Basic Electronic

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Calculate Voltage Drop in series circuit


Source voltage = 12 Volt
Resistors arrange in series = 110 Ohm, 330 Ohm, 560 Ohm

So total resistance


Rt = R1 + R2 + R3
Rt = 110 + 330 + 560 = 1000 Ohm (1 KOhm)

Therefore total current


I = V/R where I Total Current, V Voltage, Total Resistance
I = 12 V/1000 Ohms
I = 0.012 Amp or 12 mA

So we have:


Rt = 1000 Ohm or 1 KOhm
It = 0.012 Amp or 12 mA

Calculate voltage drop


First Method

Vr1 = Ir1 x R1 = 0.012 Amp x 110 Ohm = 1.32 volt drop
Vr2 = Ir1 x R2 = 0.012 Amp x 330 Ohm = 3.96 volt drop
Vr3 = Ir1 x R3 = 0.012 Amp x 560 Ohm = 6.72 volt drop
So total voltage drop = 1.32 + 3.96 + 6.72 = 12 volt same as total source

Second Method

Vr1 = (R1/Rt) x Vt = (110/1000) x 12 = 1.32
Vr2 = (R2/Rt) x Vt = (330/1000) x 12 = 3.96
Vr3 = (R3/Rt) x Vt = (560/1000) x 12 = 6.72

So total voltage drop = 12 Volt



Ohm's Law


Voltage = Current x Resistance [V = I x R]
Current = Voltage / Resistance [I = V / R]
Resistance = Voltage / Current [R = V / I]


Diode bridge rectifier to allow current flow in clockwise or anti-clockwise direction





Measuring Internal Resistance


Measuring Internal Resistance of Batteries - It's important to understand the internal resistance of batteries because it determines how much current (amp) it can draws. Old batteries for example have high internal resistance compared to brand new batteries. As a result when load is apply to the old battery the voltage will drop drastically compare to the brand new batteries

Quality Inductors


Characteristic you need to know about quality and efficiency performance of inductors are

  1. Inductive reactance (resistance in Ohm) = XL = 2πfL
  2. Current rating

High Performance Capacitors


Characteristic you need to know about quality and performance of capacitors

  1. Low ESR capacitors (Equivalent Series Resistance)
  2. Capacitance

Precision Resistors



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