \mu\text{F}} + \frac{1}{50\ \mu\text{F}}}\\ &= \frac{ 1 }{ \frac{5}{100} + \frac{2}{100}}\\ &= \frac{ 1 }{ \frac{7}{100}}\\ &= \frac{ 100 }{ 7 }\\ &= 14.3\ \mu\text{F}\\ \end{align}


Alternate Method:

When there are only two capacitors in series you can alternately use the simplified equation:

\begin{align} C &= \frac{ C_1 \times C_2 }{ C_1 + C_2 }\\ &= \frac{ 20\ \mu\text{F} \times 50\ \mu\text{F} }{ 20\ \mu\text{F} + 50\ \mu\text{F} }\\ &= \frac{ 1000 }{ 70 }\\ &= 14.3\ \mu\text{F}\\ \end{align}


Second alternate approach:

Divide each (20 and 50) into the least common multiple (100); add the results together, and then divide that sum into the same number (the LCM).

\begin{align} \frac{100}{20} &= 5\\ \frac{100}{50} &= 2\\ \text{Total sum} &= 7\\ \frac{100}{7} &= 14.3\\ \end{align}


For more info see Wikipedia: Series and parallel circuits

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Which of the following components should be added to a capacitor to increase the capacitance?
  • An inductor in series
  • A resistor in series
  • Correct Answer
    A capacitor in parallel
  • A capacitor in series

To increase the total capacitance, you should add a capacitor in parallel.

Capacitors act in an almost opposite manner than inductors and resistors. The total capacitance of capacitors in PARALLEL is the SUM of the individual capacitance values, therefore adding additional capacitors in PARALLEL will result in a larger total capacitance.

For more info see Wikipedia: Series and parallel circuits

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Which of the following components should be added to an inductor to increase the inductance?
  • A capacitor in series
  • A resistor in parallel
  • An inductor in parallel
  • Correct Answer
    An inductor in series

To increase the total inductance, add an additional inductor in series.


Inductors act in the same manner as resistors in series and parallel circuits. The total inductance in SERIES is the SUM of the individual inductance values. Therefore, adding more inductors in SERIES will increase the total inductance.

For more info, see Wikipedia: Series and parallel circuits.

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What is the total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor connected in parallel?
  • Correct Answer
    5.9 ohms
  • 0.17 ohms
  • 10000 ohms
  • 80 ohms

The total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor in parallel is 5.9 ohms.


To calculate the total resistance of \(N\) resistors in PARALLEL, remember that the reciprocal of the \(\text{Total Resistance}\) is equal to the sum of the individual resistors' reciprocal values:

\[\frac{ 1}{ R_{Total}} = \frac{ 1}{R_1} + \frac{ 1}{R_2} + \ldots + \frac{ 1}{R_N}\]

We can re-write that as:

\[R_{Total} = \frac {1}{ \frac{ 1}{R_1} + \frac{ 1}{R_2} + \ldots + \frac{ 1}{R_n}}\]

So for this question:

\begin{align} R_{Total} &= \frac {1}{ \frac{ 1}{10\ \Omega} + \frac{ 1}{20\ \Omega} + \frac{ 1}{50\ \Omega}}\\ &= \frac {1}{ \frac{ 10}{100\ \Omega} + \frac{ 5}{100\ \Omega} + \frac{ 2}{100\ \Omega}}\\ &= \frac {1}{ \frac{ 17}{100\ \Omega}}\\ &= \frac {100}{17}\ \Omega\\ &= 5.9\ \Omega \end{align}

An easy way to check yourself is to remember that total resistance in a parallel network can never be higher than the lowest of the individual resistances. Path of least resistance.


Another hint: Divide each ohm into the least common denominator (100); then add up the three results and divide the answer into the same denominator.

100 ÷ 10 = 10; 100 ÷ 20 = 5; 100 ÷ 50 = 2

Total sum= 17

100 ÷ 17 = 5.9


For more info see Wikipedia: Series and parallel circuits

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Why is the conductor of the primary winding of many voltage step-up transformers larger in diameter than the conductor of the secondary winding?
  • To improve the coupling between the primary and secondary
  • Correct Answer
    To accommodate the higher current of the primary
  • To prevent parasitic oscillations due to resistive losses in the primary
  • To ensure that the volume of the primary winding is equal to the volume of the secondary winding

To accommodate the higher current of the primary.


A step up transformer does exactly as the name implies: it "steps up" voltage. The output voltage on the secondary is higher than the input voltage. However, the Conservation of Energy means that the input power on the primary must be higher or equal to the output power on the secondary.

Recall that
\[Power = Current \times Voltage\]

Which we usually write as
\[P = IV \text{ or } P = IE\]

Since the voltage is stepped up, you need extra current to "balance out" the higher voltage on the secondary -- you are trading extra current for the higher voltage.

Higher current means there is more loss over resistive wire. Therefore, to prevent transformer winding wires from overheating, transformer manufacturers make high current winding wire thicker.

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What is the value in nanofarads (nF) of a 22,000 picofarad (pF) capacitor?
  • 0.22
  • 2.2
  • Correct Answer
    22
  • 220

A nanofarad is by definition 10^-9 farads. A picofarad is 10^-12 farads. This means that a picofarad is one thousandth of a nanofarad. 22 thousand picofarads is therefore equivalent to 22 nanofarads.

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What is the value in microfarads of a 4700 nanofarad (nF) capacitor?
  • 47
  • 0.47
  • 47,000
  • Correct Answer
    4.7

A microfarad (\(µF\)) is equivalent to \(10^{-6}\) farads, and a nanofarad is equivalent to \(10^{-9}\) farads. This means a microfarad is 1000 times larger than a nanofarad. \({4700nF \over 1000} = 4.7µF\).

  • p (pico) \(10^{-12}\)
  • n (nano) \(10^{-9}\)
  • µ (micro) \(10^{-6}\)
  • m (milli) \(10^{-3}\)
  • k (kilo) \(10^{3}\)
  • M (mega) \(10^{6}\)
  • G (giga) \(10^{9}\)
  • T (tera) \(10^{12}\)

You only have to remember that every metric prefix has three 0´s. Write three 0´s under every prefix. Make a sheet like this:

x | milli | micro | nano | pico |
x | x x x | x x 4 | 7 0 0 | x x x |

You can read it from the sheet: 4.7 Microfarad.

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