INTRODUCTION TO ELECTRONICS: Voltage Divider Principle in Series-Parallel Circuits

The voltage-divider formula is expressed as follows:

V_x = ( R_x / R_T )( V_s )

This formula is used to determine how series resistors ( R ) split voltage drops apart as current passes through them. The net voltage drop across a series circuit’s resistors is always ( ignoring small losses ) equal to the source voltage ( V_s ). This observation is an extension of the Law of Conservation of Energy. Since voltage drops across parallel circuits are equal, we can use the voltage-divider formula to determine the voltage values for a sequence of resistors that are parallel to a resistor with a known voltage value. Consider the circuit diagram below:

Within this system, two parallel circuits are in series with one another. Within the second parallel circuit, the voltage drop across resistors R₄ and R₅ will be some fraction of the voltage across R₃. Our challenge now is to confirm the validity of the voltage values in the diagram. Let’s first determine resistance between points A and B, and then between points B and C. Subsequently, we will determine the total resistance ( R_T ) of the system at hand:

R_AB = [ R₁R₂ / ( R₁ + R₂ ) ]

R_AB = [ ( 1.0 kΩ  )( 2.2 kΩ  ) / 3.2 kΩ ]

R_AB = 688 Ω

Since R₄ and R₅ are in series with one another, we add them together prior to using the double reciprocal method to add their summ to R₃:

R₄ and R₅ = ?

330 Ω + 680 Ω = 1,010 Ω = 1.01 kΩ

R_BC = [ ( R₃ )( R₄ + R₅ ) / R₃ + R₄ + R₅ ]

R_BC = [ ( 1.0 kΩ )( 1.01 kΩ ) / 2.81 kΩ ]

R_BC = 647 Ω

R_T = R_AB + R_BC

R_T = 688 Ω + 647 Ω

R_T = 1,335 Ω = 1.335 kΩ

It will now be useful to visualize our circuit as being a two-resistor series circuit. In this way, subsequent usage of the voltage-divider formula will be more intuitive:

The voltage drop across each equivalent resistor is a fraction of the 10 V source. The value of each fraction is determined by each resistance’s value as part of the total resistance of the system:

V_AB = ( R_AB / R_T )( V_s )

V_AB = ( 688 Ω / 1.335 kΩ )( 10 V )

V_AB = ( 0.515 )( 10 V )

V_AB = 515 V

Resistors R₁ and R₂ are parallel to one another, and they are between points A and B; therefore, V_R1 and V_R2 are indeed 515 V. The resistance between points B and C are as follows:

V_BC = ( R_BC / R_T )( V_s )

V_AB = ( 647 Ω / 1.335 kΩ  )( 10 V )

V_AB = ( 0.485 )( 10 V )

V_AB = 4.85 V

Since resistor R₃ is parallel to R₄ and R₅, V_R3 is indeed 4.85 V. The voltages across R₄ and R₅, however, will be some fraction of this value:

V_R4 = [ R₄ / ( R₄ + R₅ ) ]( 4.85 V )

V_R4 = ( 330 Ω / 1.01 kΩ )( 4.85 V )

V_R4 = ( 0.327 )( 4.85 V )

V_R4 = 1.59 V

We could use the double-reciprocal formula to determine V_R5 as well; however, since we know the combined value of V_R4 and V_R5 is 4.85 V, we may derive a value for V_R5 via subtraction:

V_R4 + V_R5 = 4.85 V

1.59 V + V_R5 = 4.85 V

V_R5 = 4.85 V – 1.59 V

V_R5 = 3.26 V