In this tutorial, we are going to learn about Ohm’s Law.

**Ohm’s Law** – It is a set equations which show various electrical relate to each other. This law is based on a very simple principle.

- Circuit dynamic are viewed in 1 second increment of time.
- There are direct and indirect effects between circuit elements.

** The Basic Principle of Ohm’s Law:**

- That 1 volt of electrical pressure (Voltage) can push 1(Ampere) of current flow through 1(Ohm) of resistance. Over 1 second of time this will consume 1-watt of energy in heat or light and deplete 1 joule of stored energy. Ohm’s Law is equated to the mathematical formulas of -V or E = Voltage, I = Current, P = Power, and R = Resistance.
- P (Power) = Current (I) * Voltage (V), with this formula Ohm’s law which current and voltage have a direct relationship. With an increase in voltage or current, Power will increase. When voltage doubles, power double, as current doubles, power doubles. If an increase in current is caused by an increased voltage then both will have an effect on increases in power use and the relationship will compound by how each factor increased.
- Another formula for power consumption is Power (P) = I*I*R the square of current times resistance. In this formula, the current change has a multiple affect on power consumption. For example, you are pushing 10 amps of current down a 100 foot wire with a resistance of 3 Ohms. The power consumption in the wire would be 300 watts; if you push 20 amps down the same wire then power consumption would be 1200 watts. This is four times increase of power based on doubling the current. An increase of one element such as voltage can result in increases in current flow in the circuit.
- The formula for power is:

P = I^{2} * R – current has a squaring affect with resistance

P = I * V – current has a direct affect with voltage or current.

Ohms law can be used to analyze and design circuit. An easy method to use is the Ohms law circles as shown below image. Cover up what you want and do the calculations of the other two factors.

Using these wheels, V = I* R, I = V/R, R =V/I

P = I * V, I = P/V, V = P/I in additional, remember that P = I^{2}* R

Covering up one unknown will show you the formula to calculate for its value. Using Ohm’s law you must know two circuit measurements to find all other unknowns.

**The technical’s view of Ohm’s Law –**

Ohm’s law is generally used for circuit design. The technical may use Ohm’s law to reverse engineer (design) a circuit to aid in troubleshooting circuit faults. Having knowledge of direct or indirect affects based on circuit measurement with Ohm’s law can be helpful when compared to known values located on schematic diagrams This can be assist in circuits faults . Understand that the ground ,neutral ,or returning conductor are generally at 0V,ove in some circuits it could above ground but for this section we will assume it finding circuit faults. Understand that the ground but for this section we will assume it is at 0 Volts .the supply or source is at the highest voltage measurements ,equal to the source voltage . from this standpoint ,every circle of circuit energy goes from source to load and back to the source.

Measurements are taken with electronic meters at various circuits locations ,between the circuits connection and ground across a circuit device .each locations as measured *to* ground is called a *Circuits*** Node . **If a voltage measurement is lower than expected it indicates the that the node in the circuit has moved electrically towards ground . If the voltage measurement were higher than expected, it would indicate that the node has moved electrically towards the source.

**Let’s look at a circuit node with devices connected between a node and ground**

In this circuit ,the expected measurements was 8volts; the metered measurement was 6volts.Based on Ohm ‘s law this would indicate that the resistance of load 1 increased , or the resistance of load 2 decreased .. **V= I*R , **in this circuit there is only one path for current so the current so the current must be the same for both loads .The direct variable for both devices is the load resistances .

**Below is the example Ohm’s Law:**

**A 680 – Ohm resistor has a voltage drop of 30V, How much current and power is consumed in this device?**

**Step 1 – finding current – current is equal to I = Voltage = (V)/ Resistance (R)**

**I = 30V/680Ohm = 0.0441 Amps = 44.1mA**

**P = 44.1mA * 30V = 1.3W**

**.**