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]
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