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

Last edited by whb5k. Register to edit

Tags: none

Reversing a step-down transformer yields a step-up transformer of the same ratio (or more properly, its reciprocal).

Last edited by nschroe. Register to edit

Tags: none

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.

\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

Last edited by kd7bbc. Register to edit

Tags: none

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

\(\frac{1}{100} + \frac{1}{200} = \frac{3}{200}\)

\(\frac{1}{\frac{3}{200}}=\frac{200}{3}=67 \Omega\)

Last edited by axeinator. Register to edit

Tags: none

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.

Last edited by kd7bbc. Register to edit

Tags: none

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

The following equation describes a transformer \[\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{I_s}{I_p}\] Where:

• \(N\) = Number of turns
• \(V\) = Voltage
• \(I\) = Current
• \(_p\) = Primary
• \(_s\) = Secondary

---

\[\frac{500}{1500} = \frac{120V}{xV}\] \[\frac{120V}{\frac{1}{3}} = 360V\]

Last edited by kc9zyz. Register to edit

Tags: none

The impedance ratio of a transformer is equal to the square of its turn ratio.

This means the square root of the impedance ratio of the transformer is equal to its turn ratio.

---

1. Find desired impedance ratio \[\frac{600 \Omega}{50 \Omega} = 12\]
2. Take the square root of the impedance ratio to find the turn ratio \[\sqrt{12} \approx 3.464 \approx 3.5\]

Last edited by kc9zyz. Register to edit

Tags: none

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.

Last edited by grfd711. Register to edit

Tags: none

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

Last edited by gconklin. Register to edit

Tags: none

(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

Last edited by qubit. Register to edit

Tags: none

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.

Last edited by marvsherman419. Register to edit

Tags: none

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}

---

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

---

For more info see Wikipedia: Series and parallel circuits

Last edited by kd7bbc. Register to edit

Tags: none

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

Last edited by kd7bbc. Register to edit

Tags: none

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.

Last edited by kd7bbc. Register to edit

Tags: none