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

## Resistors

The symbol for a resistor:

Real resistors:

Try wikipedia for more on resistors and for the resistor color codes.

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

${\displaystyle V=IR\,}$

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:

${\displaystyle P=IV\,}$

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:

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

or

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

## Capacitors

The symbol for a capacitor:

or

The capacitor on the right is polarized. The potential on the straight side (with the plus sign) should always be higher than the potential on the curved side.

Real capacitors:

Notice that the capacitor on the far right is polarized; the negative terminal is marked on the can with white negative signs. The polarization is also indicated by the length of the leads: the short lead is negative, the long lead is positive.

A capacitor is a device that stores electric charges. The current through a capacitor can be changed instantly, but it takes time to change the voltage across a capacitor.

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:

${\displaystyle q(t)=Cv(t)\,}$

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:

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

Rearranging and then integrating with respect to time give:

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

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

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

The energy stored in a capacitor (in joules) is given by the equation:

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

## Inductors

The symbol for an inductor:

Real inductors (and items with inductance):

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. The voltage across an inductor can be changed instantly, but an inductor will resist a change in current.

Unless we are tuning an oscillator or something, we generally don't purposefully add inductors to mechatronics circuits. However, any device with coils, such as motors or transformers, add inductance to a circuit.

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:

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

If we multiply both sides by dt, we get:

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

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

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

which is equal to:

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

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

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

The energy stored in the inductor is given by:

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

## Elements in Series and Parallel

Resistors connected in series and parallel:

Two Elements in Series and Parallel
Resistor Capacitor Inductor
Series ${\displaystyle R_{eq}=R_{1}+R_{2}\,}$ ${\displaystyle C_{eq}={\frac {C_{1}C_{2}}{C_{1}+C_{2}}}}$ ${\displaystyle L_{eq}=L_{1}+L_{2}\,}$
Parallel ${\displaystyle R_{eq}={\frac {R_{1}R_{2}}{R_{1}+R_{2}}}}$ ${\displaystyle C_{eq}=C_{1}+C_{2}\,}$

${\displaystyle L_{eq}={\frac {L_{1}L_{2}}{L_{1}+L_{2}}}}$

### More than 2 Elements in series or parallel

Here we provide the equations for calculating the equivalant resistance of 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 simplify 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:

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

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

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

#### 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:

${\displaystyle V_{s}=V_{R1}=V_{R2}\,}$   (KVL)

${\displaystyle I_{s}=I_{1}+I_{2}\,}$           (KCL)

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

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

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

## References

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