Image Courtesy: Data Sheet - Chip Resistors (Yageo / Farnell) |

As mentioned in previous post about series and parallel resistors, let us prove the formulas to begin with:

If the Resistors R1 and R2 are connected in **series**, we can apply Kirchhoff's law here:

**V = V1 + V2, and V1 = IR1, V2 = IR2**

**=> V = IR1 + IR2 = I (R1 + R2)**

**=> as per ohm's law, V = IR => IR = I (R1 + R2)**

**OR**

**R = R1 + R2, where R is the equivalent resistance for two resistors connected in series.**

Similarly,

If two or more resistors are connected in **parallel** circuit assuming it is a closed circuit with equivalent current I, then current running through R1 is I1 and current through R2 is I2.

**Thus, I = I1 + I2**

**=> V/R = V/R1 + V/R2 [as voltage remains the same across parallel resistance]**

**OR**

**1/R = 1 / R1 + 1 / R2**

** A popular "international" alternative notation replaces the decimal point with the unit multiplier, thus 4k7 or 1M0. A 2.2ohm resistor becomes 2R2. There is an analogous scheme for capacitors and inductors.*

- Conductance:
**G = 1/R** - Net conductance: G = G1 + G2 + G3 + G4 +...
- Unit of conductance: siemens, S = 1 / Ω, is also known as mho (ohm spelled backwards, symbol ℧).
- Power and Resistors:
- Power dissipation: P = IV = I²R = V²/R
__Example__: Prove the power cannot exceed 1/4 watt, R greater than 1k, battery = 15 volt, no matter how it is connected.- since, P = V²/R, max. voltage V = 15 v, and R = 1 kΩ
- thus, P = (15 V)² / 1 kΩ = 15/1000 = 0.225 W

__Example__: New York City requires 10^{10 }W power, 115 volts. Heavy power cable = 1 inch diameter. What happens if we try to supply power through a 1 foot diameter pure copper power cable . It's R = 5 x 10^{-8}Ω. Calculate:- Power lost per foot I²R losses:
- I = P/V = 10
^{10 }W / 115 V = 1000 x 10^{7}/ 115 = 86.96 MA - So, P = I²R = (86.96 x 10
^{6}A)^{2}x 5 x 10^{-8}Ω/ft = 3.78 x 10^{8}W/ft

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