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

For a DC circuit with constant voltage source V_{T} and
resistors in series, the voltage drop V_{i} in resistor R_{i}
is given by the formula:

*V*_{i} - voltage drop in resistor R_{i} in volts
[V].

*V _{T}* - the equivalent voltage source or
voltage drop in volts [V].

*R*_{i} - resistance of resistor *R*_{i} in
ohms
[Ω].

*R*_{1} - resistance of resistor *R*_{1} in
ohms
[Ω].

*R*_{2} - resistance of resistor *R*_{2} in
ohms
[Ω].

*R*_{3} - resistance of resistor *R*_{3} in
ohms
[Ω].

Voltage source of V_{T}=30V is connected to resistors in
series, R_{1}=30Ω, R_{2}=40Ω.

Find the voltage drop on resistor R_{2}.

*V*_{2} = *V*_{T} ×
*R*_{2 }/ (*R*_{1}+*R*_{2})
= 30V × 40Ω / (30Ω+40Ω)
= 17.14V

For an AC circuit with voltage source V_{T} and loads in
series, the voltage drop V_{i} in load Z_{i} is
given by the formula:

*V*_{i} - voltage drop in load Z_{i} in volts
[V].

*V _{T}* - the equivalent voltage source or
voltage drop in volts [V].

*Z*_{i} - impedance of load *Z*_{i} in
ohms
[Ω].

*Z*_{1} - impedance of load *Z*_{1} in
ohms
[Ω].

*Z*_{2} - impedance of load *Z*_{2} in
ohms
[Ω].

*Z*_{3} - impedance of load *Z*_{3} in
ohms
[Ω].

Voltage source of V_{T}=30V∟60°
is connected to loads in series, Z_{1}=30Ω∟20°,
Z_{2}=40Ω∟-50°.

Find the voltage drop in load Z_{1}.

*V*_{2} = *V*_{T} ×
*Z*_{1 }/ (*Z*_{1}+*Z*_{2})

= 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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