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| 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 1010 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 = 1010 W / 115 V = 1000 x 107 / 115 = 86.96 MA
- So, P = I²R = (86.96 x 106 A)2 x 5 x 10-8 Ω/ft = 3.78 x 108 W/ft
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