# Voltage Divider

Voltage divider rule finds the voltage over a load in electrical circuit, when the loads are connected in series.

## Voltage divider rule for DC circuit

For a DC circuit with constant voltage source VT and resistors in series, the voltage drop Vi in resistor Ri is given by the formula: Vi - voltage drop in resistor Ri in volts [V].

VT - the equivalent voltage source or voltage drop in volts [V].

Ri - resistance of resistor Ri in ohms [Ω].

R1 - resistance of resistor R1 in ohms [Ω].

R2 - resistance of resistor R2 in ohms [Ω].

R3 - resistance of resistor R3 in ohms [Ω].

#### Example

Voltage source of VT=30V is connected to resistors in series, R1=30Ω, R2=40Ω.

Find the voltage drop on resistor R2.

V2 = VT × R2 / (R1+R2) = 30V × 40Ω / (30Ω+40Ω) = 17.14V

## Voltage divider for AC circuit

For an AC circuit with voltage source VT and loads in series, the voltage drop Vi in load Zi is given by the formula: Vi - voltage drop in load Zi in volts [V].

VT - the equivalent voltage source or voltage drop in volts [V].

Zi - impedance of load Zi in ohms [Ω].

Z1 - impedance of load Z1 in ohms [Ω].

Z2 - impedance of load Z2 in ohms [Ω].

Z3 - impedance of load Z3 in ohms [Ω].

#### Example

Voltage source of VT=30V∟60° is connected to loads in series, Z1=30Ω∟20°, Z2=40Ω∟-50°.

Find the voltage drop in load Z1.

V2 = VT × Z1 / (Z1+Z2)

= 30V∟60° × 30Ω∟20° / (30Ω∟20°+40Ω∟-50°)

= 30V∟60° × 30Ω∟20° / (30cos(20)+j30sin(20)+40cos(-50)+j40sin(-50))

= 30V∟60° × 30Ω∟20° / (28.19+j10.26+25.71-j30.64)

= 30V∟60° × 30Ω∟20° / (53.9-j20.38)

= 30V∟60° × 30Ω∟20° / 57.62Ω∟-20.71°

= (30V×30Ω/57.62Ω) ∟ (60°+20°+20.71°)

= 15.62V∟100.71°

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