Resistors (Ohm's Law), Capacitors, and Inductors

Resistors

The symbol for a resistor:

The relationship between the current through a conductor with resistance and the voltage across the same conductor is described by Ohm's law:

$V=IR\,$

where V is the voltage across the conductor, I is the current through the conductor, and R is the resistance of the conductor.

The power dissipated by the resistor is equal to the voltage multiplied by the current:

$P=IV\,$

If $I$ is measured in amps and $V$ in volts, then the power $P$ is in watts.

By plugging in different forms of $V=IR$, we can rewrite $P=IV$ as:

$P=I^{2}R\,$

or

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

Capacitors

The symbol for a capacitor:

File:Capacitor symbol.jpg or File:Capacitor polarized symbol.jpg

A capacitor is a device that stores electric charges. (Do not confuse this with a battery; a battery creates an electric potential difference, but doesn't store charge. A battery is analogous to a water pump, while a capacitor is analogous to a water tank.) In its initial state, the capacitor acts like a short circuit. At its steady state,the capacitor acts like an open circuit. In this state, a capacitor has a constant voltage across it and zero current through it.

The unit of measurement for the capacitance of a capacitor is the farad, which is equal to 1 coulomb per volt.

The charge(q), voltage (v), and capacitance(C) of a capacitor are related as follows:

$q(t)=Cv(t)\,$

where q(t) and v(t) are the values for charge and voltage, expressed as a function of time.

Differentiating both sides with respect to time gives:

$i(t)=C{\frac {dv}{dt}}$

Rearranging and then integrating with respect to time give:

$v(t')={\frac {1}{C}}\int _{t_{0}}^{t'}i(t)dt+v(t_{0})$

If we assume that the charge, voltage, and current of the capacitor are zero at $t_{0}=-\infty$, our equation reduces to:

$v(t')={\frac {1}{C}}\int _{-\infty }^{t'}i(t)dt$

or

$v(t)={\frac {1}{C}}q(t)$,

which is equivalent to the first equation.

Assuming the same zero initial states as before, the energy stored in a capacitor (in joules) is given by the equation:

$w_{C}(t)={\frac {1}{2}}Cv^{2}$

Inductors

The symbol for an inductor:

An inductor stores energy in the form of a magnetic field, usually by means of a coil of wire. An inductor resists change in the current flowing through it. In its initial state, an inductor acts like an open circuit. In its steady state, it acts like a short circuit. In this state, the inductor has a constant current flowing through it and no voltage across it.

Devices with coils of wire, such as motors or transformers, will add inductance to a circuit. However, we generally don't purposefully add inductors to circuits.

The relationship between the voltage across the inductor is linearly related by a factor L, the inductance, to the time rate of change of the current through the inductor. The unit for inductance is the henry, and is equal to a volt-second per ampere.

The relationship between the voltage and the current is as follows:

$v(t)=L{\frac {di}{dt}}$

If we multiply both sides by dt, we get:

$di={\frac {1}{L}}vdt$

Integrating both sides from $t_{0}$ to $t'$ gives:

$i(t')-i(t_{0})={\frac {1}{L}}\int _{t_{0}}^{t'}vdt$

which is equal to:

$i(t')={\frac {1}{L}}\int _{t_{0}}^{t'}vdt+i(t_{0})$

assuming that the voltage, current and energy of the inductor are all zero at $t=-\infty$ reduces the equation to

$i(t')={\frac {1}{L}}\int _{-\infty }^{t'}vdt$

The energy stored in the inductor is given by:

$w_{L}(t)={\frac {1}{2}}Li^{2}$

Two Elements in Series and Parallel

Two Elements in Series and Parallel

	Resistor	Capacitor	Inductor
Series	$R_{eq}=R_{1}+R_{2}\,$	$C_{eq}={\frac {C_{1}C_{2}}{C_{1}+C_{2}}}$	$L_{eq}=L_{1}+L_{2}\,$
Parallel	$R_{eq}={\frac {R_{1}R_{2}}{R_{1}+R_{2}}}$	$C_{eq}=C_{1}+C_{2}\,$	$L_{eq}={\frac {L_{1}L_{2}}{L_{1}+L_{2}}}$

More than 2 Elements in series or parallel

Here we provide the equations for calculating more three or more resistors in parallel; the same form can be applied to the corresponding equations for capacitors and inductors. Of course, you can always just simply a network of elements by combining two at a time using the equations above.

To find the combined resistance of resistors connected in series, simply add the resistances:

$R_{eq}=R_{1}+R_{2}+\cdot \cdot \cdot +R_{n}$

If the resistors are connected in parallel, the equation is:

${\frac {1}{R_{eq}}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+\cdot \cdot \cdot +{\frac {1}{R_{n}}}$

Proof for Resistors in Parallel equation

Here we provide the derivation for the parallel resistors equation. The corresponding equations for capacitors and inductors can be derived with a similar method.

We can prove the equation for parallel resistors by using Kirchhoff's voltage and current laws:

$V_{s}=V_{R1}=V_{R2}\,$   (KVL)

$I_{s}=I_{1}+I_{2}\,$           (KCL)

Plugging in the constitutive law for resistors in the second equation yields:

$I_{s}={\frac {V_{s}}{R_{1}}}+{\frac {V_{s}}{R_{2}}}=V_{s}\left({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}\right)=V_{s}\left({\frac {R_{1}+R_{2}}{R_{1}R_{2}}}\right)$

$R_{eq}={\frac {V_{s}}{I_{s}}}={\frac {R_{1}R_{2}}{R_{1}+R_{2}}}$

References

Hayt, William H. Jr., Jack E. Kemmerly, and Steven M. Durbin. Engineering Circuit Analysis. 6th ed. New York:McGraw-Hill, 2002.