### 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

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__Quality Inductors__

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

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

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__High Performance Capacitors__

Characteristic you need to know about quality and performance of capacitors

- Low ESR capacitors (Equivalent Series Resistance)
- Capacitance

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