Mutual Inductance is what causes a voltage to appear across the secondary winding of a transformer when an AC voltage source is connected across the transformer's 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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(B). The Primary winding or coil is the part of a transformer that is normally connected to the incoming source of energy. Energy can flow either direction through a transformer, but by convention, the winding connected to the energy source is considered to be the input or primary.

For more info see Wikipedia: Transformer

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

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

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

R_total = R₁ + R₂ + R₃ + … + R_n

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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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.

another hint: 2250-turn =>2(25)0 (26).7 25 then 26

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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:
\[\text{A turns ratio of 12.2 to 1}\]

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

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(D). The equivalent capacitance of two 5000 picofarad capacitors and one 750 picofarad capacitor connected in parallel is 10750 picofarads.

Because resistors and capacitors act almost as opposites in an electrical circuit, the rules for capacitors in series and in parallel are the inverse of those for resistors. So to find the total capacitance of capacitors in parallel simply take the sum of all the capacitance values together:

C = C1 + C2 + C3 + ...

For this question:

C = 5000 pF + 5000 pF + 750 pF = 10750 pF

For more info see Wikipedia: Series and parallel circuits

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

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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}

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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}

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Silly hint: Just remember II33. Inductance Inductors 3.3. Also, there are 3 letters between I and M for “Millihenries.”

For more info see Wikipedia: Series and parallel circuits

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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.

Hint: add the values because they're in series.

For more info see Wikipedia: Series and parallel circuits.

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

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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}

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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}

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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}

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SILLY HINT: Answer will be lower than the lowest (20) in the series (and .07 is waaaayyy too low!)

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For more info see Wikipedia: Series and parallel circuits

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

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

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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.

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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.

\begin{align} 100 \div 10 &= 10\\ 100 \div 20 &= 5\\ 100 \div 50 &= 2\\ \\ \text{Total sum} &= 17 \\ 100 \div 17 &= 5.9\\ R &= 5.9 \end{align}

For more info see Wikipedia: Series and parallel circuits

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