Ohm's Law and Resistors

 Resistors and devices limit the current flow or set the voltage within the circuits. Suprisingly, some resistors can have adjustable resistances. 

If DC voltage is applied across a resistor, the amount of current that flow through the resistor can be determined by Ohm's Law. 

Ohms law states V = IR, while Ohm's power law states that P = I*V = V^2/R = I^2*R.

Resistances come with large values, whereas voltage, current, and power levels usually come up in small fractions. All resistors have maximum allowable power ratings that must not be exceeded, or you'll end up frying the resistor, in watts. If the power rating is less, there could be smoke, so select a resistor with the power rating twice the maximum value anticipated. As a resistance decreases the current and power increasess, until you find a point that burns up the resistor. 

Here is an example: 

Usually resistors are connected in a variety of ways. The voltage of resistors are the same when they are in parallel, but the current will vary. 

Here is the formula for finding resistors in parallel: 

and Itotal = I1 + I2 + ... + IN or V1/R1 + V2/R2 + V3/R3 + ... + VN/RN

So Itotal = Vtotal (1/R1 + 1/R2 + 1/R3 + ... _ (1/R4) which explains the arithmetic. 

The current divider will split into 2 currents, which are I1 and I2: 

Now we want to go over resistors in series, which is infinitely simpler. 

The total resistance of a circuit is the sum of all of the individual resistances, which is Rtotal = R1 + R2 + R3 + R4 + ... which indicates as many resistors as necessary. The applied voltage is Vtotal = V1 + V2 + V3 + ... + VN, which is Kirchoff's voltage law and applying Ohm's law with the total current you get the correct equation.

The voltage drop across a 2000Ω resistor is twice as large as the voltage across the 1000Ω resistor.

To find the voltage drop, apply Ohms Law as

V1 = I1 x R1 = 0.003A * 1000 Ω = 3V

V2 = I2 x R2 = 0.003A x 2000 Ω = 6V

V1 = IR1 = Vin x R1/(R1 + R2)

P = I x Vin = (0.003A) (9V) = 0.027W = 27mW

A large resistor dissipates twice as much power proportional to the smaller resistor.

We want to select R2 so that I2 is 10% of the desired load current, and this is called the bleeder resistance or the bleeder current, and we calculate the bleeder resistance using Ohm's law. 

The next thing I want to describe is the voltage divider circuit, and our goal is to calculate the current going through the circuit and the voltage + R2. We calculate the following: 

Now, we want to design a voltage divider circuit with an output of 6V and a current of 60mA using a 12V battery. The circuit is going to look the same, first of all, and this circuit consumes 60mA, so we need to determine the total resistance of the circuit.

Now the last question is to design a divider circuit that meets requirements using a 12V battery to have a device of 5K ohms requiring a voltage of 3V. Here's how to do it.

The next thing I want to go over is reducing a complex resistor network. I can do this by solving 2 problems: 

We want to work on a second question. The load for a voltage divider requires both voltage and current. 

The values of R4, R2, and R1 are computed exactly as was done in the Preceding example. You want R4 to be at 10% of the other currents.