# Electronics/DC Voltage and Current

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## Ohm's Law

Ohm's law describes the relationship between voltage, current, and resistance.Voltage and current are proportional to the potential difference and inversely proportional to the resistance of the circuit
    $V=I\cdot R$
    Voltage (V) is measured in volts (V); Current (I) in amperes (A); and resistance (R) in ohms (Ω).
In this example, the current going through any point in the circuit, I, will be equal to the voltage V divided by the resistance R.
In this example, the voltage across the resistor, V, will be equal to the supplied current, I, times the resistance R.
If two of the values (V, I, or R) are known, the other can be calculated using this formula.
Any more complicated circuit has an equivalent resistance that will allow us to calculate the current draw from the voltage source. Equivalent resistance is worked out using the fact that all resistors are either in parallel or series. Similarly, if the circuit only has a current source, the equivalent resistance can be used to calculate the voltage dropped across the current source.

## Kirchoff's Voltage Law

Kirchoff's Voltage Law (KVL):
    The sum of voltage drops around any loop in the circuit that starts and ends at the same place must be zero.

## Voltage as a Physical Quantity

* Voltage is the potential difference between two charged objects.
* Potentials can be added or subtracted in series to make larger or smaller potentials as is commonly done in batteries.
* Positive charge flow from areas of high potential to lower potential.
* All the components of a circuit have resistance that acts as a potential drop.

## Kirchoff's Current Law

Kirchoff's Current Law (KCL):
    The sum of all current entering a node must equal the sum of all currents leaving the node.

### KCL Example

-I$_1$ + I$_2$ + I$_3$ = 0 ↔ I$_1$ = I$_2$ + I$_3$
I$_1$ - I$_2$ - I$_3$ - I$_4$ = 0 ↔ I$_2$ + I$_3$ + I$_4$ = I$_1$
    $I_{1}=I_{2}+I_{3}+\cdots +I_{n}\,\!$
Here is more about Kirchhoff's laws, which can be integrated here

## Consequences of KVL and KCL
#### Voltage Dividers

If two circuit elements are in series, there is a voltage drop across each element, but the current through both must be the same. The voltage at any point in the chain divides according to the resistances. A simple circuit with two (or more) resistors in series with a source is called a voltage divider.
Figure A: Voltage Divider circuit.
Consider the circuit in Figure A. According to KVL the voltage $V_{in}$ is dropped across resistors $R_{1}$ and $R_{2}$. If a current i flows through the two series resistors then by Ohm's Law.
    $i={\frac {V_{in}}{R_{1}+R_{2}}}$.
So
    $V_{out}=iR_{2}\,\!$
Therefore
    $V_{out}={\frac {V_{in}R_{2}}{R_{1}+R_{2}}}$
Similary if $V_{R1}$ is the voltage across $R_{1}$ then
    $V_{R1}={\frac {V_{in}R_{1}}{R_{1}+R_{2}}}$
In general for n series resistors the voltage dropped across one of them say $R_{i}$ is
    $V_{Ri}={\frac {V_{in}R_{i}}{R_{eq}}}$
Where
    $R_{eq}=R_{1}+\cdots +R_{n}$

##### Voltage Dividers as References

Clearly voltage dividers can be used as references. If you have a 9 volt battery and you want 4.5 volts, then connect two equal valued resistors in series and take the reference across the second and ground. There are clearly other concerns though, the first concern is current draw and the effect of the source impedance. Clearly connecting two 100 ohm resistors is a bad idea if the source impedance is, say, 50 ohms. Then the current draw would be 0.036 mA which is quite large if the battery is rated, say, 200 milliampere hours. The loading is more annoying with that source impedance too, the reference voltage with that source impedance is ${\frac {9(100)}{250}}=3.6{\mbox{ V}}$. So clearly, increasing the order of the resistor to at least 1 k$\Omega$ is the way to go to reduce the current draw and the effect of loading. The other problem with these voltage divider references is that the reference cannot be loaded if we put a 100 Ω resistor in parallel with a 10 kΩ resistor. When the voltage divider is made of two 10 kΩ resistors, then the resistance of the reference resistor becomes somewhere near 100 Ω. This clearly means a terrible reference. If a 10 MΩ resistor is used for the reference resistor will still be some where around 10 kΩ but still probably less. The effect of tolerances is also a problem; if the resistors are rated 5% then the resistance of 10 kΩ resistors can vary by ±500 Ω. This means more inaccuracy with this sort of reference.

#### Current Dividers

If two elements are in parallel, the voltage across them must be the same, but the current divides according to the resistances. A simple circuit with two (or more) resistors in parallel with a source is called a current divider.
Figure B: Parallel Resistors.
If a voltage V appears across the resistors in Figure B with only $R_{1}$ and $R_{2}$ for the moment then the current flowing in the circuit, before the division, i is according to Ohms Law.
    $i={\frac {V}{R_{eq}}}$
Using the equivalent resistance for a parallel combination of resistors is
    $i={\frac {V(R_{1}+R_{2})}{R_{1}R_{2}}}$ (1)
The current through $R_{1}$ according to Ohms Law is
    $i_{1}={\frac {V}{R_{1}}}$ (2)
Dividing equation (2) by (1)
    $i_{1}={\frac {iR_{2}}{R_{2}+R_{1}}}$
Similarly
    $i_{2}={\frac {iR_{1}}{R_{2}+R_{1}}}$
In general with n Resistors the current $i_{x}$ is
    $i_{x}={\frac {iR_{1}R_{2}\cdots R_{n}}{(R_{2}\cdots R_{n}+\cdots +R_{1}\cdots R_{n-1})R_{x}}}$
Or possibly more simply
    ${\frac {i_{x}}{i}}={\frac {R_{eq}}{R_{x}}}$
Where
    $R_{eq}={\frac {R_{1}\cdots R_{n}}{R_{2}\cdots R_{n}+\cdots +R_{1}\cdots R_{n-1}}}$